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• Trying to login to openssh, permission denied

  - by noah sisk

  I have been trying to login to ssh on a ubuntu 11.04 server as root with the AllowRootLogin thing set to yes but i have been getting a "Permision denied" Heres a copy of my attempt with ssh -v: Last login: Fri Jun 8 21:07:20 on ttys000 noah-sisks-macbook-pro:~ phreshness$ ssh -v [email protected] -p 22 OpenSSH_5.6p1, OpenSSL 0.9.8r 8 Feb 2011 debug1: Reading configuration data /etc/ssh_config debug1: Applying options for * debug1: Connecting to 192.168.1.133 [192.168.1.133] port 22. debug1: Connection established. debug1: identity file /Users/phreshness/.ssh/id_rsa type -1 debug1: identity file /Users/phreshness/.ssh/id_rsa-cert type -1 debug1: identity file /Users/phreshness/.ssh/id_dsa type -1 debug1: identity file /Users/phreshness/.ssh/id_dsa-cert type -1 debug1: Remote protocol version 2.0, remote software version OpenSSH_5.9p1 Debian-5ubuntu1 debug1: match: OpenSSH_5.9p1 Debian-5ubuntu1 pat OpenSSH* debug1: Enabling compatibility mode for protocol 2.0 debug1: Local version string SSH-2.0-OpenSSH_5.6 debug1: SSH2_MSG_KEXINIT sent debug1: SSH2_MSG_KEXINIT received debug1: kex: server->client aes128-ctr hmac-md5 none debug1: kex: client->server aes128-ctr hmac-md5 none debug1: SSH2_MSG_KEX_DH_GEX_REQUEST(1024<1024<8192) sent debug1: expecting SSH2_MSG_KEX_DH_GEX_GROUP debug1: SSH2_MSG_KEX_DH_GEX_INIT sent debug1: expecting SSH2_MSG_KEX_DH_GEX_REPLY debug1: Host '192.168.1.133' is known and matches the RSA host key. debug1: Found key in /Users/phreshness/.ssh/known_hosts:6 debug1: ssh_rsa_verify: signature correct debug1: SSH2_MSG_NEWKEYS sent debug1: expecting SSH2_MSG_NEWKEYS debug1: SSH2_MSG_NEWKEYS received debug1: Roaming not allowed by server debug1: SSH2_MSG_SERVICE_REQUEST sent debug1: SSH2_MSG_SERVICE_ACCEPT received debug1: Authentications that can continue: publickey,password debug1: Next authentication method: publickey debug1: Trying private key: /Users/phreshness/.ssh/id_rsa debug1: Trying private key: /Users/phreshness/.ssh/id_dsa debug1: Next authentication method: password [email protected]'s password: debug1: Authentications that can continue: publickey,password Permission denied, please try again. [email protected]'s password:

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• How to get the second word from a String?

  - by Pentium10

  Take these examples Smith John Smith-Crane John Smith-Crane John-Henry Smith-Crane John Henry I would like to get the John The first word after the space, but it might not be until the end, it can be until a non alpha character. How would this be in Java 1.5?

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• Algorithm to compare people names to detect identicalness

  - by Pentium10

  I am working on address book synchronization algorithm. I would like to reuse some code if there exists, but couldn't find one yet. Does someone know about an algorithm that will tell me in numbers/float/procent how much two names are identical. Levenstein distance is not good in this approach, as names and our adddress books are matching the begining of each of the name sections. John Smith should match Smith Jon, Jonathan Smith, Johnny Smith

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• Exchange 2010 OWA - a few questions about using multiple mailboxes

