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  • SD Card only mounted after a reboot

    - by evothur
    Hi everyone. I have a Kingston 2GB MicroSD and I plug it in via an inconix MicroSD Adapter to the internal card reader of my Samsung N210 Netbook with Ubuntu 10.10, but it doesn't show up. Only if I reboot the system when the card's plugged in it shows up. Why does it need a reboot for mounting? sudo fdisk -l gives the output below. But I can only see the drive when I reboot the computer while the card's plugged. Disk /dev/sda: 160.0 GB, 160041885696 bytes 255 heads, 63 sectors/track, 19457 cylinders Units = cylinders of 16065 * 512 = 8225280 bytes Sector size (logical/physical): 512 bytes / 512 bytes I/O size (minimum/optimal): 512 bytes / 512 bytes Disk identifier: 0x9a5a7990 Device Boot Start End Blocks Id System /dev/sda1 1 1959 15728640 27 Unknown Partition 1 does not end on cylinder boundary. /dev/sda2 * 1959 1972 102400 7 HPFS/NTFS /dev/sda3 1972 18992 136718750 83 Linux /dev/sda4 18992 19458 3738625 5 Extended /dev/sda5 18992 19458 3738624 82 Linux swap / Solaris Disk /dev/sdb: 1973 MB, 1973420032 bytes 60 heads, 59 sectors/track, 1088 cylinders Units = cylinders of 3540 * 512 = 1812480 bytes Sector size (logical/physical): 512 bytes / 512 bytes I/O size (minimum/optimal): 512 bytes / 512 bytes Disk identifier: 0x00000000 Device Boot Start End Blocks Id System /dev/sdb1 1 1089 1927100+ 6 FAT16

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  • Are Windows partitions gone?

    - by Gigili
    I had Windows 7 on my laptop (factory setting), because of some performance issues, I decided to use recovery options to restore it to its factory condition but I don't know what has happened or what I have done that the whole operating system was gone after playing around with recovery options from the boot menu. I couldn't find Windows, so I installed Ubuntu 11.04 on my laptop. Last time I had Ubuntu on it, it was not really compatible with laptop's configuration and I had a bit of problems trying to do normal tasks I used to do on Windows. Now I want to make sure that Windows and its drivers are gone so that I can try to install a newer version of Ubuntu or Windows. I tried the command sudo fdisk -l And the result shown was: myaccount@myaccount-VPCS116FG:~$ sudo fdisk -l [sudo] password for myaccount: Disk /dev/sda: 320.1 GB, 320072933376 bytes 255 heads, 63 sectors/track, 38913 cylinders Units = cylinders of 16065 * 512 = 8225280 bytes Sector size (logical/physical): 512 bytes / 512 bytes I/O size (minimum/optimal): 512 bytes / 512 bytes Disk identifier: 0x00025b5f Device Boot Start End Blocks Id System /dev/sda1 * 1 38409 308515840 83 Linux /dev/sda2 38409 38914 4052993 5 Extended /dev/sda5 38409 38914 4052992 82 Linux swap / Solaris Disk /dev/dm-0: 4150 MB, 4150263808 bytes 255 heads, 63 sectors/track, 504 cylinders Units = cylinders of 16065 * 512 = 8225280 bytes Sector size (logical/physical): 512 bytes / 512 bytes I/O size (minimum/optimal): 512 bytes / 512 bytes Disk identifier: 0xa668cfe8 Disk /dev/dm-0 doesn't contain a valid partition table Is it gone? If not, what command should I try to have access to Windows partitions? Thank you.

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  • ubuntu boots only with usb

    - by klimat
    Just installed Ubuntu 11.04. But it boots only from usb. Seems like I didn't pay attention during selecting boot device. sudo fdisk -l [sudo] password for klim: Disk /dev/sda: 500.1 GB, 500107862016 bytes 255 heads, 63 sectors/track, 60801 cylinders Units = cylinders of 16065 * 512 = 8225280 bytes Sector size (logical/physical): 512 bytes / 4096 bytes I/O size (minimum/optimal): 4096 bytes / 4096 bytes Disk identifier: 0x000177e1 Device Boot Start End Blocks Id System /dev/sda1 1 60045 482302976 83 Linux /dev/sda2 60045 60802 6080513 5 Extended Partition 2 does not start on physical sector boundary. /dev/sda5 60045 60802 6080512 82 Linux swap / Solaris Disk /dev/sdb: 4004 MB, 4004511744 bytes 124 heads, 62 sectors/track, 1017 cylinders Units = cylinders of 7688 * 512 = 3936256 bytes Sector size (logical/physical): 512 bytes / 512 bytes I/O size (minimum/optimal): 512 bytes / 512 bytes Disk identifier: 0x000eee1a Device Boot Start End Blocks Id System /dev/sdb1 * 1 1017 3909317 b W95 FAT32 grub updating or another "grub" operations don't work as I've tried. Can I just copy whole boot folder from usb to HD or smth like that? Any kind of help is appreciated. Apologize for my newbie skills.

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  • What is an efficient algorithm for randomly assigning a pool of objects to a parent using specific rules

    - by maple_shaft
    I need some expert answers to help me determine the most efficient algorithm in this scenario. Consider the following data structures: type B { A parent; } type A { set<B> children; integer minimumChildrenAllowed; integer maximumChildrenAllowed; } I have a situation where I need to fetch all the orphan children (there could be hundreds of thousands of these) and assign them RANDOMLY to A type parents based on the following rules. At the end of the job, there should be no orphans left At the end of the job, no object A should have less children than its predesignated minimum. At the end of the job, no object A should have more children than its predesignated maximum. If we run out of A objects then we should create a new A with default values for minimum and maximum and assign remaining orphans to these objects. The distribution of children should be as evenly distributed as possible. There may already be some children assigned to A before the job starts. I was toying with how to do this but I am afraid that I would just end up looping across the parents sorted from smallest to largest, and then grab an orphan for each parent. I was wondering if there is a more efficient way to handle this?

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  • Grub not loading after Windows 8 Install

    - by RazorXsr
    My system was configured to dual boot Ubuntu 12.04.1 LTS and Windows 7. Today I got my hands on the MSDN release of Windows 8 and I installed it over my Windows 7. Now the computer just boots to Windows 8 directly without loading the GRUB screen. So I followed the steps as suggested in: https://help.ubuntu.com/community/RecoveringUbuntuAfterInstallingWindows. Running this command: ls -l /dev/disk/by-label/ gives the following output: total 0 lrwxrwxrwx 1 root root 10 Sep 11 07:51 Entertainment -> ../../sda2 lrwxrwxrwx 1 root root 10 Sep 11 02:45 PENDRIVE -> ../../sdb1 Also fdisk -l command gives this as the output: Disk /dev/sda: 320.1 GB, 320072933376 bytes 255 heads, 63 sectors/track, 38913 cylinders, total 625142448 sectors Units = sectors of 1 * 512 = 512 bytes Sector size (logical/physical): 512 bytes / 512 bytes I/O size (minimum/optimal): 512 bytes / 512 bytes Disk identifier: 0x1246aa23 Device Boot Start End Blocks Id System /dev/sda1 * 2048 319582199 159790076 7 HPFS/NTFS/exFAT /dev/sda2 319582208 602906623 141662208 7 HPFS/NTFS/exFAT /dev/sda3 602908672 625135615 11113472 83 Linux Disk /dev/sdb: 1939 MB, 1939865600 bytes 64 heads, 63 sectors/track, 939 cylinders, total 3788800 sectors Units = sectors of 1 * 512 = 512 bytes Sector size (logical/physical): 512 bytes / 512 bytes I/O size (minimum/optimal): 512 bytes / 512 bytes Disk identifier: 0xc3072e18 Device Boot Start End Blocks Id System /dev/sdb1 * 2248 3788799 1893276 c W95 FAT32 (LBA) So I assume that I have to run this: sudo grub-install /dev/sda3 to get GRUB up and running. But I am getting this error: /usr/sbin/grub-probe: error: cannot find a device for /boot/grub (is /dev mounted?). Can anyone please guide me in the right direction? The current Ubuntu installation is far too customized to my needs to lose it to a boot manager issue! Any help is much appreciated!

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  • How can I get rid of the motd message "*** /dev/sdb1 will be checked for errors at next reboot ***"? [duplicate]

    - by kmm
    This question already has an answer here: Persistent “disk will be checked…” in the message of the day (motd) even after reboot 3 answers My motd persistently has: *** /dev/sdb1 will be checked for errors at next reboot *** The problem is that I don't have /dev/sdb1 on my system. I only have /dev/sdb2 (mouted as /) and /dev/sda1 which mounts to /media/backup. I delete that line from /etc/motd, but it reappears after reboot. Here's my df output: Filesystem Size Used Avail Use% Mounted on /dev/sdb2 73G 3.7G 66G 6% / udev 490M 4.0K 490M 1% /dev tmpfs 200M 760K 199M 1% /run none 5.0M 0 5.0M 0% /run/lock none 498M 0 498M 0% /run/shm /dev/sda1 1.9T 429G 1.4T 25% /media/backup Update Here is the output of sudo fdisk -l Disk /dev/sda: 2000.4 GB, 2000398934016 bytes 255 heads, 63 sectors/track, 243201 cylinders, total 3907029168 sectors Units = sectors of 1 * 512 = 512 bytes Sector size (logical/physical): 512 bytes / 512 bytes I/O size (minimum/optimal): 512 bytes / 512 bytes Disk identifier: 0x0003dfc2 Device Boot Start End Blocks Id System /dev/sda1 63 3907024064 1953512001 83 Linux Disk /dev/sdb: 80.0 GB, 80026361856 bytes 255 heads, 63 sectors/track, 9729 cylinders, total 156301488 sectors Units = sectors of 1 * 512 = 512 bytes Sector size (logical/physical): 512 bytes / 512 bytes I/O size (minimum/optimal): 512 bytes / 512 bytes Disk identifier: 0x00049068 Device Boot Start End Blocks Id System /dev/sdb1 152301568 156301311 1999872 82 Linux swap / Solaris /dev/sdb2 * 2048 152301567 76149760 83 Linux Partition table entries are not in disk order I guess /dev/sdb1 is my swap space.

