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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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  • 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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  • 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 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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  • SQL SERVER – PAGEIOLATCH_DT, PAGEIOLATCH_EX, PAGEIOLATCH_KP, PAGEIOLATCH_SH, PAGEIOLATCH_UP – Wait Type – Day 9 of 28

    - by pinaldave
    It is very easy to say that you replace your hardware as that is not up to the mark. In reality, it is very difficult to implement. It is really hard to convince an infrastructure team to change any hardware because they are not performing at their best. I had a nightmare related to this issue in a deal with an infrastructure team as I suggested that they replace their faulty hardware. This is because they were initially not accepting the fact that it is the fault of their hardware. But it is really easy to say “Trust me, I am correct”, while it is equally important that you put some logical reasoning along with this statement. PAGEIOLATCH_XX is such a kind of those wait stats that we would directly like to blame on the underlying subsystem. Of course, most of the time, it is correct – the underlying subsystem is usually the problem. From Book On-Line: PAGEIOLATCH_DT Occurs when a task is waiting on a latch for a buffer that is in an I/O request. The latch request is in Destroy mode. Long waits may indicate problems with the disk subsystem. PAGEIOLATCH_EX Occurs when a task is waiting on a latch for a buffer that is in an I/O request. The latch request is in Exclusive mode. Long waits may indicate problems with the disk subsystem. PAGEIOLATCH_KP Occurs when a task is waiting on a latch for a buffer that is in an I/O request. The latch request is in Keep mode. Long waits may indicate problems with the disk subsystem. PAGEIOLATCH_SH Occurs when a task is waiting on a latch for a buffer that is in an I/O request. The latch request is in Shared mode. Long waits may indicate problems with the disk subsystem. PAGEIOLATCH_UP Occurs when a task is waiting on a latch for a buffer that is in an I/O request. The latch request is in Update mode. Long waits may indicate problems with the disk subsystem. PAGEIOLATCH_XX Explanation: Simply put, this particular wait type occurs when any of the tasks is waiting for data from the disk to move to the buffer cache. ReducingPAGEIOLATCH_XX wait: Just like any other wait type, this is again a very challenging and interesting subject to resolve. Here are a few things you can experiment on: Improve your IO subsystem speed (read the first paragraph of this article, if you have not read it, I repeat that it is easy to say a step like this than to actually implement or do it). This type of wait stats can also happen due to memory pressure or any other memory issues. Putting aside the issue of a faulty IO subsystem, this wait type warrants proper analysis of the memory counters. If due to any reasons, the memory is not optimal and unable to receive the IO data. This situation can create this kind of wait type. Proper placing of files is very important. We should check file system for the proper placement of files – LDF and MDF on separate drive, TempDB on separate drive, hot spot tables on separate filegroup (and on separate disk), etc. Check the File Statistics and see if there is higher IO Read and IO Write Stall SQL SERVER – Get File Statistics Using fn_virtualfilestats. It is very possible that there are no proper indexes on the system and there are lots of table scans and heap scans. Creating proper index can reduce the IO bandwidth considerably. If SQL Server can use appropriate cover index instead of clustered index, it can significantly reduce lots of CPU, Memory and IO (considering cover index has much lesser columns than cluster table and all other it depends conditions). You can refer to the two articles’ links below previously written by me that talk about how to optimize indexes. Create Missing Indexes Drop Unused Indexes Updating statistics can help the Query Optimizer to render optimal plan, which can only be either directly or indirectly. I have seen that updating statistics with full scan (again, if your database is huge and you cannot do this – never mind!) can provide optimal information to SQL Server optimizer leading to efficient plan. Checking Memory Related Perfmon Counters SQLServer: Memory Manager\Memory Grants Pending (Consistent higher value than 0-2) SQLServer: Memory Manager\Memory Grants Outstanding (Consistent higher value, Benchmark) SQLServer: Buffer Manager\Buffer Hit Cache Ratio (Higher is better, greater than 90% for usually smooth running system) SQLServer: Buffer Manager\Page Life Expectancy (Consistent lower value than 300 seconds) Memory: Available Mbytes (Information only) Memory: Page Faults/sec (Benchmark only) Memory: Pages/sec (Benchmark only) Checking Disk Related Perfmon Counters Average Disk sec/Read (Consistent higher value than 4-8 millisecond is not good) Average Disk sec/Write (Consistent higher value than 4-8 millisecond is not good) Average Disk Read/Write Queue Length (Consistent higher value than benchmark is not good) Note: The information presented here is from my experience and there is no way that I claim it to be accurate. I suggest reading Book OnLine for further clarification. All of the discussions of Wait Stats in this blog is generic and varies from system to system. It is recommended that you test this on a development server before implementing it to a production server. Reference: Pinal Dave (http://blog.SQLAuthority.com) Filed under: Pinal Dave, PostADay, SQL, SQL Authority, SQL Query, SQL Scripts, SQL Server, SQL Tips and Tricks, SQL Wait Stats, SQL Wait Types, T SQL, Technology

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  • Coherence Warnings in WLS