  - by Alexey Smolik

  We have an Exchange 2010 SP2 deployment and we need that our users could access multiple mailboxes in OWA. The problem is that a user (eg John Smith) needs to access not just somebody else's (eg Tom Anderson) mailboxes, but his OWN mailboxes, e.g. in different domains: [email protected], [email protected], [email protected], etc. Of course it is preferable for the user to work with all of his mailboxes from a single window. Such mailboxes can be added as multiple Exchange accounts in Outlook, that works almost fine. But in OWA, there are problems: 1) In the left pane - as I've learned - we can open only Inbox folders from other mailboxes. No way to view all folders like in Outlook? 2) With Send-As permissions set, when trying to send a message from another address, that message is saved in the Sent Items folder of the mailbox that is opened in OWA, and not in the mailbox the message is sent from. The same thing with the trash can. Is there a way to fix that? Also, this problem exists in desktop Outlook when mailboxes are added automatically via the Auto Mapping feature, so that we need to turn it off and add the accounts manually. Is there a simpler workaround? 3) Okay, suppose we only open Inbox folders in the left pane. The problem is that the mailbox names shown there are formed from Display Name attributes. But those names are all identical! All the mailboxes are owned by John Smith, so they should be all named John Smith - so that letter recepient sees "John Smith" in the "from" field, no matter what mailbox it is sent from. Also, the user knows what's his name - no need to tell him. He wants to know what mailbox he works with. So we need a way to either: a) customize OWA to show mailbox email address instead of user Display Name, or b) make Exchange use another attribute to put in the "from" field when sending letters 4) Okay, we can switch between mailboxes using "Open Other Mailbox" in the upper-right corner menu. But: a) To select a mailbox we need to enter its name (or first letters). It there a way to show a list of links to mailboxes the user has full access to? Eg in the page header... b) If we start entering the first letters, we see a popup list with possible mailboxes to be opened. But there are all mailboxes (apparently from GAL), not only mailboxes the user has permission to open! How to filter that popup list? c) The same problem as in (3) with mailbox naming. We can see the opened mailbox email address ONLY in the page URL, which is insufficient for many users. In the left pane we see "John Smith" which is useless. 5) Each mailbox is tied with a separate user in AD. If one has several mailboxes, we need to have additional dummy AD accounts, create additional OUs to store them, etc. That's not very nice, is there any standartized, optimal way to build such a structure? We would really appreciate any answers or additional info for any of these questions. Thank you in advance.

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• Cloud-Burst 2012&ndash;Windows Azure Developer Conference in Sweden

  - by Alan Smith

  The Sweden Windows Azure Group (SWAG) will running “Cloud-Burst 2012”, a two-day Windows Azure conference hosted at the Microsoft offices in Akalla, near Stockholm on the 27th and 28th September, with an Azure Hands-on Labs Day at AddSkills on the 29th September. The event is free to attend, and will be featuring presentations on the latest Azure technologies from Microsoft MVPs and evangelists. The following presentations will be delivered on the Thursday (27th) and Friday (29th): · Connecting Devices to Windows Azure - Windows Azure Technical Evangelist Brady Gaster · Grid Computing with 256 Windows Azure Worker Roles - Connected System Developer MVP Alan Smith · ‘Warts and all’. The truth about Windows Azure development - BizTalk MVP Charles Young · Using Azure to Integrate Applications - BizTalk MVP Charles Young · Riding the Windows Azure Service Bus: Cross-‘Anything’ Messaging - Windows Azure MVP & Regional Director Christian Weyer · Windows Azure, Identity & Access - and you - Developer Security MVP Dominick Baier · Brewing Beer with Windows Azure - Windows Azure MVP Maarten Balliauw · Architectural patterns for the cloud - Windows Azure MVP Maarten Balliauw · Windows Azure Web Sites and the Power of Continuous Delivery - Windows Azure MVP Magnus Mårtensson · Advanced SQL Azure - Analyze and Optimize Performance - Windows Azure MVP Nuno Godinho · Architect your SQL Azure Databases - Windows Azure MVP Nuno Godinho   There will be a chance to get your hands on the latest Azure bits and an Azure trial account at the Hands-on Labs Day on Saturday (29th) with Brady Gaster, Magnus Mårtensson and Alan Smith there to provide guidance, and some informal and entertaining presentations. Attendance for the conference and Hands-on Labs Day is free, but please only register if you can make it, (and cancel if you cannot). Cloud-Burst 2012 event details and registration is here: http://www.azureug.se/CloudBurst2012/ Registration for Sweden Windows Azure Group Stockholm is here: swagmembership.eventbrite.com The event has been made possible by kind contributions from our sponsors, Knowit, AddSkills and Microsoft Sweden.