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  • How do I get 12.04 to recognize swap partition so that I can hibernate?

    - by Kayla
    I justed installed 12.04 and used gparted to erase and enlarge my swap partition. When I rebooted, gparted said that the file partition for the swap was unknown. Gparted doesn't let me change the file partition to "linux-swap". It does let me change it to NTFS, but when I reboot, it goes back to "unknown". Thanks in advance for your help. Output from sudo swapon -s: Filename Type Size Used Priority /dev/mapper/cryptswap1 partition 9025532 0 -1 Output from sudo fdisk -l: Disk /dev/sda: 250.1 GB, 250059350016 bytes 255 heads, 63 sectors/track, 30401 cylinders, total 488397168 sectors Units = sectors of 1 * 512 = 512 bytes Sector size (logical/physical): 512 bytes / 512 bytes I/O size (minimum/optimal): 512 bytes / 512 bytes Disk identifier: 0x9d63ac84 Device Boot Start End Blocks Id System /dev/sda1 * 2048 2459647 1228800 7 HPFS/NTFS/exFAT /dev/sda2 2459648 197836472 97688412+ 7 HPFS/NTFS/exFAT /dev/sda3 466890752 488395119 10752184 7 HPFS/NTFS/exFAT /dev/sda4 197836798 466890751 134526977 5 Extended /dev/sda5 197836800 448837631 125500416 83 Linux /dev/sda6 448839680 466890751 9025536 82 Linux swap / Solaris Partition table entries are not in disk order Disk /dev/mapper/cryptswap1: 9242 MB, 9242148864 bytes 255 heads, 63 sectors/track, 1123 cylinders, total 18051072 sectors Units = sectors of 1 * 512 = 512 bytes Sector size (logical/physical): 512 bytes / 512 bytes I/O size (minimum/optimal): 512 bytes / 512 bytes Disk identifier: 0x951b7f53 Disk /dev/mapper/cryptswap1 doesn't contain a valid partition table

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  • Mounting a new hard drive (sda1) to my existing filesystem

    - by shank22
    I tried to read some posts regarding mounting a new hard drive, but I am facing some problem. My new hard drive is sda1. The output of sudo fdisk -l is: sudo fdisk -l Disk /dev/sdb: 999.7 GB, 999653638144 bytes 255 heads, 63 sectors/track, 121534 cylinders, total 1952448512 sectors Units = sectors of 1 * 512 = 512 bytes Sector size (logical/physical): 512 bytes / 512 bytes I/O size (minimum/optimal): 512 bytes / 512 bytes Disk identifier: 0x00016485 Device Boot Start End Blocks Id System /dev/sdb1 * 2048 1935822847 967910400 83 Linux /dev/sdb2 1935824894 1952446463 8310785 5 Extended /dev/sdb5 1935824896 1952446463 8310784 82 Linux swap / Solaris Disk /dev/sda: 1000.2 GB, 1000204886016 bytes 255 heads, 63 sectors/track, 121601 cylinders, total 1953525168 sectors Units = sectors of 1 * 512 = 512 bytes Sector size (logical/physical): 512 bytes / 4096 bytes I/O size (minimum/optimal): 4096 bytes / 4096 bytes Disk identifier: 0x78dbcdc1 Device Boot Start End Blocks Id System /dev/sda1 2048 1953521663 976759808 7 HPFS/NTFS/exFAT What should be done to add this new sda1 hard drive on booting up? What should be added in the /etc/fstab file? I have not performed any partition on the new sda1 drive. I need help on how to proceed from scratch and can't afford to take any risk. Please help!

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  • Having problem with C++ file handling

    - by caramel1991
    Our lecturer has given us a task,I've attempted it and try every single effort I can,but I still struggle with one of the problem in it,here goes the question: The company you work at receives a monthly report in a text format. The report contains the following information. • Department Name • Head of Department Name • Month • Minimum spending of the month • Maximum spending of the month Your program is to obtain the name of the input file from the user. Implement a structure to represent the data: Once the file has been read into your program, print out the following statistics for the user: • List which department has the minimum spending per month by month • List which department has the minimum spending by month by month Write the information into a file called “MaxMin.txt” Then do a processing so that the Department Name, Head of Department Name, Minimum spending and Maximum spending are written to separate files based on the month, eg Jan, Feb, March and so on. and of course our lecturer does send us a text file with the content: Engineering Bill Jan 2000 15000 IT Jack Jan 300 20000 HR Jill Jan 1500 10000 Engineering Bill Feb 5000 45000 IT Jack Feb 4500 7000 HR Jill Feb 5600 60000 Engineering Bill Mar 5000 45000 IT Jack Mar 4500 7000 HR Jill Mar 5600 60000 Engineering Bill Apr 5000 45000 IT Jack Apr 4500 7000 HR Jill Apr 5600 60000 Engineering Bill May 2000 15000 IT Jack May 300 20000 HR Jill May 1500 10000 Engineering Bill Jun 2000 15000 IT Jack Jun 300 20000 HR Jill Jun 1500 10000 and here's the c++ code I've written ue#include include include using namespace std; struct Record { string depName; string head; string month; float max; float min; string name; }myRecord[19]; int main () { string line; ofstream minmax,jan,feb,mar,apr,may,jun; char a[50]; char b[50]; int i = 0,j,k; float temp; //float maxjan=myRecord[0].max,maxfeb=myRecord[0].max,maxmar=myRecord[0].max,maxapr=myRecord[0].max,maxmay=myRecord[0].max,maxjune=myRecord[0].max; float minjan=myRecord[1].min,minfeb=myRecord[1].min,minmar=myRecord[1].min,minapr=myRecord[1].min,minmay=myRecord[1].min,minjune=myRecord[1].min; float maxjan=0,maxfeb=0,maxmar=0,maxapr=0,maxmay=0,maxjune=0; //float minjan=0,minfeb=0,minmar=0,minapr=0,minmay=0,minjune=0; string maxjanDep,maxfebDep,maxmarDep,maxaprDep,maxmayDep,maxjunDep; string minjanDep,minfebDep,minmarDep,minaprDep,minmayDep,minjunDep; cout<<"Enter file name: "; cina; ifstream myfile (a); //minmax.open ("MaxMin.txt"); if (myfile.is_open()){ while (! myfile.eof()){ myfilemyRecord[i].depNamemyRecord[i].headmyRecord[i].monthmyRecord[i].minmyRecord[i].max; cout << myRecord[i].depName<<"\t"< cout<<"Enter file name: "; cinb; ifstream myfile1 (b); minmax.open ("MaxMin.txt"); jan.open ("Jan.txt"); feb.open ("Feb.txt"); mar.open ("March.txt"); apr.open ("April.txt"); may.open ("May.txt"); jun.open ("Jun.txt"); if (myfile1.is_open()){ while (! myfile1.eof()){ myfile1myRecord[i].depNamemyRecord[i].headmyRecord[i].monthmyRecord[i].minmyRecord[i].max; if (myRecord[i].month == "Jan"){ jan<< myRecord[i].depName<<"\t"< if (maxjan< myRecord[i].max){ maxjan=myRecord[i].max; maxjanDep=myRecord[i].depName;} //if (minjan myRecord[i].min){ // minjan=myRecord[i].min; //minjanDep=myRecord[i].depName; //} for (k=1;k<=3;k++){ for (j=0;j<2;j++){ if (myRecord[j].minmyRecord[j+1].min){ temp=myRecord[j].min; myRecord[j].min=myRecord[j+1].min; myRecord[j+1].min=temp; minjanDep=myRecord[j].depName; }}} } if (myRecord[i].month == "Feb"){ feb<< myRecord[i].depName<<"\t"< //if (minfebmyRecord[i].min){ //minfeb=myRecord[i].min; //minfebDep=myRecord[i].depName; //} for (k=1;k<=3;k++){ for (j=0;j<2;j++){ if (myRecord[j].minmyRecord[j+1].min){ temp=myRecord[j].min; myRecord[j].min=myRecord[j+1].min; myRecord[j+1].min=temp; minfebDep=myRecord[j+1].depName; }}} } if (myRecord[i].month == "Mar"){ mar<< myRecord[i].depName<<"\t"< if (myRecord[i].month == "Apr"){ apr<< myRecord[i].depName<<"\t"< if (minaprmyRecord[i].min){ minapr=myRecord[i].min; minaprDep=myRecord[i].min;} } if (myRecord[i].month == "May"){ may< if (minmaymyRecord[i].min){ minmay=myRecord[i].min; minmayDep=myRecord[i].depName;} } if (myRecord[i].month == "Jun"){ jun<< myRecord[i].depName<<"\t"< if (minjunemyRecord[i].min){ minjune=myRecord[i].min; minjunDep=myRecord[i].depName;} } i++; myfile.close(); } minmax<<"department that has maximum spending at jan "< else{ cout << "Unable to open file"< } sorry inside that code ue#include should has iostream along with another two #include fstream and string,but at here it was treated as html tag,so i can't type it. my problem here is,I can't seems to get the minimum spending,I've try all I can but I'm still lingering on it,any idea??THANK YOU!

