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• Welcoming Gavin Payne to Coeo

  - by Christian

  I’m very pleased to announce that Gavin Payne starts with us today as a Senior Consultant! We’ve known Gavin for a while now through his work in the UK SQL Server Community and when a role came up in our consulting practice I took the opportunity to talk to him about it. Gavin brings a broad range of experience from his recent background as a Solution Architect and has a particular interest in virtualization which is very prominent in the work that we do so we’re thrilled to have him on board. He’s also presenting a couple of sessions at the upcoming SQLBits conference in Brighton where Coeo is once again sponsoring and exhibiting so be sure to congratulate him in person if you’re going to be there! Gavin has a prolific online presence so be sure to subscribe to his blog and follow him on twitter! Blog: http://blog.gavinpayneuk.com/ Twitter: http://twitter.com/gavinpayneuk SQLBits: http://sqlbits.com/Speakers/Gavin_Payne   Welcome Gavin! Christian Bolton  - MCA, MCM, MVP Technical Director http://coeo.com - SQL Server Consulting & Managed Services

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• Congratulations to Gavin Payne–Microsoft Certified Architect

  - by Christian

  Huge congratulations to Gavin who became the 6th person in the WORLD outside Microsoft to qualify as a Microsoft Certified Architect in SQL Server today. Gavin’s worked so hard for this since the start of the year and all that work culminated in a grilling 4 hour review board during the PASS summit in Seattle less than 2 weeks ago -- he received his official results last night. To put things into perspective, there are only 25 people on the planet that are qualified to this level in SQL Server; only 6 of those don’t work for Microsoft; and 2 of those work for Coeo. Coeo is the only partner in the UK to have an MCA in SQL Server, and now we have 2! Well done Gavin, we’re all very proud of what you’ve achieved!   Christian Bolton - MCA, MCM, MVP Technical Director http://coeo.com - SQL Server Consulting & Managed Services You can read more about the Certified Architect program on Microsoft’s website here: http://bit.ly/4ar5QP

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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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• PASS Summit book launch and meet the authors - Professional SQL Server 2012 Internals & Troubleshooting

  - by Christian

  I’m very pleased to announce that we’ll be officially launching our new book, Professional SQL Server 2012 Internals and Troubleshooting at the PASS Summit in Seattle tomorrow. In partnership with our great friends at SQL Sentry we’ll have most of the authors at the SQL Sentry exhibitors stand from 12:30 on Thursday 8th November for a book signing event which will give you a rare opportunity to meet with the authors and contributors, many of which have flown in from around the world. SQL Sentry also have lots and lots of copies to give away for free so be sure to drop by their stand and ask about it! If you really can’t wait or run the risk of not getting a copy then the PASS bookstore has a few copies for sale but don’t expect them to be there for long! You can also order it from your favourite online retailer: amazon.com: http://amzn.to/U9IlPV barnesandnoble.com: http://bitly.com/Ux1gog amazon.co.uk: http://bitly.com/WBJ18l I’ll be writing a follow-up post very soon explaining why I think you should buy this book so look out for it!   Christian Bolton - MCA, MCM, MVP Technical Director http://coeo.com - SQL Server Consulting & Managed Services

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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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• TechDays 2011 Sweden Videos

  - by Your DisplayName here!

  All the videos from the excellent Örebro event are now online. Dominick Baier: A Technical Introduction to the Windows Identity Foundation (watch) Dominick Baier & Christian Weyer: Securing REST-Services and Web APIs on the Windows Azure Platform (watch) Christian Weyer: Real World Azure - Elasticity from on-premise to the cloud and back (watch) Our interview with Robert (watch)

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• Enterprise 2.0 Conference: Building Social Business

