Re: [PHP] Difficulty with SQL LIKE clause

- Original Message - From: David Freedman [EMAIL PROTECTED] To: php-general@lists.php.net Sent: Friday, February 25, 2005 11:27 PM Subject: [PHP] Difficulty with SQL LIKE clause When I use this query in PHP it works, and I get all things with the YEAR of 1977, as I expected. $query=

RE: [PHP] Difficulty with SQL LIKE clause

Not exactly clear what database you're using. In SQL Server the syntax is SELECT [options] FROM [table] WHERE [field] LIKE '%whatever value you're looking for%' The % means any string of characters, so it doesn't have to be exact Best of luck Nate Tobik (412)661-5700 x206 VigilantMinds

Re: [PHP] Difficulty with SQL LIKE clause

Hello David, Friday, February 25, 2005, 1:27:54 PM, you wrote: D Can the '*' be used? What am I doing wrong. It's the % symbol, not the *. -- Leif (TB lists moderator and fellow end user). Using The Bat! 3.0.2.3 Rush under Windows XP 5.1 Build 2600 Service Pack 2 on a Pentium 4 2GHz with

RE: [PHP] Difficulty with SQL LIKE clause

You should use a _ instead. Please have a look at the mySQL manual: http://dev.mysql.com/doc/mysql/en/string-comparison-functions.html Christian When I use this query in PHP it works, and I get all things with the YEAR of 1977, as I expected. $query= SELECT * FROM my_table WHERE Year LIKE 1977 ;