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  • OPN Oracle ECM 10g R3 Implementation Boot Camp - (12-14/Abr/10)

    - by Claudia Costa
    É com entusiasmo que lhe anunciamos o bootcamp de Oracle ECM 10g R3 Implementation que irá realizar nos dias 12-14 de Abril  que abordará os tópicos abaixo descritos. Com o objectivo de ajudar os parceiros a desenvolver competências, a Oracle University e a Oracle Alliances&Channel, desenharam este bootcamp, compactando os conteúdos e reduzindo assim os custos. Preço por participante (3 dias) - 1.250 Eur + Iva  Oracle offers the most unified, usable enterprise content management platform in today's market. With centralized control across single or multiple repositories, common core functionality, and easily scalable content management capabilities, Oracle provides content management solutions for many content types and users-wherever they work in the enterprise.   The Oracle Enterprise Content Management (ECM) Implementation Boot Camp examines the fundamental concepts, techniques, and architecture of Oracle's ECM technologies. Join this training to learn how you can manage and maintain unstructured content   Target Audience:  The Oracle ECM Implementation Boot Camp is designed for architects, technical consultants, team/project leaders and functional consultants of our system integrator partners who want to ramp-up on ECM technology.   Contents:  The ECM Implementation Boot Camp is a three-day hands-on workshop, designed for Oracle Partners who are new to ECM, and will provide implementation instruction on the ECM technology offered by Oracle. The boot camp will: • Provide hands-on experience in implementing Oracle's truly unified, open and standard base ECM technology • Provide the strategic direction about Oracle's Fusion Middleware/Enterprise 2.0 and its role in composite application development • Expose broad set of Oracle's ECM technologies.   Objectives: The Oracle ECM Implementation Boot Camp is primarily focused on the Oracle's ECM offering to manage and maintain unstructured content and covers Universal Content Management (UCM), Image and Process Management (IPM), Universal Records Management (URM), and Information Rights Management (IRM):   Topics Covered • Introduction to Oracle UCM o UCM Overview o UCM Architecture Overview • Content Server and Document Management basics o Installation and Administration Skills § User and Security Admin § Configuration (metadata, DCLs, profiles, rules, etc.) § Workflow Admin § System Properties and Component Manager § Managing Subscriptions o Contributing Content § Browser form § WebDAV folder § Desktop Integration o Searching • Web Content Management o Site Studio • Universal Records Management • Information Right Management (IRM) • Image & Process Management (IPM) • Oracle Document Capture • Oracle eMail Archive Service. Labs • Content Server Installation • Use and Administration of Content Server • Introduction to Site Studio • Use and Administration of Records Manager Demo: The R&D Group and the New Patent Focus: Information Rights Management, Knowledge Management, Accounts Payable Image Automation, Imaging and Process Management Case Study Use Case 1: Enable City of Xalco to streamline internal processes by empowering city employees to quickly and efficiently manage and publish information on their employee intranet and eventually public Web site. Use Case 2: Help Acme & Co in archiving its goal is to become "paperless" by managing all of their company's business content in a central, Web-based repository. Acme's business content ranges from policies and procedures to Employee listings and marketing materials.   Agenda: Day 1 ·         ECM Overview & Content Server ·         ECM Overview ·         ECM Architecture and Installation ·         UCM and Digital Asset Management DEMO ·         Lab 1 - Content Server Installation ·         Lab 2 - Use and Administration of Content Server   Day 2 ·         Web Content Management ·         Lab 2 - Use and Administration of Content ·         Server (continued) ·         Introduction to Web Content Management ·         Lab 3 - Site Studio   Day 3 ·         URM/IRM/IPM ·         Introduction to Universal Records Management ·         Lab 4 - URM ·         Introduction to Information Rights Management ·         Information Rights Management DEMO ·         Introduction to Image and Process Management ·         Image and Process Management Demo ·         Oracle Document Capture ·         Oracle eMail Archive   Material needed for Bootcamp: This Boot camp requires attendees to provide their own laptops for this class. Attendee laptops must meet the following minimum hardware/software requirements: Hardware • RAM: 2GB RM minimum (1 GB RAM is not enough) • HDD: 15 GB free HDD space   Pre requistes: To ensure a valuable learning experience, participation in this boot camp requires completing the prerequisite courses and successfully passing the prerequisite assessment test that is mapped into the Oracle Enterprise Content Management Implementation Boot Camp guided learning path. At a minimum, participants with equivalent skills and background should review the guided learning path and successfully pass the prerequisite assessment test to ensure they possess the background necessary to benefit from participation in the boot Camp.   ---------------------------------------------------------------------   Para mais informações/inscrições, contacte: Mónica Pires  21 423 51 44 Horário e Local 9:30h - 12:30h e 14:00h - 17:00 ( 6 horas/dia )Oracle, Porto Salvo - Oeiras.

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  • Juniper Strategy, LLC is hiring SharePoint Developers&hellip;

    - by Mark Rackley
    Isn’t everybody these days? It seems as though there are definitely more jobs than qualified devs these days, but yes, we are looking for a few good devs to help round out our burgeoning SharePoint team. Juniper Strategy is located in the DC area, however we will consider remote devs for the right fit. This is your chance to get in on the ground floor of a bright company that truly “gets it” when it comes to SharePoint, Project Management, and Information Assurance. We need like-minded people who “get it”, enjoy it, and who are looking for more than just a job. We have government and commercial opportunities as well as our own internal product that has a bright future of its own. Our immediate needs are for SharePoint .NET developers, but feel free to submit your resume for us to keep on file as it looks as though we’ll need several people in the coming months. Please email us your resume and salary requirements to [email protected] Below are our official job postings. Thanks for stopping by, we look forward to  hearing from you. Senior SharePoint .NET Developer Senior developer will focus on design and coding of custom, end-to-end business process solutions within the SharePoint framework. Senior developer with the ability to serve as a senior developer/mentor and manage day-to-day development tasks. Work with business consultants and clients to gather requirements to prepare business functional specifications. Analyze and recommend technical/development alternative paths based on business functional specifications. For selected development path, prepare technical specification and build the solution. Assist project manager with defining development task schedule and level-of-effort. Lead technical solution deployment. Job Requirements Minimum of 7 years experience in agile development, with at least 3 years of SharePoint-related development experience (SPS, SharePoint 2007/2010, WSS2-4). Thorough understanding of and demonstrated experience in development under the SharePoint Object Model, with focus on the WSS 3.0 foundation (MOSS 2007 Standard/Enterprise, Project Server 2007). Experience with using multiple data sources/repositories for database CRUD activities, including relational databases, SAP, Oracle e-Business. Experience with designing and deploying performance-based solutions in SharePoint for business processes that involve a very large number of records. Experience designing dynamic dashboards and mashups with data from multiple sources (internal to SharePoint as well as from external sources). Experience designing custom forms to facilitate user data entry, both with and without leveraging Forms Services. Experience building custom web part solutions. Experience with designing custom solutions for processing underlying business logic requirements including, but not limited to, SQL stored procedures, C#/ASP.Net workflows/event handlers (including timer jobs) to support multi-tiered decision trees and associated computations. Ability to create complex solution packages for deployment (e.g., feature-stapled site definitions). Must have impeccable communication skills, both written and verbal. Seeking a "tinkerer"; proactive with a thirst for knowledge (and a sense of humor). A US Citizen is required, and need to be able to pass NAC/E-Verify. An active Secret clearance is preferred. Applicants must pass a skills assessment test. MCP/MCTS or comparable certification preferred. Salary & Travel Negotiable SharePoint Project Lead Define project task schedule, work breakdown structure and level-of-effort. Serve as principal liaison to the customer to manage deliverables and expectations. Day-to-day project and team management, including preparation and maintenance of project plans, budgets, and status reports. Prepare technical briefings and presentation decks, provide briefs to C-level stakeholders. Work with business consultants and clients to gather requirements to prepare business functional specifications. Analyze and recommend technical/development alternative paths based on business functional specifications. The SharePoint Project Lead will be working with SharePoint architects and system owners to perform requirements/gap analysis and develop the underlying functional specifications. Once we have functional specifications as close to "final" as possible, the Project Lead will be responsible for preparation of the associated technical specification/development blueprint, along with assistance in preparing IV&V/test plan materials with support from other team members. This person will also be responsible for day-to-day management of "developers", but is also expected to engage in development directly as needed.  Job Requirements Minimum 8 years of technology project management across the software development life-cycle, with a minimum of 3 years of project management relating specifically to SharePoint (SPS 2003, SharePoint2007/2010) and/or Project Server. Thorough understanding of and demonstrated experience in development under the SharePoint Object Model, with focus on the WSS 3.0 foundation (MOSS 2007 Standard/Enterprise, Project Server 2007). Ability to interact and collaborate effectively with team members and stakeholders of different skill sets, personalities and needs. General "development" skill set required is a fundamental understanding of MOSS 2007 Enterprise, SP1/SP2, from the top-level of skinning to the core of the SharePoint object model. Impeccable communication skills, both written and verbal, and a sense of humor are required. The projects will require being at a client site at least 50% of the time in Washington DC (NW DC) and Maryland (near Suitland). A US Citizen is required, and need to be able to pass NAC/E-Verify. An active Secret clearance is preferred. PMP certification, PgMP preferred. Salary & Travel Negotiable

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  • Login failed for user 'sa' because the account is currently locked out. The system administrator can