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• SQL Query to retrieve highest item up to a point in a group

  - by James

  The best way of describing this is I have a table of people with their names and ages. Assume that people with the same surname are from the same family. I need a query in oracle which will retrieve a list of the oldest person in each family, but not older than a certain age. Table: person name surname age =============================== James Smith 23 Sarah Powell 17 Barry Smith 31 Mark Smith 35 Mary Smith 18 Bob Powell 30 How do I retrieve the oldest person in each family under 30? Results I'm after name surname age =============================== James Smith 23 Sarah Powell 17

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• SQL Query that can return intersecting data

  - by Alex

  I have a hard time finding a good question title - let me just show you what I have and what the desired outcome is. I hope this can be done in SQL (I have SQL Server 2008). 1) I have a table called Contacts and in that table I have fields like these: FirstName, LastName, CompanyName 2) Some demo data: FirstName LastName CompanyName John Smith Smith Corp Paul Wade Marc Andrews Microsoft Bill Gates Microsoft Steve Gibbs Smith Corp Diane Rowe ABC Inc. 3) I want to get an intersecting list of people and companies, but companies only once. This would look like this: Name ABC Inc. Bill Gates Diane Rowe John Smith Marc Andrews Microsoft Smith Corp Steve Gibbs Paul Wade Can I do this with SQL? How?

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• Can this method to convert a name to proper case be improved?

  - by Kelsey

  I am writing a basic function to convert millions of names (one time batch process) from their current form, which is all upper case, to a proper mixed case. I came up with the following so far: public string ConvertToProperNameCase(string input) { TextInfo textInfo = new CultureInfo("en-US", false).TextInfo; char[] chars = textInfo.ToTitleCase(input.ToLower()).ToCharArray(); for (int i = 0; i + 1 < chars.Length; i++) { if ((chars[i].Equals('\'')) || (chars[i].Equals('-'))) { chars[i + 1] = Char.ToUpper(chars[i + 1]); } } return new string(chars);; } It works in most cases such as: JOHN SMITH - John Smith SMITH, JOHN T - Smith, John T JOHN O'BRIAN - John O'Brian JOHN DOE-SMITH - John Doe-Smith There are some edge cases that do no work like: JASON MCDONALD - Jason Mcdonald (Correct: Jason McDonald) OSCAR DE LA HOYA - Oscar De La Hoya (Correct: Oscar de la Hoya) MARIE DIFRANCO - Marie Difranco (Correct: Marie DiFranco) These are not captured and I am not sure if I can handle all these odd edge cases. Can anyone think of anything I could change or add to capture more edge case? I am sure there are tons of edge cases I am not even thinking of as well. All casing should following North American conventions too meaning that if certain countries expect a specific capitalization format, and that differs from the North American format, then the North American format takes precedence.

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• Checking data of all same class elements

  - by Tiffani

  I need the code to check the data-name value of all instances of .account-select. Right now it just checks the first .account-select element and not any subsequent ones. The function right now is on click of an element such as John Smith, it checks the data-name of the .account-select lis. If the data-names are the same, it does not create a new li with the John Smith data. If no data-names are equal to John Smith, then it adds an li with John Smith. This is the JS-Fiddle I made for it so you can see what I am referring to: http://jsfiddle.net/rsxavior/vDCNy/22/ Any help would be greatly appreciated. This is the Jquery Code I am using right now. $('.account').click(function () { var acc = $(this).data("name"); var sel = $('.account-select').data("name"); if (acc === sel) { } else { $('.account-hidden-li').append('<li class="account-select" data-name="'+ $(this).data("name") +'">' + $(this).data("name") + '<a class="close bcn-close" data-dismiss="alert" href="#">&times;</a></li>'); } }); And the HTML: <ul> <li><a class="account" data-name="All" href="#">All</a></li> <li><a class="account" data-name="John Smith" href="#">John Smith</a></li> </ul> <ul class="account-hidden-li"> <ul>

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• Rewriting from headers in Postfix

  - by inxilpro

  I want to configure Postfix to replace the 'From' header in all forwarded/aliased messages with a custom email address, and the 'Reply-To' header with the original sender's address. Is that something that can be done with a simple configuration change, or am I looking at a more complex problem? For example: Original Message: From: "John Smith" <[email protected]> To: "Jane Rice" <[email protected]> Would get translated to: From: "My Email Forwarding Service" <[email protected]> Reply-To: "John Smith" <[email protected]> To: "Jane Rice" <[email protected]> Ideally, I would also have it rewrite the message body (adding something about how the message was forwarded for them), but I know that's much more difficult. We have a number of email aliases, and everytime someone reports spam they received through their alias, our server gets flagged. I'm trying to minimize that damage as much as possible. Any help is greatly appreciated!