    - by john.graves(at)oracle.com
    With 11g (10.3.4 WLS), coherence is now built into many applications.  I’ve been noticing errors in my OSB logs like these:####<10/03/2011 10:45:40 AM EST> <Warning> <Coherence> <osb-jeos> <osb_server1> <Logger@324239121 3.6.0.4> <<anonymous>> <> <583c1 0bfdbd326ba:-8c38159:12e9d02c829:-8000-0000000000000003> <1299714340643> <BEA-000000> <Oracle Coherence 3.6.0.4 (member=n/a): Unic astUdpSocket failed to set receive buffer size to 714 packets (1023KB); actual size is 12%, 89 packets (127KB). Consult your OS do cumentation regarding increasing the maximum socket buffer size. Proceeding with the actual value may cause sub-optimal performanc e.> ####<10/03/2011 10:45:40 AM EST> <Warning> <Coherence> <osb-jeos> <osb_server1> <Logger@324239121 3.6.0.4> <<anonymous>> <> <583c1 0bfdbd326ba:-8c38159:12e9d02c829:-8000-0000000000000003> <1299714340650> <BEA-000000> <Oracle Coherence 3.6.0.4 (member=n/a): Pref erredUnicastUdpSocket failed to set receive buffer size to 1428 packets (1.99MB); actual size is 6%, 89 packets (127KB). Consult y our OS documentation regarding increasing the maximum socket buffer size. Proceeding with the actual value may cause sub-optimal p erformance.> ####<10/03/2011 10:45:40 AM EST> <Warning> <Coherence> <osb-jeos> <osb_server1> <Logger@324239121 3.6.0.4> <<anonymous>> <> <583c1 0bfdbd326ba:-8c38159:12e9d02c829:-8000-0000000000000003> <1299714340659> <BEA-000000> <Oracle Coherence 3.6.0.4 (member=n/a): Mult icastUdpSocket failed to set receive buffer size to 714 packets (1023KB); actual size is 12%, 89 packets (127KB). Consult your OS documentation regarding increasing the maximum socket buffer size. Proceeding with the actual value may cause sub-optimal performa nce.> I was able to “fix” this on my ubuntu system by adding the following lines to the /etc/sysctl.conf file:# Setup networking for coherence # maximum receive socket buffer size, default 131071 net.core.rmem_max = 2000000 # maximum send socket buffer size, default 131071 net.core.wmem_max = 1000000 # default receive socket buffer size, default 65535 net.core.rmem_default = 2524287 # default send socket buffer size, default 65535 net.core.wmem_default = 2524287 .csharpcode, .csharpcode pre { font-size: small; color: black; font-family: consolas, "Courier New", courier, monospace; background-color: #ffffff; /*white-space: pre;*/ } .csharpcode pre { margin: 0em; } .csharpcode .rem { color: #008000; } .csharpcode .kwrd { color: #0000ff; } .csharpcode .str { color: #006080; } .csharpcode .op { color: #0000c0; } .csharpcode .preproc { color: #cc6633; } .csharpcode .asp { background-color: #ffff00; } .csharpcode .html { color: #800000; } .csharpcode .attr { color: #ff0000; } .csharpcode .alt { background-color: #f4f4f4; width: 100%; margin: 0em; } .csharpcode .lnum { color: #606060; }   .csharpcode, .csharpcode pre { font-size: small; color: black; font-family: consolas, "Courier New", courier, monospace; background-color: #ffffff; /*white-space: pre;*/ } .csharpcode pre { margin: 0em; } .csharpcode .rem { color: #008000; } .csharpcode .kwrd { color: #0000ff; } .csharpcode .str { color: #006080; } .csharpcode .op { color: #0000c0; } .csharpcode .preproc { color: #cc6633; } .csharpcode .asp { background-color: #ffff00; } .csharpcode .html { color: #800000; } .csharpcode .attr { color: #ff0000; } .csharpcode .alt { background-color: #f4f4f4; width: 100%; margin: 0em; } .csharpcode .lnum { color: #606060; }

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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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  • Plan Caching and Query Memory Part II (Hash Match) – When not to use stored procedure - Most common performance mistake SQL Server developers make.