    - by cabhilash
    Login failed for user 'sa' because the account is currently locked out. The system administrator can unlock it. (Microsoft SQL Server, Error: 18486) SQL server has local password policies. If policy is enabled which locks down the account after X number of failed attempts then the account is automatically locked down.This error with 'sa' account is very common. sa is default administartor login available with SQL server. So there are chances that an ousider has tried to bruteforce your system. (This can cause even if a legitimate tries to access the account with wrong password.Sometimes a user would have changed the password without informing others. So the other users would try to lo) You can unlock the account with the following options (use another admin account or connect via windows authentication) Alter account & unlock ALTER LOGIN sa WITH PASSWORD='password' UNLOCK Use another account Almost everyone is aware of the sa account. This can be the potential security risk. Even if you provide strong password hackers can lock the account by providing the wrong password. ( You can provide extra security by installing firewall or changing the default port but these measures are not always practical). As a best practice you can disable the sa account and use another account with same privileges.ALTER LOGIN sa DISABLE You can edit the lock-ot options using gpedit.msc( in command prompt type gpedit.msc and press enter). Navigate to Account Lokout policy as shown in the figure The Following options are available Account lockout threshold This security setting determines the number of failed logon attempts that causes a user account to be locked out. A locked-out account cannot be used until it is reset by an administrator or until the lockout duration for the account has expired. You can set a value between 0 and 999 failed logon attempts. If you set the value to 0, the account will never be locked out. Failed password attempts against workstations or member servers that have been locked using either CTRL+ALT+DELETE or password-protected screen savers count as failed logon attempts. Account lockout duration This security setting determines the number of minutes a locked-out account remains locked out before automatically becoming unlocked. The available range is from 0 minutes through 99,999 minutes. If you set the account lockout duration to 0, the account will be locked out until an administrator explicitly unlocks it. If an account lockout threshold is defined, the account lockout duration must be greater than or equal to the reset time. Default: None, because this policy setting only has meaning when an Account lockout threshold is specified. Reset account lockout counter after This security setting determines the number of minutes that must elapse after a failed logon attempt before the failed logon attempt counter is reset to 0 bad logon attempts. The available range is 1 minute to 99,999 minutes. If an account lockout threshold is defined, this reset time must be less than or equal to the Account lockout duration. Default: None, because this policy setting only has meaning when an Account lockout threshold is specified.When creating SQL user you can set CHECK_POLICY=on which will enforce the windows password policy on the account. The following policies will be applied Define the Enforce password history policy setting so that several previous passwords are remembered. With this policy setting, users cannot use the same password when their password expires.  Define the Maximum password age policy setting so that passwords expire as often as necessary for your environment, typically, every 30 to 90 days. With this policy setting, if an attacker cracks a password, the attacker only has access to the network until the password expires.  Define the Minimum password age policy setting so that passwords cannot be changed until they are more than a certain number of days old. This policy setting works in combination with the Enforce password historypolicy setting. If a minimum password age is defined, users cannot repeatedly change their passwords to get around the Enforce password history policy setting and then use their original password. Users must wait the specified number of days to change their passwords.  Define a Minimum password length policy setting so that passwords must consist of at least a specified number of characters. Long passwords--seven or more characters--are usually stronger than short ones. With this policy setting, users cannot use blank passwords, and they have to create passwords that are a certain number of characters long.  Enable the Password must meet complexity requirements policy setting. This policy setting checks all new passwords to ensure that they meet basic strong password requirements.  Password must meet the following complexity requirement, when they are changed or created: Not contain the user's entire Account Name or entire Full Name. The Account Name and Full Name are parsed for delimiters: commas, periods, dashes or hyphens, underscores, spaces, pound signs, and tabs. If any of these delimiters are found, the Account Name or Full Name are split and all sections are verified not to be included in the password. There is no check for any character or any three characters in succession. Contain characters from three of the following five categories:  English uppercase characters (A through Z) English lowercase characters (a through z) Base 10 digits (0 through 9) Non-alphabetic characters (for example, !, $, #, %) A catch-all category of any Unicode character that does not fall under the previous four categories. This fifth category can be regionally specific.

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  • Memory Efficient Windows SOA Server

    - by Antony Reynolds
    Installing a Memory Efficient SOA Suite 11.1.1.6 on Windows Server Well 11.1.1.6 is now available for download so I thought I would build a Windows Server environment to run it.  I will minimize the memory footprint of the installation by putting all functionality into the Admin Server of the SOA Suite domain. Required Software 64-bit JDK SOA Suite If you want 64-bit then choose “Generic” rather than “Microsoft Windows 32bit JVM” or “Linux 32bit JVM” This has links to all the required software. If you choose “Generic” then the Repository Creation Utility link does not show, you still need this so change the platform to “Microsoft Windows 32bit JVM” or “Linux 32bit JVM” to get the software. Similarly if you need a database then you need to change the platform to get the link to XE for Windows or Linux. If possible I recommend installing a 64-bit JDK as this allows you to assign more memory to individual JVMs. Windows XE will work, but it is better if you can use a full Oracle database because of the limitations on XE that sometimes cause it to run out of space with large or multiple SOA deployments. Installation Steps The following flow chart outlines the steps required in installing and configuring SOA Suite. The steps in the diagram are explained below. 64-bit? Is a 64-bit installation required?  The Windows & Linux installers will install 32-bit versions of the Sun JDK and JRockit.  A separate JDK must be installed for 64-bit. Install 64-bit JDK The 64-bit JDK can be either Hotspot or JRockit.  You can choose either JDK 1.7 or 1.6. Install WebLogic If you are using 64-bit then install WebLogic using “java –jar wls1036_generic.jar”.  Make sure you include Coherence in the installation, the easiest way to do this is to accept the “Typical” installation. SOA Suite Required? If you are not installing SOA Suite then you can jump straight ahead and create a WebLogic domain. Install SOA Suite Run the SOA Suite installer and point it at the existing Middleware Home created for WebLogic.  Note to run the SOA installer on Windows the user must have admin privileges.  I also found that on Windows Server 2008R2 I had to start the installer from a command prompt with administrative privileges, granting it privileges when it ran caused it to ignore the jreLoc parameter. Database Available? Do you have access to a database into which you can install the SOA schema.  SOA Suite requires access to an Oracle database (it is supported on other databases but I would always use an oracle database). Install Database I use an 11gR2 Oracle database to avoid XE limitations.  Make sure that you set the database character set to be unicode (AL32UTF8).  I also disabled the new security settings because they get in the way for a developer database.  Don’t forget to check that number of processes is at least 150 and number of sessions is not set, or is set to at least 200 (in the DB init parameters). Run RCU The SOA Suite database schemas are created by running the Repository Creation Utility.  Install the “SOA and BPM Infrastructure” component to support SOA Suite.  If you keep the schema prefix as “DEV” then the config wizard is easier to complete. Run Config Wizard The Config wizard creates the domain which hosts the WebLogic server instances.  To get a minimum footprint SOA installation choose the “Oracle Enterprise Manager” and “Oracle SOA Suite for developers” products.  All other required products will be automatically selected. The “for developers” installs target the appropriate components at the AdminServer rather than creating a separate managed server to house them.  This reduces the number of JVMs required to run the system and hence the amount of memory required.  This is not suitable for anything other than a developer environment as it mixes the admin and runtime functions together in a single server.  It also takes a long time to load all the required modules, making start up a slow process. If it exists I would recommend running the config wizard found in the “oracle_common/common/bin” directory under the middleware home.  This should have access to all the templates, including SOA. If you also want to run BAM in the same JVM as everything else then you need to “Select Optional Configuration” for “Managed Servers, Clusters and Machines”. To target BAM at the AdminServer delete the “bam_server1” managed server that is created by default.  This will result in BAM being targeted at the AdminServer. Installation Issues I had a few problems when I came to test everything in my mega-JVM. Following applications were not targeted and so I needed to target them at the AdminServer: b2bui composer Healthcare UI FMW Welcome Page Application (11.1.0.0.0) How Memory Efficient is It? On a Windows 2008R2 Server running under VirtualBox I was able to bring up both the 11gR2 database and SOA/BPM/BAM in 3G memory.  I allocated a minimum 512M to the PermGen and a minimum of 1.5G for the heap.  The setting from setSOADomainEnv are shown below: set DEFAULT_MEM_ARGS=-Xms1536m -Xmx2048m set PORT_MEM_ARGS=-Xms1536m -Xmx2048m set DEFAULT_MEM_ARGS=%DEFAULT_MEM_ARGS% -XX:PermSize=512m -XX:MaxPermSize=768m set PORT_MEM_ARGS=%PORT_MEM_ARGS% -XX:PermSize=512m -XX:MaxPermSize=768m I arrived at these numbers by monitoring JVM memory usage in JConsole. Task Manager showed total system memory usage at 2.9G – just below the 3G I allocated to the VM. Performance is not stellar but it runs and I could run JDeveloper alongside it on my 8G laptop, so in that sense it was a result!