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• Resize Debian in VirtualBox

  - by Poni

  I have a VM with one HD of size 3GB and I'd like to enlarge its HD to 7GB. So I execute this command on the host (while guest is shutdown): VBoxManage modifyhd debian.vdi --resize 7168 Then I run the guest, Debian 6, and then: smith@debian6:~$ df -h Filesystem Size Used Avail Use% Mounted on /dev/sda1 2.8G 2.6G 60M 98% / tmpfs 61M 0 61M 0% /lib/init/rw udev 57M 160K 57M 1% /dev tmpfs 61M 0 61M 0% /dev/shm smith@debian6:~$ sudo parted /dev/sda print Model: ATA VBOX HARDDISK (scsi) Disk /dev/sda: 3221MB Sector size (logical/physical): 512B/512B Partition Table: msdos Number Start End Size Type File system Flags 1 1049kB 3035MB 3034MB primary ext3 boot 2 3036MB 3220MB 185MB extended 5 3036MB 3220MB 185MB logical linux-swap(v1) smith@debian6:~$ cat /proc/partitions major minor #blocks name 8 0 3145728 sda 8 1 2962432 sda1 8 2 1 sda2 8 5 180224 sda5 So, no automatic resizing (detection) of the HD/partition (while VirtualBox, in the host, shows it's 7GB now). Ok... Then I do: smith@debian6:~$ sudo resize2fs /dev/sda1 resize2fs 1.41.12 (17-May-2010) The filesystem is already 740608 blocks long. Nothing to do! smith@debian6:~$ sudo parted GNU Parted 2.3 Using /dev/sda Welcome to GNU Parted! Type 'help' to view a list of commands. (parted) select /dev/sda1 Using /dev/sda1 (parted) resize WARNING: you are attempting to use parted to operate on (resize) a file system. parted's file system manipulation code is not as robust as what you'll find in dedicated, file-system-specific packages like e2fsprogs. We recommend you use parted only to manipulate partition tables, whenever possible. Support for performing most operations on most types of file systems will be removed in an upcoming release. Partition number? 1 Start? 0 End? [3034MB]? Here I'm stuck. At the above parted it asks me to resize to 3GB. No point in that, right.. What should I do in order to enlarge this partition?

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• OpenLDAP Authentication UID vs CN issues

  - by user145457

  I'm having trouble authenticating services using uid for authentication, which I thought was the standard method for authentication on the user. So basically, my users are added in ldap like this: # jsmith, Users, example.com dn: uid=jsmith,ou=Users,dc=example,dc=com uidNumber: 10003 loginShell: /bin/bash sn: Smith mail: [email protected] homeDirectory: /home/jsmith displayName: John Smith givenName: John uid: jsmith gecos: John Smith gidNumber: 10000 cn: John Smith title: System Administrator But when I try to authenticate using typical webapps or services like this: jsmith password I get: ldapsearch -x -h ldap.example.com -D "cn=jsmith,ou=Users,dc=example,dc=com" -W -b "dc=example,dc=com" Enter LDAP Password: ldap_bind: Invalid credentials (49) But if I use: ldapsearch -x -h ldap.example.com -D "uid=jsmith,ou=Users,dc=example,dc=com" -W -b "dc=example,dc=com" It works. HOWEVER...most webapps and authentication methods seem to use another method. So on a webapp I'm using, unless I specify the user as: uid=smith,ou=users,dc=example,dc=com Nothing works. In the webapp I just need users to put: jsmith in the user field. Keep in mind my ldap is using the "new" cn=config method of storing settings. So if someone has an obvious ldif I'm missing please provide. Let me know if you need further info. This is openldap on ubuntu 12.04. Thanks, Dave

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• Should be simple: existing laptop with local user and outlook 2007 migrate on same computer to domain user with outlook 2007 emails intact