    - by sqlworkshops
    SQL Server estimates Memory requirement at compile time, when stored procedure or other plan caching mechanisms like sp_executesql or prepared statement are used, the memory requirement is estimated based on first set of execution parameters. This is a common reason for spill over tempdb and hence poor performance. Common memory allocating queries are that perform Sort and do Hash Match operations like Hash Join or Hash Aggregation or Hash Union. This article covers Hash Match operations with examples. It is recommended to read Plan Caching and Query Memory Part I before this article which covers an introduction and Query memory for Sort. In most cases it is cheaper to pay for the compilation cost of dynamic queries than huge cost for spill over tempdb, unless memory requirement for a query does not change significantly based on predicates.   This article covers underestimation / overestimation of memory for Hash Match operation. Plan Caching and Query Memory Part I covers underestimation / overestimation for Sort. It is important to note that underestimation of memory for Sort and Hash Match operations lead to spill over tempdb and hence negatively impact performance. Overestimation of memory affects the memory needs of other concurrently executing queries. In addition, it is important to note, with Hash Match operations, overestimation of memory can actually lead to poor performance.   To read additional articles I wrote click here.   The best way to learn is to practice. To create the below tables and reproduce the behavior, join the mailing list by using this link: www.sqlworkshops.com/ml and I will send you the table creation script. Most of these concepts are also covered in our webcasts: www.sqlworkshops.com/webcasts  Let’s create a Customer’s State table that has 99% of customers in NY and the rest 1% in WA.Customers table used in Part I of this article is also used here.To observe Hash Warning, enable 'Hash Warning' in SQL Profiler under Events 'Errors and Warnings'. --Example provided by www.sqlworkshops.com drop table CustomersState go create table CustomersState (CustomerID int primary key, Address char(200), State char(2)) go insert into CustomersState (CustomerID, Address) select CustomerID, 'Address' from Customers update CustomersState set State = 'NY' where CustomerID % 100 != 1 update CustomersState set State = 'WA' where CustomerID % 100 = 1 go update statistics CustomersState with fullscan go   Let’s create a stored procedure that joins customers with CustomersState table with a predicate on State. --Example provided by www.sqlworkshops.com create proc CustomersByState @State char(2) as begin declare @CustomerID int select @CustomerID = e.CustomerID from Customers e inner join CustomersState es on (e.CustomerID = es.CustomerID) where es.State = @State option (maxdop 1) end go  Let’s execute the stored procedure first with parameter value ‘WA’ – which will select 1% of data. set statistics time on go --Example provided by www.sqlworkshops.com exec CustomersByState 'WA' goThe stored procedure took 294 ms to complete.  The stored procedure was granted 6704 KB based on 8000 rows being estimated.  The estimated number of rows, 8000 is similar to actual number of rows 8000 and hence the memory estimation should be ok.  There was no Hash Warning in SQL Profiler. To observe Hash Warning, enable 'Hash Warning' in SQL Profiler under Events 'Errors and Warnings'.   Now let’s execute the stored procedure with parameter value ‘NY’ – which will select 99% of data. -Example provided by www.sqlworkshops.com exec CustomersByState 'NY' go  The stored procedure took 2922 ms to complete.   The stored procedure was granted 6704 KB based on 8000 rows being estimated.    The estimated number of rows, 8000 is way different from the actual number of rows 792000 because the estimation is based on the first set of parameter value supplied to the stored procedure which is ‘WA’ in our case. This underestimation will lead to spill over tempdb, resulting in poor performance.   There was Hash Warning (Recursion) in SQL Profiler. To observe Hash Warning, enable 'Hash Warning' in SQL Profiler under Events 'Errors and Warnings'.   Let’s recompile the stored procedure and then let’s first execute the stored procedure with parameter value ‘NY’.  In a production instance it is not advisable to use sp_recompile instead one should use DBCC FREEPROCCACHE (plan_handle). This is due to locking issues involved with sp_recompile, refer to our webcasts, www.sqlworkshops.com/webcasts for further details.   exec sp_recompile CustomersByState go --Example provided by www.sqlworkshops.com exec CustomersByState 'NY' go  Now the stored procedure took only 1046 ms instead of 2922 ms.   The stored procedure was granted 146752 KB of memory. The estimated number of rows, 792000 is similar to actual number of rows of 792000. Better performance of this stored procedure execution is due to better estimation of memory and avoiding spill over tempdb.   There was no Hash Warning in SQL Profiler.   Now let’s execute the stored procedure with parameter value ‘WA’. --Example provided by www.sqlworkshops.com exec CustomersByState 'WA' go  The stored procedure took 351 ms to complete, higher than the previous execution time of 294 ms.    This stored procedure was granted more memory (146752 KB) than necessary (6704 KB) based on parameter value ‘NY’ for estimation (792000 rows) instead of parameter value ‘WA’ for estimation (8000 rows). This is because the estimation is based on the first set of parameter value supplied to the stored procedure which is ‘NY’ in this case. This overestimation leads to poor performance of this Hash Match operation, it might also affect the performance of other concurrently executing queries requiring memory and hence overestimation is not recommended.     The estimated number of rows, 792000 is much more than the actual number of rows of 8000.  Intermediate Summary: This issue can be avoided by not caching the plan for memory allocating queries. Other possibility is to use recompile hint or optimize for hint to allocate memory for predefined data range.Let’s recreate the stored procedure with recompile hint. --Example provided by www.sqlworkshops.com drop proc CustomersByState go create proc CustomersByState @State char(2) as begin declare @CustomerID int select @CustomerID = e.CustomerID from Customers e inner join CustomersState es on (e.CustomerID = es.CustomerID) where es.State = @State option (maxdop 1, recompile) end go  Let’s execute the stored procedure initially with parameter value ‘WA’ and then with parameter value ‘NY’. --Example provided by www.sqlworkshops.com exec CustomersByState 'WA' go exec CustomersByState 'NY' go  The stored procedure took 297 ms and 1102 ms in line with previous optimal execution times.   The stored procedure with parameter value ‘WA’ has good estimation like before.   Estimated number of rows of 8000 is similar to actual number of rows of 8000.   The stored procedure with parameter value ‘NY’ also has good estimation and memory grant like before because the stored procedure was recompiled with current set of parameter values.  Estimated number of rows of 792000 is similar to actual number of rows of 792000.    The compilation time and compilation CPU of 1 ms is not expensive in this case compared to the performance benefit.   There was no Hash Warning in SQL Profiler.   Let’s recreate the stored procedure with optimize for hint of ‘NY’. --Example provided by www.sqlworkshops.com drop proc CustomersByState go create proc CustomersByState @State char(2) as begin declare @CustomerID int select @CustomerID = e.CustomerID from Customers e inner join CustomersState es on (e.CustomerID = es.CustomerID) where es.State = @State option (maxdop 1, optimize for (@State = 'NY')) end go  Let’s execute the stored procedure initially with parameter value ‘WA’ and then with parameter value ‘NY’. --Example provided by www.sqlworkshops.com exec CustomersByState 'WA' go exec CustomersByState 'NY' go  The stored procedure took 353 ms with parameter value ‘WA’, this is much slower than the optimal execution time of 294 ms we observed previously. This is because of overestimation of memory. The stored procedure with parameter value ‘NY’ has optimal execution time like before.   The stored procedure with parameter value ‘WA’ has overestimation of rows because of optimize for hint value of ‘NY’.   Unlike before, more memory was estimated to this stored procedure based on optimize for hint value ‘NY’.    The stored procedure with parameter value ‘NY’ has good estimation because of optimize for hint value of ‘NY’. Estimated number of rows of 792000 is similar to actual number of rows of 792000.   Optimal amount memory was estimated to this stored procedure based on optimize for hint value ‘NY’.   There was no Hash Warning in SQL Profiler.   This article covers underestimation / overestimation of memory for Hash Match operation. Plan Caching and Query Memory Part I covers underestimation / overestimation for Sort. It is important to note that underestimation of memory for Sort and Hash Match operations lead to spill over tempdb and hence negatively impact performance. Overestimation of memory affects the memory needs of other concurrently executing queries. In addition, it is important to note, with Hash Match operations, overestimation of memory can actually lead to poor performance.   Summary: Cached plan might lead to underestimation or overestimation of memory because the memory is estimated based on first set of execution parameters. It is recommended not to cache the plan if the amount of memory required to execute the stored procedure has a wide range of possibilities. One can mitigate this by using recompile hint, but that will lead to compilation overhead. However, in most cases it might be ok to pay for compilation rather than spilling sort over tempdb which could be very expensive compared to compilation cost. The other possibility is to use optimize for hint, but in case one sorts more data than hinted by optimize for hint, this will still lead to spill. On the other side there is also the possibility of overestimation leading to unnecessary memory issues for other concurrently executing queries. In case of Hash Match operations, this overestimation of memory might lead to poor performance. When the values used in optimize for hint are archived from the database, the estimation will be wrong leading to worst performance, so one has to exercise caution before using optimize for hint, recompile hint is better in this case.   I explain these concepts with detailed examples in my webcasts (www.sqlworkshops.com/webcasts), I recommend you to watch them. The best way to learn is to practice. To create the above tables and reproduce the behavior, join the mailing list at www.sqlworkshops.com/ml and I will send you the relevant SQL Scripts.  Register for the upcoming 3 Day Level 400 Microsoft SQL Server 2008 and SQL Server 2005 Performance Monitoring & Tuning Hands-on Workshop in London, United Kingdom during March 15-17, 2011, click here to register / Microsoft UK TechNet.These are hands-on workshops with a maximum of 12 participants and not lectures. For consulting engagements click here.   Disclaimer and copyright information:This article refers to organizations and products that may be the trademarks or registered trademarks of their various owners. Copyright of this article belongs to R Meyyappan / www.sqlworkshops.com. You may freely use the ideas and concepts discussed in this article with acknowledgement (www.sqlworkshops.com), but you may not claim any of it as your own work. This article is for informational purposes only; you use any of the suggestions given here entirely at your own risk.   R Meyyappan [email protected] LinkedIn: http://at.linkedin.com/in/rmeyyappan

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  • image focus calculation

    - by Oren Mazor
    Hi folks, I'm trying to develop an image focusing algorithm for some test automation work. I've chosen to use AForge.net, since it seems like a nice mature .net friendly system. Unfortunately, I can't seem to find information on building autofocus algorithms from scratch, so I've given it my best try: take image. apply sobel edge detection filter, which generates a greyscale edge outline. generate a histogram and save the standard dev. move camera one step closer to subject and take another picture. if the standard dev is smaller than previous one, we're getting more in focus. otherwise, we've past the optimal distance to be taking pictures. is there a better way? update: HUGE flaw in this, by the way. as I get past the optimal focus point, my "image in focus" value continues growing. you'd expect a parabolic-ish function looking at distance/focus-value, but in reality you get something that's more logarithmic

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  • Complicated idea - how to create car racing for my RPG game's players

    - by Donator
    So, I want to create car racing for my RPG game's players. Player can create race and choose how many participants can participate in race. After race is being created, other people can join it. When the maximum participants are collected, race begins. My idea, when the last participant joins, then instantly choose the winner (who's car is the best, that person wins), but how can I do it? If I choose to pick the winner after the last participant joins, then I have to put many queries in one page (select data from table, then delete the race, then select players' cars' statistics and pick the winner and then again, using mysql, send message to everyone). But this idea is really not optimal and it will lag cruelly for that last person. Maybe you have any ideas how I can avoid lag and make it more optimal. Thank you very much.