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  • Getting started with Oracle Database In-Memory Part III - Querying The IM Column Store

    - by Maria Colgan
    In my previous blog posts, I described how to install, enable, and populate the In-Memory column store (IM column store). This weeks post focuses on how data is accessed within the IM column store. Let’s take a simple query “What is the most expensive air-mail order we have received to date?” SELECT Max(lo_ordtotalprice) most_expensive_order FROM lineorderWHERE  lo_shipmode = 5; The LINEORDER table has been populated into the IM column store and since we have no alternative access paths (indexes or views) the execution plan for this query is a full table scan of the LINEORDER table. You will notice that the execution plan has a new set of keywords “IN MEMORY" in the access method description in the Operation column. These keywords indicate that the LINEORDER table has been marked for INMEMORY and we may use the IM column store in this query. What do I mean by “may use”? There are a small number of cases were we won’t use the IM column store even though the object has been marked INMEMORY. This is similar to how the keyword STORAGE is used on Exadata environments. You can confirm that the IM column store was actually used by examining the session level statistics, but more on that later. For now let's focus on how the data is accessed in the IM column store and why it’s faster to access the data in the new column format, for analytical queries, rather than the buffer cache. There are four main reasons why accessing the data in the IM column store is more efficient. 1. Access only the column data needed The IM column store only has to scan two columns – lo_shipmode and lo_ordtotalprice – to execute this query while the traditional row store or buffer cache has to scan all of the columns in each row of the LINEORDER table until it reaches both the lo_shipmode and the lo_ordtotalprice column. 2. Scan and filter data in it's compressed format When data is populated into the IM column it is automatically compressed using a new set of compression algorithms that allow WHERE clause predicates to be applied against the compressed formats. This means the volume of data scanned in the IM column store for our query will be far less than the same query in the buffer cache where it will scan the data in its uncompressed form, which could be 20X larger. 3. Prune out any unnecessary data within each column The fastest read you can execute is the read you don’t do. In the IM column store a further reduction in the amount of data accessed is possible due to the In-Memory Storage Indexes(IM storage indexes) that are automatically created and maintained on each of the columns in the IM column store. IM storage indexes allow data pruning to occur based on the filter predicates supplied in a SQL statement. An IM storage index keeps track of minimum and maximum values for each column in each of the In-Memory Compression Unit (IMCU). In our query the WHERE clause predicate is on the lo_shipmode column. The IM storage index on the lo_shipdate column is examined to determine if our specified column value 5 exist in any IMCU by comparing the value 5 to the minimum and maximum values maintained in the Storage Index. If the value 5 is outside the minimum and maximum range for an IMCU, the scan of that IMCU is avoided. For the IMCUs where the value 5 does fall within the min, max range, an additional level of data pruning is possible via the metadata dictionary created when dictionary-based compression is used on IMCU. The dictionary contains a list of the unique column values within the IMCU. Since we have an equality predicate we can easily determine if 5 is one of the distinct column values or not. The combination of the IM storage index and dictionary based pruning, enables us to only scan the necessary IMCUs. 4. Use SIMD to apply filter predicates For the IMCU that need to be scanned Oracle takes advantage of SIMD vector processing (Single Instruction processing Multiple Data values). Instead of evaluating each entry in the column one at a time, SIMD vector processing allows a set of column values to be evaluated together in a single CPU instruction. The column format used in the IM column store has been specifically designed to maximize the number of column entries that can be loaded into the vector registers on the CPU and evaluated in a single CPU instruction. SIMD vector processing enables the Oracle Database In-Memory to scan billion of rows per second per core versus the millions of rows per second per core scan rate that can be achieved in the buffer cache. I mentioned earlier in this post that in order to confirm the IM column store was used; we need to examine the session level statistics. You can monitor the session level statistics by querying the performance views v$mystat and v$statname. All of the statistics related to the In-Memory Column Store begin with IM. You can see the full list of these statistics by typing: display_name format a30 SELECT display_name FROM v$statname WHERE  display_name LIKE 'IM%'; If we check the session statistics after we execute our query the results would be as follow; SELECT Max(lo_ordtotalprice) most_expensive_order FROM lineorderWHERE lo_shipmode = 5; SELECT display_name FROM v$statname WHERE  display_name IN ('IM scan CUs columns accessed',                        'IM scan segments minmax eligible',                        'IM scan CUs pruned'); As you can see, only 2 IMCUs were accessed during the scan as the majority of the IMCUs (44) in the LINEORDER table were pruned out thanks to the storage index on the lo_shipmode column. In next weeks post I will describe how you can control which queries use the IM column store and which don't. +Maria Colgan

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  • SQL SERVER – Weekly Series – Memory Lane – #053 – Final Post in Series

    - by Pinal Dave
    It has been a fantastic journey to write memory lane series for an entire year. This series gave me the opportunity to go back and see what I have contributed to this blog throughout the last 7 years. This was indeed fantastic series as this provided me the opportunity to witness how technology has grown throughout the year and how I have progressed in my career while writing this blog post. This series was indeed fantastic experience readers as many joined during the last few years and were not sure what they have missed in recent years. Let us continue with the final episode of the Memory Lane Series. Here is the list of selected articles of SQLAuthority.com across all these years. Instead of just listing all the articles I have selected a few of my most favorite articles and have listed them here with additional notes below it. Let me know which one of the following is your favorite article from memory lane. 2007 Get Current User – Get Logged In User Here is the straight script which list logged in SQL Server users. Disable All Triggers on a Database – Disable All Triggers on All Servers Question : How to disable all the triggers for a database? Additionally, how to disable all the triggers for all servers? For answer execute the script in the blog post. Importance of Master Database for SQL Server Startup I have received following questions many times. I will list all the questions here and answer them together. What is the purpose of Master database? Should our backup Master database? Which database is must have database for SQL Server for startup? Which are the default system database created when SQL Server 2005 is installed for the first time? What happens if Master database is corrupted? Answers to all of the questions are very much related. 2008 DECLARE Multiple Variables in One Statement SQL Server is a great product and it has many features which are very unique to SQL Server. Regarding feature of SQL Server where multiple variable can be declared in one statement, it is absolutely possible to do. 2009 How to Enable Index – How to Disable Index – Incorrect syntax near ‘ENABLE’ Many times I have seen that the index is disabled when there is a large update operation on the table. Bulk insert of very large file updates in any table using SSIS is usually preceded by disabling the index and followed by enabling the index. I have seen many developers running the following query to disable the index. 2010 List of all the Views from Database Many emails I received suggesting that they have hundreds of the view and now have no clue what is going on and how many of them have indexes and how many does not have an index. Some even asked me if there is any way they can get a list of the views with the property of Index along with it. Here is the quick script which does exactly the same. You can also include many other columns from the same view. Minimum Maximum Memory – Server Memory Options I was recently reading about SQL Server Memory Options over here. While reading this one line really caught my attention is minimum value allowed for maximum memory options. The default setting for min server memory is 0, and the default setting for max server memory is 2147483647. The minimum amount of memory you can specify for max server memory is 16 megabytes (MB). 2011 Fundamentals of Columnstore Index There are two kinds of storage in a database. Row Store and Column Store. Row store does exactly as the name suggests – stores rows of data on a page – and column store stores all the data in a column on the same page. These columns are much easier to search – instead of a query searching all the data in an entire row whether the data are relevant or not, column store queries need only to search a much lesser number of the columns. How to Ignore Columnstore Index Usage in Query In summary the question in simple words “How can we ignore using the column store index in selective queries?” Very interesting question – you can use I can understand there may be the cases when the column store index is not ideal and needs to be ignored the same. You can use the query hint IGNORE_NONCLUSTERED_COLUMNSTORE_INDEX to ignore the column store index. The SQL Server Engine will use any other index which is best after ignoring the column store index. 2012 Storing Variable Values in Temporary Array or Temporary List SQL Server does not support arrays or a dynamic length storage mechanism like list. Absolutely there are some clever workarounds and few extra-ordinary solutions but everybody can;t come up with such solution. Additionally, sometime the requirements are very simple that doing extraordinary coding is not required. Here is the simple case. Move Database Files MDF and LDF to Another Location It is not common to keep the Database on the same location where OS is installed. Usually Database files are in SAN, Separate Disk Array or on SSDs. This is done usually for performance reason and manageability perspective. Now the challenges comes up when database which was installed at not preferred default location and needs to move to a different location. Here is the quick tutorial how you can do it. UNION ALL and ORDER BY – How to Order Table Separately While Using UNION ALL If your requirement is such that you want your top and bottom query of the UNION resultset independently sorted but in the same result set you can add an additional static column and order by that column. Let us re-create the same scenario. Copy Data from One Table to Another Table – SQL in Sixty Seconds #031 – Video http://www.youtube.com/watch?v=FVWIA-ACMNo Reference: Pinal Dave (http://blog.sqlauthority.com)Filed under: Memory Lane, PostADay, SQL, SQL Authority, SQL Query, SQL Server, SQL Tips and Tricks, T SQL, Technology

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  • Mounting ddrescue image after recovery (in over my head)