  - by bifpowell

  I have Dell Laptop with windows 7 64 bit and for the last year it's been just a machine with an account like: machine\john there are files in folders and stuff in c:\users\john and john uses outlook 2007 as a pop3 client and has identifiable local appdata pst files. Now I installed a server and want to have everything be domain-centric so I added this laptop to the domain with admin credentials and then logged in as a domain user as: domain\john.smith Now I want to duplicate machine\john (outlook emails mostly) to domain\john.smith. In the past I used the Files and Settings Xfer Wizard and done. I tried that here and it crunched away for a while, made the file, but the restore had no effect - it ran for a while, had a progress bar, but it's like nothing happened at all afterwards. I've rebooted the machine, logged in as domain administrator as the first user to log on after the restart and tried: c:\users\john xcopy c:\users\john c:\users\john.smith /V /C /F /H /K /Y /E ...and it copies some of it, but when it gets to c:\users\john.smith\appdata\local\application data it chokes "Access denied, unable to create directory" I also tried logging in as domain\john.smith and copying the entire directory that the PSTs are in from machine\john and a lot of the mail was there when I launched outlook after replacing the PSTs, but not all of them??? I got errors about files in use when doing this method, which I figure must be why not all the old emails are in the inbox?... There must be some extremely simple way to do what must be a very common requirement. Any guidance appreciated.

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• Comparing columns in Excel

  - by Regan

  I needed to take columns A, B, and C and compare D, E and F. Here's an example: A B C D E F Jump Smith 5 Jump Smith 8 Run Naylor 2 Swim Fran 4 Swim Fran 7 Jog Dylan 1 Jump Fran 3 Jog Smith 4 So I want to match column A and B with D and E but still have both number related C for 2011 and F for 2012. Can anyone please help with that formula? My data is from A3-C4344 and D3 - D4470.

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• Varnish: User specific pages

  - by jchong0707

  I'm new to Varnish and am interested in using it to speed up my web application I wanted to know if Varnish can handle caching and serving user specific content. For example if I have a page say for example /welcome which is dynamically generated in the backend and is user specific So if User John Smith shows up to /welcome it'll show in the page itself 'Welcome John Smith' and if Bob Smith shows up to /welcome it'll show 'Welcome Bob Smith' Ideally both of those /welcome pages will be cached for each unique User, is this something Varnish can do? (is this even a good Use Case of Varnish?) Thanks!

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• Copying content on webpages in safari. To HTML

  - by Carl Smith

  Hi, is there an easier way to copy and paste website content in html? Want to copy and look like this. Product Information: Length: S / M / L Material: Polyester and Elasthane Brand: Roxana Exclusive Style: Basque But when i paste it into my content box it looks like this- Product Information Length: S / M / L Material: Polyester and Elasthane Brand: Roxana Exclusive Style: Basque Then i need to edit it in the html editor to rearrange it. Is the some sort of app or plugin that i can get so i can turn the text of the page into html so it looks right straight away when i copy it into my content box? If that makes any sense? Thanks Carl Smith :-)

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• Gateway IP Returns to Zero

  - by Robert Smith

  When you set a static IP under Ubuntu 12.04.1, you must supply the desired machine IP and the gateway IP, all using the Network Manager. When I first entered them and rebooted, everything worked great. On the second boot, however, Firefox could find no Web page. Upon checking, I discovered that the gateway IP had returned to zero. Now, no matter how often I resupply it, it returns to zero immediately after NM "saves" it: that is, appears as zero when redisplayed. The only way I can get to the Internet is to restore DHCP operation. I need to use static IP for access to my home network. Would appreciate any suggestion. --Robert Smith

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• Ways to ensure unique instances of a class?

  - by Peanut

  I'm looking for different ways to ensure that each instance of a given class is a uniquely identifiable instance. For example, I have a Name class with the field name. Once I have a Name object with name initialised to John Smith I don't want to be able to instantiate a different Name object also with the name as John Smith, or if instantiation does take place I want a reference to the orginal object to be passed back rather than a new object. I'm aware that one way of doing this is to have a static factory that holds a Map of all the current Name objects and the factory checks that an object with John Smith as the name doesn't already exist before passing back a reference to a Name object. Another way I could think of off the top of my head is having a static Map in the Name class and when the constructor is called throwing an exception if the value passed in for name is already in use in another object, however I'm aware throwing exceptions in a constructor is generally a bad idea. Are there other ways of achieving this?