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  • Most Efficient Alternative Method of Storing Settings for iPhone Apps

    - by JPK
    I am not using the Settings bundle to store the settings for my app, as I prefer to allow the user to access the settings within the app (they may be changed fairly often). I do realize that there is the option to do both, but for now, I am trying to find the most optimal place to store the settings within the app. I have a good number of settings (from what I have read, probably too many for NSUserDefaults), and the two main options I am considering are: 1) storing the settings in a dictionary in the plist, loading the settings into a NSDictionary property in the app delegate and accessing them via the sharedDelegate 2) storing the settings in a Core Data entity (1 row on Settings entity), loading the settings into a Settings object in the app delegate and accessing them via the sharedDelegate Of these two, which would be the optimal method, performance wise?

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  • Programming languages: out of the box legibility and extensibility

    - by sova
    Two excellent results of SOLID development ideology are - Legibility - Extensibility over the life of a project (http://en.m.wikipedia.org/wiki/Solid_(object-oriented_design) Although SOLID is a set of language-agnostic design ideas, some languages inherently support these ideas better than others. Out-of-the-box or after various customizations, in your opinion which language is best-suited to be both easily readable and easy to extend functionality in? Some definitions to pre-empt biases and flamewars: Legibility: amount of thinking done to understand the code proportional to the amount of code: (amount_think-energy / amount_code) is fairly constant and as low as possible in the optimal case. Extensibility: Addition of X amount of functionality requires a change in code or code additions in proportion to X (amount_added_functionality / amount_added_code) is fairly constant and as high as possible in the optimal case. Supporting information and tutorials encouraged. Code snippets welcome.

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  • jQuery Validation Plugin: Packer undefined error?

    - by Rosarch
    I'm using the jQuery validation plugin from bassistance.de. It works fine. From <head>: <script type="text/javascript" src="/static/JQuery.js"></script> <script type="text/javascript" src="/static/js-lib/jquery.validate.pack.js"></script> <script type="text/javascript" src="/static/js-lib/jquery.validate.additional-methods.js"></script> At first, this was the only validation code I had, and it worked: $("form").validate(); $("#form-username").rules("add", { required: true, email: true, }); It was validating this HTML: <form id="form-username-form" action="api/user_of_email" method="get"> <p> <label for="form-username">Email:</label> <input type="text" name="email" id="form-username" /> <input type="submit" value="Submit" id="form-submit" /> </p> </form> Great, everything works. But then I add this JS: $("#form-choose-options input[type='text']").rules("add", { number: true, }); to validate this markup: <form id="form-choose-options" action="api/set_options" method="get"> <p> <label for="form-min-credits">Min credits per term:</label><input type="text" name="min_credits" id="form-min-credits" /> <br /> <label for="form-optimal-credits">Optimal credits per term:</label><input type="text" name="optimal_credits" id="form-optimal-credits" /> <br /> <label for="form-max-credits">Max credits per term:</label><input type="text" name="max_credits" id="form-max-credits" /> <br /> <label for="form-low-GPA">Lowest acceptable GPA:</label><input type="text" name="low_GPA" id="form-low-GPA" /> <br /> <label for="form-high-GPA">Highest realistic GPA:</label><input type="text" name="high_GPA" id="form-high-GPA" /> <br /> <input type="hidden" class="user-pk" name="pk"/> <input type="submit" value="Submit" /> </p> </form> This causes a javascript error on document load: $.data(f.form, "validator") is undefined The error is from the packer function. What am I doing wrong?

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  • Finding good heuristic for A* search

    - by Martin
    I'm trying to find the optimal solution for a little puzzle game called Twiddle (an applet with the game can be found here). The game has a 3x3 matrix with the number from 1 to 9. The goal is to bring the numbers in the correct order using the minimum amount of moves. In each move you can rotate a 2x2 square either clockwise or counterclockwise. I.e. if you have this state 6 3 9 8 7 5 1 2 4 and you rotate the upper left 2x2 square clockwise you get 8 6 9 7 3 5 1 2 4 I'm using a A* search to find the optimal solution. My f() is simply the number of rotations need. My heuristic function already leads to the optimal solution but I don't think it's the best one you can find. My current heuristic takes each corner, looks at the number at the corner and calculates the manhatten distance to the position this number will have in the solved state (which gives me the number of rotation needed to bring the number to this postion) and sums all these values. I.e. You take the above example: 6 3 9 8 7 5 1 2 4 and this end state 1 2 3 4 5 6 7 8 9 then the heuristic does the following 6 is currently at index 0 and should by at index 5: 3 rotations needed 9 is currently at index 2 and should by at index 8: 2 rotations needed 1 is currently at index 6 and should by at index 0: 2 rotations needed 4 is currently at index 8 and should by at index 3: 3 rotations needed h = 3 + 2 + 2 + 3 = 10 But there is the problem, that you rotate 4 elements at once. So there a rare cases where you can do two (ore more) of theses estimated rotations in one move. This means theses heuristic overestimates the distance to the solution. My current workaround is, to simply excluded one of the corners from the calculation which solves this problem at least for my test-cases. I've done no research if really solves the problem or if this heuristic still overestimates in same edge-cases. So my question is: What is the best heuristic you can come up with? (Disclaimer: This is for a university project, so this is a bit of homework. But I'm free to use any resource if can come up with, so it's okay to ask you guys. Also I will credit Stackoverflow for helping me ;) )

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  • concurrent doubly-linked list (1 writer, n-readers)

    - by Arne
    Hi guys, I am back in the field of programming for my Diploma-thesis now and stumbled over the following issue: I need to implement a thread-safe doubly-linked list for one thread writing the list at any position (delete, insert, mutate node data) and one to many threads traversing and reading the list. I am well aware that mutexes can be used to serialize access to the list, still I presume that a naive lock around any write operation will be less than optimal. I am wondering whether there are better variants. (I am well aware that 'optimal' has not much of a practical meaning as long as no exact measure/profiling are available but this is an academic thesis after all..) I am very gratefull for code-samples as well as references to academic granted these have at least a tiny bit of practical relevance. Thanks at lot

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  • Basics of Join Predicate Pushdown in Oracle