    - by BorgDomination
    I'm having problems mounting the recovery image. I've tried to mount the image multiple ways. quark@DS9 ~ $ sudo mount -t ext4 /media/jump1/1recover/sdb1.img /mnt mount: wrong fs type, bad option, bad superblock on /dev/loop0, missing codepage or helper program, or other error In some cases useful info is found in syslog - try dmesg | tail or so quark@DS9 ~ $ sudo mount -r -o loop /media/jump1/1recover/sdb1.img recover mount: you must specify the filesystem type quark@DS9 ~ $ sudo mount /media/jump1/1recover/sdb1.img mnt mount: you must specify the filesystem type It doesn't even give me detailed information on the file I just made, nautilus says it's 160gb. quark@DS9 ~ $ file /media/jump1/1recover/sdb1.img /media/jump1/1recover/sdb1.img: data quark@DS9 ~ $ mmls /media/jump1/1recover/sdb1.img Cannot determine partition type I'm not sure what I'm doing wrong or if I started this process incorrectly from the beginning. I've outlined what I've done so far below. I'm clueless, I'd appreciate if someone had some input for me. What I have done from the beginning My laptop has two hard drives. One has the dual boot Win7 / Linux Mint system files. Secondary one contained my /home folder. The laptop was jarred and the /home disk was broken. I tried a LiveCD recovery, it failed. Wouldn't even load a Live session with the disk installed. So I turned to ddrescue. quark@DS9 ~ $ sudo fdisk -l Disk /dev/sda: 160.0 GB, 160041885696 bytes 255 heads, 63 sectors/track, 19457 cylinders, total 312581808 sectors Units = sectors of 1 * 512 = 512 bytes Sector size (logical/physical): 512 bytes / 512 bytes I/O size (minimum/optimal): 512 bytes / 512 bytes Disk identifier: 0x0009fc18 Device Boot Start End Blocks Id System /dev/sda1 * 2048 112642047 56320000 7 HPFS/NTFS/exFAT /dev/sda2 138033152 312580095 87273472 83 Linux /dev/sda3 112644094 138033151 12694529 5 Extended /dev/sda5 112644096 132173823 9764864 83 Linux /dev/sda6 132175872 138033151 2928640 82 Linux swap / Solaris Partition table entries are not in disk order Disk /dev/sdb: 160.0 GB, 160041885696 bytes 255 heads, 63 sectors/track, 19457 cylinders, total 312581808 sectors Units = sectors of 1 * 512 = 512 bytes Sector size (logical/physical): 512 bytes / 512 bytes I/O size (minimum/optimal): 512 bytes / 512 bytes Disk identifier: 0x0002a8ea Device Boot Start End Blocks Id System /dev/sdb1 * 63 312576704 156288321 83 Linux Disk /dev/sdc: 1000.2 GB, 1000204886016 bytes 255 heads, 63 sectors/track, 121601 cylinders, total 1953525168 sectors Units = sectors of 1 * 512 = 512 bytes Sector size (logical/physical): 512 bytes / 512 bytes I/O size (minimum/optimal): 512 bytes / 512 bytes Disk identifier: 0xed6d054b Device Boot Start End Blocks Id System /dev/sdc1 63 1953520064 976760001 7 HPFS/NTFS/exFAT sda - 160g internal, holds all system files and all computer functions. sdb - 160g internal, BROKEN, contains about 140g of data I'd like to recover. sdc - 1T external, contains recovery image. Only place that has space to do all this. From this site, https://apps.education.ucsb.edu/wiki/Ddrescue I used this script to create an image of the broken hard drive. I changed the destination to the external USB drive. #!/bin/sh prt=sdb1 src=/dev/$prt dst=/media/jump1/1recover/$prt.img log=$dst.log sudo time ddrescue --no-split $src $dst $log sudo time ddrescue --direct --max-retries=3 $src $dst $log sudo time ddrescue --direct --retrim --max-retries=3 $src $dst $log Everything looked like it came off without a hitch: quark@DS9 ~ $ sudo bash recover1 Press Ctrl-C to interrupt Initial status (read from logfile) rescued: 0 B, errsize: 0 B, errors: 0 Current status rescued: 160039 MB, errsize: 4096 B, current rate: 35588 B/s ipos: 3584 B, errors: 1, average rate: 22859 kB/s opos: 3584 B, time from last successful read: 0 s Finished 12.78user 1060.42system 1:56:41elapsed 15%CPU (0avgtext+0avgdata 4944maxresident)k 312580958inputs+0outputs (1major+601minor)pagefaults 0swaps Press Ctrl-C to interrupt Initial status (read from logfile) rescued: 160039 MB, errsize: 4096 B, errors: 1 Current status rescued: 160039 MB, errsize: 1024 B, current rate: 0 B/s ipos: 1536 B, errors: 1, average rate: 13 B/s opos: 1536 B, time from last successful read: 1.3 m Finished 0.00user 0.00system 3:43.95elapsed 0%CPU (0avgtext+0avgdata 4944maxresident)k 238inputs+0outputs (3major+374minor)pagefaults 0swaps Press Ctrl-C to interrupt Initial status (read from logfile) rescued: 160039 MB, errsize: 1024 B, errors: 1 Current status rescued: 160039 MB, errsize: 1024 B, current rate: 0 B/s ipos: 1536 B, errors: 1, average rate: 0 B/s opos: 1536 B, time from last successful read: 3.7 m Finished 0.00user 0.00system 3:43.56elapsed 0%CPU (0avgtext+0avgdata 4944maxresident)k 8inputs+0outputs (0major+376minor)pagefaults 0swaps It looks like, from where I'm standing it worked perfectly. Here's the log: # Rescue Logfile. Created by GNU ddrescue version 1.14 # Command line: ddrescue --direct --retrim --max-retries=3 /dev/sdb1 /media/jump1/1recover/sdb1.img /media/jump1/1recover/sdb1.img.log # current_pos current_status 0x00000600 + # pos size status 0x00000000 0x00000400 + 0x00000400 0x00000400 - 0x00000800 0x254314FC00 + I'm not sure how to proceed. Does this mean all of my data is lost???????? Appreciate ANY input!

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  • Help analyzing traceroute

    - by Abdulla
    Hello, my name is Abdulla and I'm from Kuwait. Sorry for my question as I know its not technically challenging. I'm facing some problems with my internet connection. My company has a DSL 2mb connection. My main problem is latency, in the morning its good but after that its gets really bad. My Internet provider says there's nothing wrong and that everything is working perfectly. I tried to explain to them the latency issue but they say that as long as I'm getting the download speed there isn't anything I can do about it. I only want to know if this is true and that the company can't do anything before I change my internet provider, as I feel that the guys at the contact center might getting back to me without asking tech support. Below are 2 traces I made, one in the morning and the other in the afternoon: This was taken around 17:00 Microsoft Windows XP [Version 5.1.2600] (C) Copyright 1985-2001 Microsoft Corp. C:\Documents and Settings\Administrator>ping google.com Pinging google.com [66.102.9.104] with 32 bytes of data: Reply from 66.102.9.104: bytes=32 time=387ms TTL=49 Reply from 66.102.9.104: bytes=32 time=388ms TTL=49 Reply from 66.102.9.104: bytes=32 time=375ms TTL=49 Reply from 66.102.9.104: bytes=32 time=375ms TTL=49 Ping statistics for 66.102.9.104: Packets: Sent = 4, Received = 4, Lost = 0 (0% loss), Approximate round trip times in milli-seconds: Minimum = 375ms, Maximum = 388ms, Average = 381ms C:\Documents and Settings\Administrator>ping google.com /t Pinging google.com [66.102.9.104] with 32 bytes of data: Reply from 66.102.9.104: bytes=32 time=376ms TTL=49 Reply from 66.102.9.104: bytes=32 time=382ms TTL=49 Reply from 66.102.9.104: bytes=32 time=371ms TTL=49 Reply from 66.102.9.104: bytes=32 time=378ms TTL=49 Reply from 66.102.9.104: bytes=32 time=374ms TTL=49 Reply from 66.102.9.104: bytes=32 time=371ms TTL=49 Reply from 66.102.9.104: bytes=32 time=365ms TTL=49 Reply from 66.102.9.104: bytes=32 time=366ms TTL=49 Reply from 66.102.9.104: bytes=32 time=353ms TTL=49 Reply from 66.102.9.104: bytes=32 time=331ms TTL=49 Reply from 66.102.9.104: bytes=32 time=333ms TTL=49 Reply from 66.102.9.104: bytes=32 time=348ms TTL=49 Reply from 66.102.9.104: bytes=32 time=365ms TTL=49 Reply from 66.102.9.104: bytes=32 time=346ms TTL=49 Reply from 66.102.9.104: bytes=32 time=335ms TTL=49 Reply from 66.102.9.104: bytes=32 time=340ms TTL=49 Reply from 66.102.9.104: bytes=32 time=344ms TTL=49 Reply from 66.102.9.104: bytes=32 time=333ms TTL=49 Reply from 66.102.9.104: bytes=32 time=328ms TTL=49 Reply from 66.102.9.104: bytes=32 time=332ms TTL=49 Reply from 66.102.9.104: bytes=32 time=326ms TTL=49 Reply from 66.102.9.104: bytes=32 time=333ms TTL=49 Reply from 66.102.9.104: bytes=32 time=325ms TTL=49 Reply from 66.102.9.104: bytes=32 time=333ms TTL=49 Reply from 66.102.9.104: bytes=32 time=338ms TTL=49 Reply from 66.102.9.104: bytes=32 time=341ms TTL=49 Ping statistics for 66.102.9.104: Packets: Sent = 26, Received = 26, Lost = 0 (0% loss), Approximate round trip times in milli-seconds: Minimum = 325ms, Maximum = 382ms, Average = 348ms Control-C ^C C:\Documents and Settings\Administrator>travert google.com 'travert' is not recognized as an internal or external command, operable program or batch file. C:\Documents and Settings\Administrator>tracert google.com Tracing route to google.com [66.102.9.104] over a maximum of 30 hops: 1 <1 ms <1 ms <1 ms 192.168.0.1 2 6 ms 6 ms 6 ms 80-184-31-1.adsl.kems.net [80.184.31.1] 3 7 ms 7 ms 8 ms 168.187.0.226 4 7 ms 8 ms 9 ms 168.187.0.125 5 180 ms 187 ms 188 ms if-11-2.core1.RSD-Riyad.as6453.net [116.0.78.89] 6 209 ms 222 ms 204 ms 195.219.167.57 7 541 ms 536 ms 540 ms 195.219.167.42 8 553 ms 552 ms 538 ms Vlan1102.icore1.PVU-Paris.as6453.net [195.219.24 1.109] 9 547 ms 543 ms 542 ms xe-9-1-0.edge4.paris1.level3.net [4.68.110.213] 10 540 ms 523 ms 531 ms ae-33-51.ebr1.Paris1.Level3.net [4.69.139.193] 11 755 ms 761 ms 695 ms ae-45-45.ebr1.London1.Level3.net [4.69.143.101] 12 271 ms 263 ms 400 ms ae-11-51.car1.London1.Level3.net [4.69.139.66] 13 701 ms 730 ms 742 ms 195.50.118.210 14 659 ms 641 ms 660 ms 209.85.255.76 15 280 ms 283 ms 292 ms 209.85.251.190 16 308 ms 293 ms 296 ms 72.14.232.239 17 679 ms 700 ms 721 ms 64.233.174.18 18 268 ms 281 ms 269 ms lm-in-f104.1e100.net [66.102.9.104] Trace complete. C:\Documents and Settings\Administrator> This was taken at 10:00am Microsoft Windows XP [Version 5.1.2600] (C) Copyright 1985-2001 Microsoft Corp. C:\Documents and Settings\Administrator>ping google.com Pinging google.com [66.102.9.106] with 32 bytes of data: Reply from 66.102.9.106: bytes=32 time=110ms TTL=49 Reply from 66.102.9.106: bytes=32 time=111ms TTL=49 Reply from 66.102.9.106: bytes=32 time=112ms TTL=49 Reply from 66.102.9.106: bytes=32 time=120ms TTL=49 Ping statistics for 66.102.9.106: Packets: Sent = 4, Received = 4, Lost = 0 (0% loss), Approximate round trip times in milli-seconds: Minimum = 110ms, Maximum = 120ms, Average = 113ms C:\Documents and Settings\Administrator>ping google.com /t Pinging google.com [66.102.9.106] with 32 bytes of data: Reply from 66.102.9.106: bytes=32 time=109ms TTL=49 Reply from 66.102.9.106: bytes=32 time=110ms TTL=49 Reply from 66.102.9.106: bytes=32 time=111ms TTL=49 Reply from 66.102.9.106: bytes=32 time=111ms TTL=49 Reply from 66.102.9.106: bytes=32 time=112ms TTL=49 Reply from 66.102.9.106: bytes=32 time=112ms TTL=49 Reply from 66.102.9.106: bytes=32 time=116ms TTL=49 Reply from 66.102.9.106: bytes=32 time=110ms TTL=49 Reply from 66.102.9.106: bytes=32 time=109ms TTL=49 Reply from 66.102.9.106: bytes=32 time=110ms TTL=49 Reply from 66.102.9.106: bytes=32 time=109ms TTL=49 Reply from 66.102.9.106: bytes=32 time=110ms TTL=49 Reply from 66.102.9.106: bytes=32 time=112ms TTL=49 Reply from 66.102.9.106: bytes=32 time=109ms TTL=49 Reply from 66.102.9.106: bytes=32 time=110ms TTL=49 Reply from 66.102.9.106: bytes=32 time=115ms TTL=49 Reply from 66.102.9.106: bytes=32 time=110ms TTL=49 Reply from 66.102.9.106: bytes=32 time=109ms TTL=49 Reply from 66.102.9.106: bytes=32 time=110ms TTL=49 Reply from 66.102.9.106: bytes=32 time=113ms TTL=49 Reply from 66.102.9.106: bytes=32 time=115ms TTL=49 Reply from 66.102.9.106: bytes=32 time=109ms TTL=49 Reply from 66.102.9.106: bytes=32 time=110ms TTL=49 Ping statistics for 66.102.9.106: Packets: Sent = 32, Received = 32, Lost = 0 (0% loss), Approximate round trip times in milli-seconds: Minimum = 109ms, Maximum = 135ms, Average = 112ms Control-C ^C C:\Documents and Settings\Administrator>tracert google.com Tracing route to google.com [66.102.9.104] over a maximum of 30 hops: 1 <1 ms <1 ms <1 ms 192.168.0.1 2 6 ms 6 ms 6 ms 80-184-31-1.adsl.kems.net [80.184.31.1] 3 8 ms 7 ms 6 ms 168.187.0.226 4 6 ms 7 ms 7 ms 168.187.0.125 5 20 ms 20 ms 18 ms if-11-2.core1.RSD-Riyad.as6453.net [116.0.78.89] 6 171 ms 205 ms 215 ms 195.219.167.57 7 191 ms 215 ms 226 ms 195.219.167.42 8 * 103 ms 94 ms Vlan1102.icore1.PVU-Paris.as6453.net [195.219.24 1.109] 9 94 ms 95 ms 97 ms xe-9-1-0.edge4.paris1.level3.net [4.68.110.213] 10 94 ms 94 ms 94 ms ae-33-51.ebr1.Paris1.Level3.net [4.69.139.193] 11 101 ms 101 ms 101 ms ae-48-48.ebr1.London1.Level3.net [4.69.143.113] 12 102 ms 102 ms 101 ms ae-11-51.car1.London1.Level3.net [4.69.139.66] 13 103 ms 102 ms 103 ms 195.50.118.210 14 137 ms 103 ms 100 ms 209.85.255.76 15 130 ms 124 ms 124 ms 209.85.251.190 16 114 ms 116 ms 116 ms 72.14.232.239 17 135 ms 113 ms 126 ms 64.233.174.18 18 126 ms 125 ms 127 ms lm-in-f104.1e100.net [66.102.9.104] Trace complete. C:\Documents and Settings\Administrator>