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• Streaming files from EventMachine handler?

  - by Noah

  I am creating a streaming eventmachine server. I'm concerned about avoiding blocking IO or doing anything else to muck up the event loop. From what I've read, ruby's non-blocking IO can be used to stream files in a non-blocking way, or I can call next_tick, but I'm a little unclear about which of these approaches is preferable. Part of the problem is that I have not found a good explanation of non-blocking IO library functions in ruby. Short version: Assuming a long-lived network IO operation, several wall clock minutes of streaming per file, transfer, what is the best way to do this in eventmachine without gumming up the event loop? while 1 do file.read do |bytes| @conn.send_data bytes end end I understand that the above code will block and I'm wondering what to put in its place. Also, I cannot use the FileStreamer class that is part of eventmachine as is, because I need to manipulate the data after it's read but before it's sent. Thanks, Noah

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• What is the effect of this order_by clause?

  - by bread

  I don't understand what this order_by clause is doing and whether I need it or not: select c.customerid, c.firstname, c.lastname, i.order_date, i.item, i.price from items_ordered i, customers c where i.customerid = c.customerid group by c.customerid, i.item, i.order_date order by i.order_date desc; This produces this data: 10330 Shawn Dalton 30-Jun-1999 Pogo stick 28.00 10101 John Gray 30-Jun-1999 Raft 58.00 10410 Mary Ann Howell 30-Jan-2000 Unicycle 192.50 10101 John Gray 30-Dec-1999 Hoola Hoop 14.75 10449 Isabela Moore 29-Feb-2000 Flashlight 4.50 10410 Mary Ann Howell 28-Oct-1999 Sleeping Bag 89.22 10339 Anthony Sanchez 27-Jul-1999 Umbrella 4.50 10449 Isabela Moore 22-Dec-1999 Canoe 280.00 10298 Leroy Brown 19-Sep-1999 Lantern 29.00 10449 Isabela Moore 19-Mar-2000 Canoe paddle 40.00 10413 Donald Davids 19-Jan-2000 Lawnchair 32.00 10330 Shawn Dalton 19-Apr-2000 Shovel 16.75 10439 Conrad Giles 18-Sep-1999 Tent 88.00 10298 Leroy Brown 18-Mar-2000 Pocket Knife 22.38 10299 Elroy Keller 18-Jan-2000 Inflatable Mattress 38.00 10438 Kevin Smith 18-Jan-2000 Tent 79.99 10101 John Gray 18-Aug-1999 Rain Coat 18.30 10449 Isabela Moore 15-Dec-1999 Bicycle 380.50 10439 Conrad Giles 14-Aug-1999 Ski Poles 25.50 10449 Isabela Moore 13-Aug-1999 Unicycle 180.79 10101 John Gray 08-Mar-2000 Sleeping Bag 88.70 10299 Elroy Keller 06-Jul-1999 Parachute 1250.00 10438 Kevin Smith 02-Nov-1999 Pillow 8.50 10101 John Gray 02-Jan-2000 Lantern 16.00 10315 Lisa Jones 02-Feb-2000 Compass 8.00 10449 Isabela Moore 01-Sep-1999 Snow Shoes 45.00 10438 Kevin Smith 01-Nov-1999 Umbrella 6.75 10298 Leroy Brown 01-Jul-1999 Skateboard 33.00 10101 John Gray 01-Jul-1999 Life Vest 125.00 10330 Shawn Dalton 01-Jan-2000 Flashlight 28.00 10298 Leroy Brown 01-Dec-1999 Helmet 22.00 10298 Leroy Brown 01-Apr-2000 Ear Muffs 12.50 While if I remove the order_by clause completely, as in this query: select c.customerid, c.firstname, c.lastname, i.order_date, i.item, i.price from items_ordered i, customers c where i.customerid = c.customerid group by c.customerid, i.item, i.order_date; I get these results: 10101 John Gray 30-Dec-1999 Hoola Hoop 14.75 10101 John Gray 02-Jan-2000 Lantern 16.00 10101 John Gray 01-Jul-1999 Life Vest 125.00 10101 John Gray 30-Jun-1999 Raft 58.00 10101 John Gray 18-Aug-1999 Rain Coat 18.30 10101 John Gray 08-Mar-2000 Sleeping Bag 88.70 10298 Leroy Brown 01-Apr-2000 Ear Muffs 12.50 10298 Leroy Brown 01-Dec-1999 Helmet 22.00 10298 Leroy Brown 19-Sep-1999 Lantern 29.00 10298 Leroy Brown 18-Mar-2000 Pocket Knife 22.38 10298 Leroy Brown 01-Jul-1999 Skateboard 33.00 10299 Elroy Keller 18-Jan-2000 Inflatable Mattress 38.00 10299 Elroy Keller 06-Jul-1999 Parachute 1250.00 10315 Lisa Jones 02-Feb-2000 Compass 8.00 10330 Shawn Dalton 01-Jan-2000 Flashlight 28.00 10330 Shawn Dalton 30-Jun-1999 Pogo stick 28.00 10330 Shawn Dalton 19-Apr-2000 Shovel 16.75 10339 Anthony Sanchez 27-Jul-1999 Umbrella 4.50 10410 Mary Ann Howell 28-Oct-1999 Sleeping Bag 89.22 10410 Mary Ann Howell 30-Jan-2000 Unicycle 192.50 10413 Donald Davids 19-Jan-2000 Lawnchair 32.00 10438 Kevin Smith 02-Nov-1999 Pillow 8.50 10438 Kevin Smith 18-Jan-2000 Tent 79.99 10438 Kevin Smith 01-Nov-1999 Umbrella 6.75 10439 Conrad Giles 14-Aug-1999 Ski Poles 25.50 10439 Conrad Giles 18-Sep-1999 Tent 88.00 10449 Isabela Moore 15-Dec-1999 Bicycle 380.50 10449 Isabela Moore 22-Dec-1999 Canoe 280.00 10449 Isabela Moore 19-Mar-2000 Canoe paddle 40.00 10449 Isabela Moore 29-Feb-2000 Flashlight 4.50 10449 Isabela Moore 01-Sep-1999 Snow Shoes 45.00 10449 Isabela Moore 13-Aug-1999 Unicycle 180.79 I'm not sure what the order_by is doing here and if it's having the intended effects.