    - by Maria Colgan
    Happy New Year to all of our readers! We hope you all had a great holiday season. We start the new year by continuing our series on Optimizer transformations. This time it is the turn of Predicate Pushdown. I would like to thank Rafi Ahmed for the content of this blog.Normally, a view cannot be joined with an index-based nested loop (i.e., index access) join, since a view, in contrast with a base table, does not have an index defined on it. A view can only be joined with other tables using three methods: hash, nested loop, and sort-merge joins. Introduction The join predicate pushdown (JPPD) transformation allows a view to be joined with index-based nested-loop join method, which may provide a more optimal alternative. In the join predicate pushdown transformation, the view remains a separate query block, but it contains the join predicate, which is pushed down from its containing query block into the view. The view thus becomes correlated and must be evaluated for each row of the outer query block. These pushed-down join predicates, once inside the view, open up new index access paths on the base tables inside the view; this allows the view to be joined with index-based nested-loop join method, thereby enabling the optimizer to select an efficient execution plan. The join predicate pushdown transformation is not always optimal. The join predicate pushed-down view becomes correlated and it must be evaluated for each outer row; if there is a large number of outer rows, the cost of evaluating the view multiple times may make the nested-loop join suboptimal, and therefore joining the view with hash or sort-merge join method may be more efficient. The decision whether to push down join predicates into a view is determined by evaluating the costs of the outer query with and without the join predicate pushdown transformation under Oracle's cost-based query transformation framework. The join predicate pushdown transformation applies to both non-mergeable views and mergeable views and to pre-defined and inline views as well as to views generated internally by the optimizer during various transformations. The following shows the types of views on which join predicate pushdown is currently supported. UNION ALL/UNION view Outer-joined view Anti-joined view Semi-joined view DISTINCT view GROUP-BY view Examples Consider query A, which has an outer-joined view V. The view cannot be merged, as it contains two tables, and the join between these two tables must be performed before the join between the view and the outer table T4. A: SELECT T4.unique1, V.unique3 FROM T_4K T4,            (SELECT T10.unique3, T10.hundred, T10.ten             FROM T_5K T5, T_10K T10             WHERE T5.unique3 = T10.unique3) VWHERE T4.unique3 = V.hundred(+) AND       T4.ten = V.ten(+) AND       T4.thousand = 5; The following shows the non-default plan for query A generated by disabling join predicate pushdown. When query A undergoes join predicate pushdown, it yields query B. Note that query B is expressed in a non-standard SQL and shows an internal representation of the query. B: SELECT T4.unique1, V.unique3 FROM T_4K T4,           (SELECT T10.unique3, T10.hundred, T10.ten             FROM T_5K T5, T_10K T10             WHERE T5.unique3 = T10.unique3             AND T4.unique3 = V.hundred(+)             AND T4.ten = V.ten(+)) V WHERE T4.thousand = 5; The execution plan for query B is shown below. In the execution plan BX, note the keyword 'VIEW PUSHED PREDICATE' indicates that the view has undergone the join predicate pushdown transformation. The join predicates (shown here in red) have been moved into the view V; these join predicates open up index access paths thereby enabling index-based nested-loop join of the view. With join predicate pushdown, the cost of query A has come down from 62 to 32.  As mentioned earlier, the join predicate pushdown transformation is cost-based, and a join predicate pushed-down plan is selected only when it reduces the overall cost. Consider another example of a query C, which contains a view with the UNION ALL set operator.C: SELECT R.unique1, V.unique3 FROM T_5K R,            (SELECT T1.unique3, T2.unique1+T1.unique1             FROM T_5K T1, T_10K T2             WHERE T1.unique1 = T2.unique1             UNION ALL             SELECT T1.unique3, T2.unique2             FROM G_4K T1, T_10K T2             WHERE T1.unique1 = T2.unique1) V WHERE R.unique3 = V.unique3 and R.thousand < 1; The execution plan of query C is shown below. In the above, 'VIEW UNION ALL PUSHED PREDICATE' indicates that the UNION ALL view has undergone the join predicate pushdown transformation. As can be seen, here the join predicate has been replicated and pushed inside every branch of the UNION ALL view. The join predicates (shown here in red) open up index access paths thereby enabling index-based nested loop join of the view. Consider query D as an example of join predicate pushdown into a distinct view. We have the following cardinalities of the tables involved in query D: Sales (1,016,271), Customers (50,000), and Costs (787,766).  D: SELECT C.cust_last_name, C.cust_city FROM customers C,            (SELECT DISTINCT S.cust_id             FROM sales S, costs CT             WHERE S.prod_id = CT.prod_id and CT.unit_price > 70) V WHERE C.cust_state_province = 'CA' and C.cust_id = V.cust_id; The execution plan of query D is shown below. As shown in XD, when query D undergoes join predicate pushdown transformation, the expensive DISTINCT operator is removed and the join is converted into a semi-join; this is possible, since all the SELECT list items of the view participate in an equi-join with the outer tables. Under similar conditions, when a group-by view undergoes join predicate pushdown transformation, the expensive group-by operator can also be removed. With the join predicate pushdown transformation, the elapsed time of query D came down from 63 seconds to 5 seconds. Since distinct and group-by views are mergeable views, the cost-based transformation framework also compares the cost of merging the view with that of join predicate pushdown in selecting the most optimal execution plan. Summary We have tried to illustrate the basic ideas behind join predicate pushdown on different types of views by showing example queries that are quite simple. Oracle can handle far more complex queries and other types of views not shown here in the examples. Again many thanks to Rafi Ahmed for the content of this blog post.

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  • SQL SERVER – How to Recover SQL Database Data Deleted by Accident