  - by kellsey.ruppel

  The way we work is changing rapidly, offering an enormous competitive advantage to those who embrace the new tools that enable contextual, agile and simplified information exchange and collaboration to distributed workforces and networks of partners and customers. As many of you are aware, Enterprise 2.0 is the term for the technologies and business practices that liberate the workforce from the constraints of legacy communication and productivity tools like email. It provides business managers with access to the right information at the right time through a web of inter-connected applications, services and devices. Enterprise 2.0 makes accessible the collective intelligence of many, translating to a huge competitive advantage in the form of increased innovation, productivity and agility. The Enterprise 2.0 Conference takes a strategic perspective, emphasizing the bigger picture implications of the technology and the exploration of what is at stake for organizations trying to change not only tools, but also culture and process. Beyond discussion of the "why", there will also be in-depth opportunities for learning the "how" that will help you bring Enterprise 2.0 to your business.You won't want to miss this opportunity to learn and hear from leading experts in the fields of technology for business, collaboration, culture change and collective intelligence. Oracle is a proud Gold sponsor of the Enterprise 2.0 Conference, taking place this week in Boston. Come and learn about Oracle at the following panel sessions and Market Leaders Theater Sessions. Tuesday, June 19, 2012 at 1:30 p.m. Market Theater Presentation Into the Activity Stream, and Beyond! Introducing Oracle Social Network Oracle Speaker: Christian Finn, Senior Director of Evangelism, Oracle WebCenter Tuesday, June 19, 2012 at 2:30 p.m.  Panel Session Innovation versus Integration Oracle Panel Speaker: Christian Finn, Senior Director of Evangelism, Oracle WebCenter Wednesday, June 20, 2012 at 1:30 p.m. Business Leadership Roundtable Oracle Panel Speaker: Christian Finn, Senior Director of Evangelism, Oracle WebCenter Wednesday, June 20, 2012 at 3:00 p.m. Market Theater Presentation Into the Activity Stream, and Beyond! Introducing Oracle Social Network Oracle Speaker: Christian Finn, Senior Director of Evangelism, Oracle WebCenter Thursday, June 21, 2012 at 8:30 a.m. Panel Session Collecting and Processing Big Data: Architecting Systems that Scale Oracle Panel Speaker: Ashok Joshi, Senior Director, Berkeley DB Development Thursday, June 21, 2012 at 11:00 a.m. Panel Session The Future of Big Data: What's Next Oracle Panel Speaker: Ashok Joshi, Senior Director, Berkeley DB Development Be sure to stop by and visit Oracle booth #501, to see live demonstrations of Oracle Social Network and Oracle WebCenter!

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• How to change default boot order ubuntu 10.04 ?

  - by Sako Christian

  Hello, How can I change default boot order in Ubuntu 10.04 from Ubuntu to Windows7? However, I already checked sudo gedit /etc/default/grub and modify the grub file to be GRUB_DEFAULT=4 and update the grup sudo update-grub I even install graph software to re order the book sudo startupmanager But still after restart the default choose for boot is Ubuntu ... Thank you, Sako Christian P.S: I am using Ubuntu 10.04 with grub version 1.98

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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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• Parsing complex string using regex

  - by wojtek_z

  My regex skills are not very good and recently a new data element has thrown my parser into a loop Take the following string "+USER=Bob Smith-GROUP=Admin+FUNCTION=Read/FUNCTION=Write" Previously I had the following for my regex : [+\\-/] Which would turn the result into USER=Bob Smith GROUP=Admin FUNCTION=Read FUNCTION=Write FUNCTION=Read But now I have values with dashes in them which is causing bad output New string looks like "+USER=Bob Smith-GROUP=Admin+FUNCTION=Read/FUNCTION=Write/FUNCTION=Read-Write" Which gives me the following result , and breaks the key = value structure. USER=Bob Smith GROUP=Admin FUNCTION=Read FUNCTION=Write FUNCTION=Read Write Can someone help me formulate a valid regex for handling this or point me to some key / value examples. Basically I need to be able to handle + - / signs in order to get combinations.

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• Are there any well known algorithms to detect the presence of names?

  - by Rhubarb

  For example, given a string: "Bob went fishing with his friend Jim Smith." Bob and Jim Smith are both names, but bob and smith are both words. Weren't for them being uppercase, there would be less indication of this outside of our knowledge of the sentence. Without doing grammar analysis, are there any well known algorithms for detecting the presence of names, at least Western names?

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• Need help joining tables...

  - by yuudachi

  I am a MySQL newbie, so sorry if this is a dumb question.. These are my tables. student table: SID (primary) student_name advisor (foreign key to faculty.facultyID) requested_advisor (foreign key to faculty.facultyID) faculty table: facultyID (primary key) advisor_name I want to query a table that shows everything in the student table, but I want advisor and requested_advisor to show up as names, not the ID numbers. so like it displays like this on the webpage: Student Name: Jane Smith SID: 860123456 Current Advisor: John Smith Requested advisor: James Smith not like this Student Name: Jane Smith SID: 860123456 Current Advisor: 1 Requested advisor: 2 SELECT student.student_name, SID, student_email, faculty.advisor_name FROM student INNER JOIN faculty ON student.advisor = faculty.facultyID; this comes out close, but I don't know how to get the requested_advisor to show up as a name.

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• Setting variables in shell script by running commands

  - by rajya vardhan

  >cat /tmp/list1 john jack >cat /tmp/list2 smith taylor It is guaranteed that list1 and list2 will have equal number of lines. f(){ i=1 while read line do var1 = `sed -n '$ip' /tmp/list1` var2 = `sed -n '$ip' /tmp/list2` echo $i,$var1,$var2 i=`expr $i+1` echo $i,$var1,$var2 done < $INFILE } So output of f() should be: 1,john,smith 2,jack,taylor But getting 1,p,p 1+1,p,p If i replace following: var1 = `sed -n '$ip' /tmp/list1` var2 = `sed -n '$ip' /tmp/list2` with this: var1=`head -$i /tmp/vip_list|tail -1` var2=`head -$i /tmp/lb_list|tail -1` Then output: 1,john,smith 1,john,smith Not an expert of shell, so please excuse if sounds childish :)

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