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• Tricky SQL query involving consecutive values

  - by Gabriel

  I need to perform a relatively easy to explain but (given my somewhat limited skills) hard to write SQL query. Assume we have a table similar to this one: exam_no | name | surname | result | date ---------+------+---------+--------+------------ 1 | John | Doe | PASS | 2012-01-01 1 | Ryan | Smith | FAIL | 2012-01-02 <-- 1 | Ann | Evans | PASS | 2012-01-03 1 | Mary | Lee | FAIL | 2012-01-04 ... | ... | ... | ... | ... 2 | John | Doe | FAIL | 2012-02-01 <-- 2 | Ryan | Smith | FAIL | 2012-02-02 2 | Ann | Evans | FAIL | 2012-02-03 2 | Mary | Lee | PASS | 2012-02-04 ... | ... | ... | ... | ... 3 | John | Doe | FAIL | 2012-03-01 3 | Ryan | Smith | FAIL | 2012-03-02 3 | Ann | Evans | PASS | 2012-03-03 3 | Mary | Lee | FAIL | 2012-03-04 <-- Note that exam_no and date aren't necessarily related as one might expect from the kind of example I chose. Now, the query that I need to do is as follows: From the latest exam (exam_no = 3) find all the students that have failed (John Doe, Ryan Smith and Mary Lee). For each of these students find the date of the first of the batch of consecutively failing exams. Another way to put it would be: for each of these students find the date of the first failing exam that comes after their last passing exam. (Look at the arrows in the table). The resulting table should be something like this: name | surname | date_since_failing ------+---------+-------------------- John | Doe | 2012-02-01 Ryan | Smith | 2012-01-02 Mary | Lee | 2012-01-04 Ann | Evans | 2012-02-03 How can I perform such a query? Thank you for your time.