    - by Pinal Dave
    In Repair a SQL Server database using a transaction log explorer, I showed how to use ApexSQL Log, a SQL Server transaction log viewer, to recover a SQL Server database after a disaster. In this blog, I’ll show you how to use another SQL Server disaster recovery tool from ApexSQL in a situation when data is accidentally deleted. You can download ApexSQL Recover here, install, and play along. With a good SQL Server disaster recovery strategy, data recovery is not a problem. You have a reliable full database backup with valid data, a full database backup and subsequent differential database backups, or a full database backup and a chain of transaction log backups. But not all situations are ideal. Here we’ll address some sub-optimal scenarios, where you can still successfully recover data. If you have only a full database backup This is the least optimal SQL Server disaster recovery strategy, as it doesn’t ensure minimal data loss. For example, data was deleted on Wednesday. Your last full database backup was created on Sunday, three days before the records were deleted. By using the full database backup created on Sunday, you will be able to recover SQL database records that existed in the table on Sunday. If there were any records inserted into the table on Monday or Tuesday, they will be lost forever. The same goes for records modified in this period. This method will not bring back modified records, only the old records that existed on Sunday. If you restore this full database backup, all your changes (intentional and accidental) will be lost and the database will be reverted to the state it had on Sunday. What you have to do is compare the records that were in the table on Sunday to the records on Wednesday, create a synchronization script, and execute it against the Wednesday database. If you have a full database backup followed by differential database backups Let’s say the situation is the same as in the example above, only you create a differential database backup every night. Use the full database backup created on Sunday, and the last differential database backup (created on Tuesday). In this scenario, you will lose only the data inserted and updated after the differential backup created on Tuesday. If you have a full database backup and a chain of transaction log backups This is the SQL Server disaster recovery strategy that provides minimal data loss. With a full chain of transaction logs, you can recover the SQL database to an exact point in time. To provide optimal results, you have to know exactly when the records were deleted, because restoring to a later point will not bring back the records. This method requires restoring the full database backup first. If you have any differential log backup created after the last full database backup, restore the most recent one. Then, restore transaction log backups, one by one, it the order they were created starting with the first created after the restored differential database backup. Now, the table will be in the state before the records were deleted. You have to identify the deleted records, script them and run the script against the original database. Although this method is reliable, it is time-consuming and requires a lot of space on disk. How to easily recover deleted records? The following solution enables you to recover SQL database records even if you have no full or differential database backups and no transaction log backups. To understand how ApexSQL Recover works, I’ll explain what happens when table data is deleted. Table data is stored in data pages. When you delete table records, they are not immediately deleted from the data pages, but marked to be overwritten by new records. Such records are not shown as existing anymore, but ApexSQL Recover can read them and create undo script for them. How long will deleted records stay in the MDF file? It depends on many factors, as time passes it’s less likely that the records will not be overwritten. The more transactions occur after the deletion, the more chances the records will be overwritten and permanently lost. Therefore, it’s recommended to create a copy of the database MDF and LDF files immediately (if you cannot take your database offline until the issue is solved) and run ApexSQL Recover on them. Note that a full database backup will not help here, as the records marked for overwriting are not included in the backup. First, I’ll delete some records from the Person.EmailAddress table in the AdventureWorks database.   I can delete these records in SQL Server Management Studio, or execute a script such as DELETE FROM Person.EmailAddress WHERE BusinessEntityID BETWEEN 70 AND 80 Then, I’ll start ApexSQL Recover and select From DELETE operation in the Recovery tab.   In the Select the database to recover step, first select the SQL Server instance. If it’s not shown in the drop-down list, click the Server icon right to the Server drop-down list and browse for the SQL Server instance, or type the instance name manually. Specify the authentication type and select the database in the Database drop-down list.   In the next step, you’re prompted to add additional data sources. As this can be a tricky step, especially for new users, ApexSQL Recover offers help via the Help me decide option.   The Help me decide option guides you through a series of questions about the database transaction log and advises what files to add. If you know that you have no transaction log backups or detached transaction logs, or the online transaction log file has been truncated after the data was deleted, select No additional transaction logs are available. If you know that you have transaction log backups that contain the delete transactions you want to recover, click Add transaction logs. The online transaction log is listed and selected automatically.   Click Add if to add transaction log backups. It would be best if you have a full transaction log chain, as explained above. The next step for this option is to specify the time range.   Selecting a small time range for the time of deletion will create the recovery script just for the accidentally deleted records. A wide time range might script the records deleted on purpose, and you don’t want that. If needed, you can check the script generated and manually remove such records. After that, for all data sources options, the next step is to select the tables. Be careful here, if you deleted some data from other tables on purpose, and don’t want to recover them, don’t select all tables, as ApexSQL Recover will create the INSERT script for them too.   The next step offers two options: to create a recovery script that will insert the deleted records back into the Person.EmailAddress table, or to create a new database, create the Person.EmailAddress table in it, and insert the deleted records. I’ll select the first one.   The recovery process is completed and 11 records are found and scripted, as expected.   To see the script, click View script. ApexSQL Recover has its own script editor, where you can review, modify, and execute the recovery script. The insert into statements look like: INSERT INTO Person.EmailAddress( BusinessEntityID, EmailAddressID, EmailAddress, rowguid, ModifiedDate) VALUES( 70, 70, N'[email protected]' COLLATE SQL_Latin1_General_CP1_CI_AS, 'd62c5b4e-c91f-403f-b630-7b7e0fda70ce', '20030109 00:00:00.000' ); To execute the script, click Execute in the menu.   If you want to check whether the records are really back, execute SELECT * FROM Person.EmailAddress WHERE BusinessEntityID BETWEEN 70 AND 80 As shown, ApexSQL Recover recovers SQL database data after accidental deletes even without the database backup that contains the deleted data and relevant transaction log backups. ApexSQL Recover reads the deleted data from the database data file, so this method can be used even for databases in the Simple recovery model. Besides recovering SQL database records from a DELETE statement, ApexSQL Recover can help when the records are lost due to a DROP TABLE, or TRUNCATE statement, as well as repair a corrupted MDF file that cannot be attached to as SQL Server instance. You can find more information about how to recover SQL database lost data and repair a SQL Server database on ApexSQL Solution center. There are solutions for various situations when data needs to be recovered. Reference: Pinal Dave (http://blog.sqlauthority.com)Filed under: PostADay, SQL, SQL Authority, SQL Backup and Restore, SQL Query, SQL Server, SQL Tips and Tricks, T SQL

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  • Edinburgh this Thurs (25th) - Rob Carrol talks about how to build a high performance, scalable repor

    - by tonyrogerson
    Scottish Area SQL Server User Group Meeting, Edinburgh - Thursday 25th March An evening of SQL Server 2008 Reporting Services Scalability and Performance with Rob Carrol, see how to build a high performance, scalable reporting platform and the tuning techniques required to ensure that report performance remains optimal as your platform grows. Pizza and drinks will be provided! Register at http://www.sqlserverfaq.com/events/221/SQL-Server-2008-Reporting-Services-Scalability-and-Performance.aspx...(read more)

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  • SQL SERVER – Tricks to Replace SELECT * with Column Names – SQL in Sixty Seconds #017 – Video

    - by pinaldave
    You might have heard many times that one should not use SELECT * as there are many disadvantages to the usage of the SELECT *. I also believe that there are always rare occasion when we need every single column of the query. In most of the cases, we only need a few columns of the query and we should retrieve only those columns. SELECT * has many disadvantages. Let me list a few and remaining you can add as a comment.  Retrieves unnecessary columns and increases network traffic When a new columns are added views needs to be refreshed manually Leads to usage of sub-optimal execution plan Uses clustered index in most of the cases instead of using optimal index It is difficult to debug. There are two quick tricks I have discussed in the video which explains how users can avoid using SELECT * but instead list the column names. 1) Drag the columns folder from SQL Server Management Studio to Query Editor 2) Right Click on Table Name >> Script TAble AS >> SELECT To… >> Select option It is extremely easy to list the column names in the table. In today’s sixty seconds video, you will notice that I was able to demonstrate both the methods very quickly. From now onwards there should be no excuse for not listing ColumnName. Let me ask a question back – is there ever a reason to SELECT *? If yes, would you please share that as a comment. More on SELECT *: SQL SERVER – Solution – Puzzle – SELECT * vs SELECT COUNT(*) SQL SERVER – Puzzle – SELECT * vs SELECT COUNT(*) SQL SERVER – SELECT vs. SET Performance Comparison I encourage you to submit your ideas for SQL in Sixty Seconds. We will try to accommodate as many as we can. If we like your idea we promise to share with you educational material. Reference: Pinal Dave (http://blog.sqlauthority.com) Filed under: Database, Pinal Dave, PostADay, SQL, SQL Authority, SQL in Sixty Seconds, SQL Query, SQL Scripts, SQL Server, SQL Server Management Studio, SQL Tips and Tricks, T SQL, Technology, Video