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  • Hide ticks at Min and Max in WPF Slider

    - by gehho
    Hi, I want to display a Slider ranging from 0.5 to 1.5 with only one tick mark at 1.0 to mark the center and default value. I have defined a Slider as follows: <Slider Minimum="0.5" Maximum="1.5" IsMoveToPointEnabled="True" IsSnapToTickEnabled="False" Orientation="Horizontal" Ticks="1.0" TickPlacement="BottomRight" Value="{Binding SomeProperty, Mode=TwoWay}"/> However, besides a tick mark at 0.0 this Slider also shows tick marks at 0.5 and 1.5, i.e. the Minimum and Maximum values. Is there a way to hide these min/max tick marks?! I checked all properties and tried changing some of them, but did not have success so far. Thanks, gehho.

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  • Please bear with me, can someone analyze this trace route please

    - by Abdulla
    Hello, my name is Abdulla and I'm from Kuwait. Sorry for my question as I know its not technically challenging. I'm facing some problems with my internet connection while gaming, I have DSL 2mb connection. My main problem is latency, in the morning its good but after that its gets really bad. My internet provider says there's nothing wrong and that everything is working perfectly. I tried to explain to them the latency issue but they say that as long as I'm getting the download speed there isn't anything I can do about it. I only want to know if this is true and that the company can't do anything before I change my internet provider, as I feel that the guys at the contact center might getting back to me without asking tech support. Below are 2 traces I made, one in the morning and the other in the afternoon: This was taken around 17:00 Microsoft Windows XP [Version 5.1.2600] (C) Copyright 1985-2001 Microsoft Corp. C:\Documents and Settings\Administrator>ping google.com Pinging google.com [66.102.9.104] with 32 bytes of data: Reply from 66.102.9.104: bytes=32 time=387ms TTL=49 Reply from 66.102.9.104: bytes=32 time=388ms TTL=49 Reply from 66.102.9.104: bytes=32 time=375ms TTL=49 Reply from 66.102.9.104: bytes=32 time=375ms TTL=49 Ping statistics for 66.102.9.104: Packets: Sent = 4, Received = 4, Lost = 0 (0% loss), Approximate round trip times in milli-seconds: Minimum = 375ms, Maximum = 388ms, Average = 381ms C:\Documents and Settings\Administrator>ping google.com /t Pinging google.com [66.102.9.104] with 32 bytes of data: Reply from 66.102.9.104: bytes=32 time=376ms TTL=49 Reply from 66.102.9.104: bytes=32 time=382ms TTL=49 Reply from 66.102.9.104: bytes=32 time=371ms TTL=49 Reply from 66.102.9.104: bytes=32 time=378ms TTL=49 Reply from 66.102.9.104: bytes=32 time=374ms TTL=49 Reply from 66.102.9.104: bytes=32 time=371ms TTL=49 Reply from 66.102.9.104: bytes=32 time=365ms TTL=49 Reply from 66.102.9.104: bytes=32 time=366ms TTL=49 Reply from 66.102.9.104: bytes=32 time=353ms TTL=49 Reply from 66.102.9.104: bytes=32 time=331ms TTL=49 Reply from 66.102.9.104: bytes=32 time=333ms TTL=49 Reply from 66.102.9.104: bytes=32 time=348ms TTL=49 Reply from 66.102.9.104: bytes=32 time=365ms TTL=49 Reply from 66.102.9.104: bytes=32 time=346ms TTL=49 Reply from 66.102.9.104: bytes=32 time=335ms TTL=49 Reply from 66.102.9.104: bytes=32 time=340ms TTL=49 Reply from 66.102.9.104: bytes=32 time=344ms TTL=49 Reply from 66.102.9.104: bytes=32 time=333ms TTL=49 Reply from 66.102.9.104: bytes=32 time=328ms TTL=49 Reply from 66.102.9.104: bytes=32 time=332ms TTL=49 Reply from 66.102.9.104: bytes=32 time=326ms TTL=49 Reply from 66.102.9.104: bytes=32 time=333ms TTL=49 Reply from 66.102.9.104: bytes=32 time=325ms TTL=49 Reply from 66.102.9.104: bytes=32 time=333ms TTL=49 Reply from 66.102.9.104: bytes=32 time=338ms TTL=49 Reply from 66.102.9.104: bytes=32 time=341ms TTL=49 Ping statistics for 66.102.9.104: Packets: Sent = 26, Received = 26, Lost = 0 (0% loss), Approximate round trip times in milli-seconds: Minimum = 325ms, Maximum = 382ms, Average = 348ms Control-C ^C C:\Documents and Settings\Administrator>travert google.com 'travert' is not recognized as an internal or external command, operable program or batch file. C:\Documents and Settings\Administrator>tracert google.com Tracing route to google.com [66.102.9.104] over a maximum of 30 hops: 1 <1 ms <1 ms <1 ms 192.168.0.1 2 6 ms 6 ms 6 ms 80-184-31-1.adsl.kems.net [80.184.31.1] 3 7 ms 7 ms 8 ms 168.187.0.226 4 7 ms 8 ms 9 ms 168.187.0.125 5 180 ms 187 ms 188 ms if-11-2.core1.RSD-Riyad.as6453.net [116.0.78.89] 6 209 ms 222 ms 204 ms 195.219.167.57 7 541 ms 536 ms 540 ms 195.219.167.42 8 553 ms 552 ms 538 ms Vlan1102.icore1.PVU-Paris.as6453.net [195.219.24 1.109] 9 547 ms 543 ms 542 ms xe-9-1-0.edge4.paris1.level3.net [4.68.110.213] 10 540 ms 523 ms 531 ms ae-33-51.ebr1.Paris1.Level3.net [4.69.139.193] 11 755 ms 761 ms 695 ms ae-45-45.ebr1.London1.Level3.net [4.69.143.101] 12 271 ms 263 ms 400 ms ae-11-51.car1.London1.Level3.net [4.69.139.66] 13 701 ms 730 ms 742 ms 195.50.118.210 14 659 ms 641 ms 660 ms 209.85.255.76 15 280 ms 283 ms 292 ms 209.85.251.190 16 308 ms 293 ms 296 ms 72.14.232.239 17 679 ms 700 ms 721 ms 64.233.174.18 18 268 ms 281 ms 269 ms lm-in-f104.1e100.net [66.102.9.104] Trace complete. C:\Documents and Settings\Administrator> This was taken at 10:00am Microsoft Windows XP [Version 5.1.2600] (C) Copyright 1985-2001 Microsoft Corp. C:\Documents and Settings\Administrator>ping google.com Pinging google.com [66.102.9.106] with 32 bytes of data: Reply from 66.102.9.106: bytes=32 time=110ms TTL=49 Reply from 66.102.9.106: bytes=32 time=111ms TTL=49 Reply from 66.102.9.106: bytes=32 time=112ms TTL=49 Reply from 66.102.9.106: bytes=32 time=120ms TTL=49 Ping statistics for 66.102.9.106: Packets: Sent = 4, Received = 4, Lost = 0 (0% loss), Approximate round trip times in milli-seconds: Minimum = 110ms, Maximum = 120ms, Average = 113ms C:\Documents and Settings\Administrator>ping google.com /t Pinging google.com [66.102.9.106] with 32 bytes of data: Reply from 66.102.9.106: bytes=32 time=109ms TTL=49 Reply from 66.102.9.106: bytes=32 time=110ms TTL=49 Reply from 66.102.9.106: bytes=32 time=111ms TTL=49 Reply from 66.102.9.106: bytes=32 time=111ms TTL=49 Reply from 66.102.9.106: bytes=32 time=112ms TTL=49 Reply from 66.102.9.106: bytes=32 time=112ms TTL=49 Reply from 66.102.9.106: bytes=32 time=116ms TTL=49 Reply from 66.102.9.106: bytes=32 time=110ms TTL=49 Reply from 66.102.9.106: bytes=32 time=109ms TTL=49 Reply from 66.102.9.106: bytes=32 time=110ms TTL=49 Reply from 66.102.9.106: bytes=32 time=109ms TTL=49 Reply from 66.102.9.106: bytes=32 time=110ms TTL=49 Reply from 66.102.9.106: bytes=32 time=112ms TTL=49 Reply from 66.102.9.106: bytes=32 time=109ms TTL=49 Reply from 66.102.9.106: bytes=32 time=110ms TTL=49 Reply from 66.102.9.106: bytes=32 time=115ms TTL=49 Reply from 66.102.9.106: bytes=32 time=110ms TTL=49 Reply from 66.102.9.106: bytes=32 time=109ms TTL=49 Reply from 66.102.9.106: bytes=32 time=110ms TTL=49 Reply from 66.102.9.106: bytes=32 time=113ms TTL=49 Reply from 66.102.9.106: bytes=32 time=115ms TTL=49 Reply from 66.102.9.106: bytes=32 time=109ms TTL=49 Reply from 66.102.9.106: bytes=32 time=110ms TTL=49 Ping statistics for 66.102.9.106: Packets: Sent = 32, Received = 32, Lost = 0 (0% loss), Approximate round trip times in milli-seconds: Minimum = 109ms, Maximum = 135ms, Average = 112ms Control-C ^C C:\Documents and Settings\Administrator>tracert google.com Tracing route to google.com [66.102.9.104] over a maximum of 30 hops: 1 <1 ms <1 ms <1 ms 192.168.0.1 2 6 ms 6 ms 6 ms 80-184-31-1.adsl.kems.net [80.184.31.1] 3 8 ms 7 ms 6 ms 168.187.0.226 4 6 ms 7 ms 7 ms 168.187.0.125 5 20 ms 20 ms 18 ms if-11-2.core1.RSD-Riyad.as6453.net [116.0.78.89] 6 171 ms 205 ms 215 ms 195.219.167.57 7 191 ms 215 ms 226 ms 195.219.167.42 8 * 103 ms 94 ms Vlan1102.icore1.PVU-Paris.as6453.net [195.219.24 1.109] 9 94 ms 95 ms 97 ms xe-9-1-0.edge4.paris1.level3.net [4.68.110.213] 10 94 ms 94 ms 94 ms ae-33-51.ebr1.Paris1.Level3.net [4.69.139.193] 11 101 ms 101 ms 101 ms ae-48-48.ebr1.London1.Level3.net [4.69.143.113] 12 102 ms 102 ms 101 ms ae-11-51.car1.London1.Level3.net [4.69.139.66] 13 103 ms 102 ms 103 ms 195.50.118.210 14 137 ms 103 ms 100 ms 209.85.255.76 15 130 ms 124 ms 124 ms 209.85.251.190 16 114 ms 116 ms 116 ms 72.14.232.239 17 135 ms 113 ms 126 ms 64.233.174.18 18 126 ms 125 ms 127 ms lm-in-f104.1e100.net [66.102.9.104] Trace complete. C:\Documents and Settings\Administrator>

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  • Flex ProgressBar component problem

    - by Abhinav
    I am trying to use the ProgressBar Flex component inside a custom Actionscript 3.0 component derived from the UIComponent class. I have set the minimum and maximum values etc. _progressBar = new ProgressBar(); _progressBar.label = "Loading"; _progressBar.minimum = 0; _progressBar.maximum = 100; _progressBar.direction = ProgressBarDirection.RIGHT; _progressBar.mode = ProgressBarMode.MANUAL; The component shows the "Loading" text but not the loading bar. Anything like _progressBar.setProgress(20, 100) does not have any effect on the code. Any ideas why this is not working?

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  • How can I bind a winforms Opacity to a TrackBar (slider)

    - by Allen
    I've got a winform with a BindingSource that has an int property named Opacity in its DataSource. I also have a TrackBar on the winform that I want to use to control the Opacity of the winform. I've bound the Value property on the TrackBar to the Opacity and that functions just fine, sliding the TrackBar will change the variable from TrackBar.Minimum (0) to TrackBar.Maximum (1). I've also bound the Opacity property of the winform to this value, however, since the TrackBar's values only go from Minimum to Maximum in +/-1 rather than +/- .1 or so (like Opacity does), it doesn't properly fade the winform. Instead, 0 will turn it opaque and 1 will turn it fully visible. I need a way to work within the architecture described above, but get the TrackBar to change its value from 0 to 1 in defined increments smaller than 1.

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  • Help with password complexity regex

    - by Alex
    I'm using the following regex to validate password complexity: /^.*(?=.{6,12})(?=.*[0-9]{2})(?=.*[A-Z]{2})(?=.*[a-z]{2}).*$/ In a nutshell: 2 lowercase, 2 uppercase, 2 numbers, min length is 6 and max length is 12. It works perfectly, except for the maximum length, when I'm using a minimum length as well. For example: /^.*(?=.{6,})(?=.*[0-9]{2})(?=.*[A-Z]{2})(?=.*[a-z]{2}).*$/ This correctly requires a minimum length of 6! And this: /^.*(?=.{,12})(?=.*[0-9]{2})(?=.*[A-Z]{2})(?=.*[a-z]{2}).*$/ Correctly requires a maximum length of 12. However, when I pair them together as in the first example, it just doesn't work!! What gives? Thanks!

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  • MySQL Student database beginner SQL employment expectations

    - by sammysmall
    Background, Student pursuing BSIT degree, employer expectations, entry level. I would like to solicit opinions from this forum as to what your professional expectations are as regards an entry level position in working in the database realm... I see many job opportunities that require a minimum of 2 or more years experience, how does one go about obtaining this experience (I have tried very hard to maintain a minimum 3.70+ GPA) but have ZERO work experience in this field... I spend non school time working in VB and SQL to try and increase my proficiency. I have considered postponing my job search until I obtain certifications from brainbench etc... Any criticism and or advise is welcomed... Again I have no experience in database other than undergrad work in class. Thank you for your time sammysmall

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  • computing "node closure" of graph with removal

    - by Fakrudeen
    Given a directed graph, the goal is to combine the node with the nodes it is pointing to and come up with minimum number of these [lets give the name] super nodes. The catch is once you combine the nodes you can't use those nodes again. [first node as well as all the combined nodes - that is all the members of one super node] The greedy approach would be to pick the node with maximum out degree and combine that node with nodes it is pointing to and remove all of them. Do this every time with the nodes which are not removed yet from graph. The greedy is O(V), but this won't necessarily output minimum number super nodes. So what is the best algorithm to do this?

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