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• Odd SQL Results

  - by Ryan Burnham

  So i have the following query Select id, [First], [Last] , [Business] as contactbusiness, (Case When ([Business] != '' or [Business] is not null) Then [Business] Else 'No Phone Number' END) from contacts The results look like id First Last contactbusiness (No column name) 2 John Smith 3 Sarah Jane 0411 111 222 0411 111 222 6 John Smith 0411 111 111 0411 111 111 8 NULL No Phone Number 11 Ryan B 08 9999 9999 08 9999 9999 14 David F NULL No Phone Number I'd expect record 2 to also show No Phone Number If i change the "[Business] is not null" to [Business] != null then i get the correct results id First Last contactbusiness (No column name) 2 John Smith No Phone Number 3 Sarah Jane 0411 111 222 0411 111 222 6 John Smith 0411 111 111 0411 111 111 8 NULL No Phone Number 11 Ryan B 08 9999 9999 08 9999 9999 14 David F NULL No Phone Number Normally you need to use is not null rather than != null. whats going on here?

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• Creating an excel macro to sum lines with duplicate values

  - by john

  I need a macro to look at the list of data below, provide a number of instances it appears and sum the value of each of them. I know a pivot table or series of forumlas could work but i'm doing this for a coworker and it has to be a 'one click here' kinda deal. The data is as follows. A B Smith 200.00 Dean 100.00 Smith 100.00 Smith 50.00 Wilson 25.00 Dean 25.00 Barry 100.00 The end result would look like this Smith 3 350.00 Dean 2 125.00 Wilson 1 25.00 Barry 1 100.00 Thanks in advance for any help you can offer!

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• Simple Select Statement on MySQL Database Hanging

  - by AlishahNovin

  I have a very simple sql select statement on a very large table, that is non-normalized. (Not my design at all, I'm just trying to optimize while simultaneously trying to convince the owners of a redesign) Basically, the statement is like this: SELECT FirstName, LastName, FullName, State FROM Activity Where (FirstName=@name OR LastName=@name OR FullName=@name) AND State=@state; Now, FirstName, LastName, FullName and State are all indexed as BTrees, but without prefix - the whole column is indexed. State column is a 2 letter state code. What I'm finding is this: When @name = 'John Smith', and @state = '%' the search is really fast and yields results immediately. When @name = 'John Smith', and @state = 'FL' the search takes 5 minutes (and usually this means the web service times out...) When I remove the FirstName and LastName comparisons, and only use the FullName and State, both cases above work very quickly. When I replace FirstName, LastName, FullName, and State searches, but use LIKE for each search, it works fast for @name='John Smith%' and @state='%', but slow for @name='John Smith%' and @state='FL' When I search against 'John Sm%' and @state='FL' the search finds results immediately When I search against 'John Smi%' and @state='FL' the search takes 5 minutes. Now, just to reiterate - the table is not normalized. The John Smith appears many many times, as do many other users, because there is no reference to some form of users/people table. I'm not sure how many times a single user may appear, but the table itself has 90 Million records. Again, not my design... What I'm wondering is - though there are many many problems with this design, what is causing this specific problem. My guess is that the index trees are just too large that it just takes a very long time traversing the them. (FirstName, LastName, FullName) Anyway, I appreciate anyone's help with this. Like I said, I'm working on convincing them of a redesign, but in the meantime, if I someone could help me figure out what the exact problem is, that'd be fantastic.

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• Is it possible to use ContainsTable to get results for more than one column?

  - by LockeCJ

  Consider the following table: People FirstName nvarchar(50) LastName nvarchar(50) Let's assume for the moment that this table has a full-text index on it for both columns. Let's suppose that I wanted to find all of the people named "John Smith" in this table. The following query seems like a perfectly rational way to accomplish this: SELECT * from People p INNER JOIN CONTAINSTABLE(People,*,'"John*" AND "Smith*"') Unfortunately, this will return no results, assuming that there is no record in the People table that contains both "John" and "Smith" in either the FirstName or LastName columns. It will not match a record with "John" in the FirstName column, and "Smith" in the LastName column, or vice-versa. My question is this: How does one accomplish what I'm trying to do above? Please consider that the example above is simplified. The real table I'm working with has ten columns and the input I'm receiving is a single string which is split up based on standard word breakers (space, dash, etc.)

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