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  • Fun with Aggregates

    - by Paul White
    There are interesting things to be learned from even the simplest queries.  For example, imagine you are given the task of writing a query to list AdventureWorks product names where the product has at least one entry in the transaction history table, but fewer than ten. One possible query to meet that specification is: SELECT p.Name FROM Production.Product AS p JOIN Production.TransactionHistory AS th ON p.ProductID = th.ProductID GROUP BY p.ProductID, p.Name HAVING COUNT_BIG(*) < 10; That query correctly returns 23 rows (execution plan and data sample shown below): The execution plan looks a bit different from the written form of the query: the base tables are accessed in reverse order, and the aggregation is performed before the join.  The general idea is to read all rows from the history table, compute the count of rows grouped by ProductID, merge join the results to the Product table on ProductID, and finally filter to only return rows where the count is less than ten. This ‘fully-optimized’ plan has an estimated cost of around 0.33 units.  The reason for the quote marks there is that this plan is not quite as optimal as it could be – surely it would make sense to push the Filter down past the join too?  To answer that, let’s look at some other ways to formulate this query.  This being SQL, there are any number of ways to write logically-equivalent query specifications, so we’ll just look at a couple of interesting ones.  The first query is an attempt to reverse-engineer T-SQL from the optimized query plan shown above.  It joins the result of pre-aggregating the history table to the Product table before filtering: SELECT p.Name FROM ( SELECT th.ProductID, cnt = COUNT_BIG(*) FROM Production.TransactionHistory AS th GROUP BY th.ProductID ) AS q1 JOIN Production.Product AS p ON p.ProductID = q1.ProductID WHERE q1.cnt < 10; Perhaps a little surprisingly, we get a slightly different execution plan: The results are the same (23 rows) but this time the Filter is pushed below the join!  The optimizer chooses nested loops for the join, because the cardinality estimate for rows passing the Filter is a bit low (estimate 1 versus 23 actual), though you can force a merge join with a hint and the Filter still appears below the join.  In yet another variation, the < 10 predicate can be ‘manually pushed’ by specifying it in a HAVING clause in the “q1” sub-query instead of in the WHERE clause as written above. The reason this predicate can be pushed past the join in this query form, but not in the original formulation is simply an optimizer limitation – it does make efforts (primarily during the simplification phase) to encourage logically-equivalent query specifications to produce the same execution plan, but the implementation is not completely comprehensive. Moving on to a second example, the following query specification results from phrasing the requirement as “list the products where there exists fewer than ten correlated rows in the history table”: SELECT p.Name FROM Production.Product AS p WHERE EXISTS ( SELECT * FROM Production.TransactionHistory AS th WHERE th.ProductID = p.ProductID HAVING COUNT_BIG(*) < 10 ); Unfortunately, this query produces an incorrect result (86 rows): The problem is that it lists products with no history rows, though the reasons are interesting.  The COUNT_BIG(*) in the EXISTS clause is a scalar aggregate (meaning there is no GROUP BY clause) and scalar aggregates always produce a value, even when the input is an empty set.  In the case of the COUNT aggregate, the result of aggregating the empty set is zero (the other standard aggregates produce a NULL).  To make the point really clear, let’s look at product 709, which happens to be one for which no history rows exist: -- Scalar aggregate SELECT COUNT_BIG(*) FROM Production.TransactionHistory AS th WHERE th.ProductID = 709;   -- Vector aggregate SELECT COUNT_BIG(*) FROM Production.TransactionHistory AS th WHERE th.ProductID = 709 GROUP BY th.ProductID; The estimated execution plans for these two statements are almost identical: You might expect the Stream Aggregate to have a Group By for the second statement, but this is not the case.  The query includes an equality comparison to a constant value (709), so all qualified rows are guaranteed to have the same value for ProductID and the Group By is optimized away. In fact there are some minor differences between the two plans (the first is auto-parameterized and qualifies for trivial plan, whereas the second is not auto-parameterized and requires cost-based optimization), but there is nothing to indicate that one is a scalar aggregate and the other is a vector aggregate.  This is something I would like to see exposed in show plan so I suggested it on Connect.  Anyway, the results of running the two queries show the difference at runtime: The scalar aggregate (no GROUP BY) returns a result of zero, whereas the vector aggregate (with a GROUP BY clause) returns nothing at all.  Returning to our EXISTS query, we could ‘fix’ it by changing the HAVING clause to reject rows where the scalar aggregate returns zero: SELECT p.Name FROM Production.Product AS p WHERE EXISTS ( SELECT * FROM Production.TransactionHistory AS th WHERE th.ProductID = p.ProductID HAVING COUNT_BIG(*) BETWEEN 1 AND 9 ); The query now returns the correct 23 rows: Unfortunately, the execution plan is less efficient now – it has an estimated cost of 0.78 compared to 0.33 for the earlier plans.  Let’s try adding a redundant GROUP BY instead of changing the HAVING clause: SELECT p.Name FROM Production.Product AS p WHERE EXISTS ( SELECT * FROM Production.TransactionHistory AS th WHERE th.ProductID = p.ProductID GROUP BY th.ProductID HAVING COUNT_BIG(*) < 10 ); Not only do we now get correct results (23 rows), this is the execution plan: I like to compare that plan to quantum physics: if you don’t find it shocking, you haven’t understood it properly :)  The simple addition of a redundant GROUP BY has resulted in the EXISTS form of the query being transformed into exactly the same optimal plan we found earlier.  What’s more, in SQL Server 2008 and later, we can replace the odd-looking GROUP BY with an explicit GROUP BY on the empty set: SELECT p.Name FROM Production.Product AS p WHERE EXISTS ( SELECT * FROM Production.TransactionHistory AS th WHERE th.ProductID = p.ProductID GROUP BY () HAVING COUNT_BIG(*) < 10 ); I offer that as an alternative because some people find it more intuitive (and it perhaps has more geek value too).  Whichever way you prefer, it’s rather satisfying to note that the result of the sub-query does not exist for a particular correlated value where a vector aggregate is used (the scalar COUNT aggregate always returns a value, even if zero, so it always ‘EXISTS’ regardless which ProductID is logically being evaluated). The following query forms also produce the optimal plan and correct results, so long as a vector aggregate is used (you can probably find more equivalent query forms): WHERE Clause SELECT p.Name FROM Production.Product AS p WHERE ( SELECT COUNT_BIG(*) FROM Production.TransactionHistory AS th WHERE th.ProductID = p.ProductID GROUP BY () ) < 10; APPLY SELECT p.Name FROM Production.Product AS p CROSS APPLY ( SELECT NULL FROM Production.TransactionHistory AS th WHERE th.ProductID = p.ProductID GROUP BY () HAVING COUNT_BIG(*) < 10 ) AS ca (dummy); FROM Clause SELECT q1.Name FROM ( SELECT p.Name, cnt = ( SELECT COUNT_BIG(*) FROM Production.TransactionHistory AS th WHERE th.ProductID = p.ProductID GROUP BY () ) FROM Production.Product AS p ) AS q1 WHERE q1.cnt < 10; This last example uses SUM(1) instead of COUNT and does not require a vector aggregate…you should be able to work out why :) SELECT q.Name FROM ( SELECT p.Name, cnt = ( SELECT SUM(1) FROM Production.TransactionHistory AS th WHERE th.ProductID = p.ProductID ) FROM Production.Product AS p ) AS q WHERE q.cnt < 10; The semantics of SQL aggregates are rather odd in places.  It definitely pays to get to know the rules, and to be careful to check whether your queries are using scalar or vector aggregates.  As we have seen, query plans do not show in which ‘mode’ an aggregate is running and getting it wrong can cause poor performance, wrong results, or both. © 2012 Paul White Twitter: @SQL_Kiwi email: [email protected]

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  • Changing the default installation path to a newly installed hard disk

    - by mgj
    Hi, I am currently working on a dual-booted PC. I am using Windows XP and Ubuntu 10.04 Lucid Lynx released in April 2010. The allocated partition to Ubuntu that I am making use of has almost exhausted. Current memory allocations on the PC wrt Ubuntu OS looks like this: bodhgaya@pc146724-desktop:~$ df -h Filesystem Size Used Avail Use% Mounted on /dev/sda2 8.6G 8.0G 113M 99% / none 998M 268K 998M 1% /dev none 1002M 580K 1002M 1% /dev/shm none 1002M 100K 1002M 1% /var/run none 1002M 0 1002M 0% /var/lock none 1002M 0 1002M 0% /lib/init/rw /dev/sda1 25G 16G 9.8G 62% /media/C /dev/sdb1 37G 214M 35G 1% /media/ubuntulinuxstore bodhgaya@pc146724-desktop:~$ cd /tmp I am trying to mount a 40GB(/dev/sdb1 - given below) new hard disk along with my existing Ubuntu system to overcome with hard disk space related issues. I referred to the following tutorial to mount a new hard disk onto the system:- http://www.smorgasbord.net/how-to-in...untu-linux%20/ I was able to successfully mount this hard disk for Ubuntu 0S. I have this new hard disk setup in /media/ubuntulinuxstore directory. The current partition in my system looks like this: bodhgaya@pc146724-desktop:/media/ubuntulinuxstore$ sudo fdisk -l [sudo] password for bodhgaya: Disk /dev/sda: 40.0 GB, 40000000000 bytes 255 heads, 63 sectors/track, 4863 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: 0x446eceb5 Device Boot Start End Blocks Id System /dev/sda1 * 2 3264 26210047+ 7 HPFS/NTFS /dev/sda2 3265 4385 9004432+ 83 Linux /dev/sda3 4386 4863 3839535 82 Linux swap / Solaris Disk /dev/sdb: 40.0 GB, 40000000000 bytes 255 heads, 63 sectors/track, 4863 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: 0xfa8afa8a Device Boot Start End Blocks Id System /dev/sdb1 1 4862 39053983+ 7 HPFS/NTFS bodhgaya@pc146724-desktop:/media/ubuntulinuxstore$ Now, I have a concern wrt the "location" where the new softwares will be installed. Generally softwares are installed via the terminal and by default a fixed path is used to where the post installation set up files can be found (I am talking in context of the drive). This is like the typical case of Windows, where softwares by default are installed in the C: drive. These days people customize their installations to a drive which they find apt to serve their purpose (generally based on availability of hard disk space). I am trying to figure out how to customize the same for Ubuntu. As we all know the most softwares are installed via commands given from the Terminal. My road block is how do I redirect the default path set on the terminal where files get installed to this new hard disk. This if done will help me overcome space constraints I am currently facing wrt the partition on which my Ubuntu is initially installed. I would also by this, save time on not formatting my system and reinstalling Ubuntu and other softwares all over again. Please help me with this, your suggestions are much appreciated.

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  • Improving performance for web scraping code

    - by Pankaj Upadhyay
    I have a website in which the code scrapes other websites for getting the accurate data. While the code works good but there a decent lag in performance because the code firsts downloads the html stream from various sites(some times 9 websites), extracts the relative part and then renders the html page. What should I do to get an optimal performance. Should I change from shared hosting (godaddy) to my own server or it has nothing to do with my hosting and I need to make changes to my code?

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  • How to repair an external harddrive?

    - by dodohjk
    I would like to reformat my hard disk, and if possible recover the (somewhat unimportant) contents if possible. I have a Western Digital 1TB hard drive which had a NTFS partition. I unplugged the drive without safely removing it first. At first a pop up was asking me to use a Windows OS to run the chkdsk /f command, however, in the effort to keep using a Linux OS I used the ntfsfix command on the ubuntu terminal Now, when I try to access the hard drive, it doesn't show up anymore in Nautilus. I tried reformatting it using Disk Utility, but it gives me an error message, and Gparted would hang on the "Scanning devices" step infinitely. Please comment any output that you would like to see and I will add it to my question. EDIT disk utility tells me is on /dev/sdb the command sudo fdisk -l gives dodohjk@DodosPC:~$ sudo fdisk -l [sudo] password for dodohjk: 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: 0x0006fa8c Device Boot Start End Blocks Id System /dev/sda1 * 4094 482344959 241170433 5 Extended /dev/sda2 482344960 488396799 3025920 82 Linux swap / Solaris /dev/sda5 4096 31461127 15728516 83 Linux /dev/sda6 31463424 52434943 10485760 83 Linux /dev/sda7 52436992 62923320 5243164+ 83 Linux /dev/sda8 62924800 482344959 209710080 83 Linux Disk /dev/sdb: 1000.2 GB, 1000202043392 bytes 255 heads, 63 sectors/track, 121600 cylinders, total 1953519616 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: 0x6e697373 This doesn't look like a partition table Probably you selected the wrong device. Device Boot Start End Blocks Id System /dev/sdb1 ? 1936269394 3772285809 918008208 4f QNX4.x 3rd part /dev/sdb2 ? 1917848077 2462285169 272218546+ 73 Unknown /dev/sdb3 ? 1818575915 2362751050 272087568 2b Unknown /dev/sdb4 ? 2844524554 2844579527 27487 61 SpeedStor Partition table entries are not in disk order I wrote something wrong here, however here the output of fsck /dev/sbd is dodohjk@DodosPC:~$ sudo fsck /dev/sdb fsck from util-linux 2.20.1 e2fsck 1.42.5 (29-Jul-2012) ext2fs_open2: Bad magic number in super-block fsck.ext2: Superblock invalid, trying backup blocks... fsck.ext2: Bad magic number in super-block while trying to open /dev/sdb The superblock could not be read or does not describe a correct ext2 filesystem. If the device is valid and it really contains an ext2 filesystem (and not swap or ufs or something else), then the superblock is corrupt, and you might try running e2fsck with an alternate superblock: e2fsck -b 8193 <device&gt;

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