> I tried to use LIKE: > SELECT URL, Name > FROM websites > WHERE 'http://www.microsoft.com/kb/knowledgeb.asp?id=3&strse=12' > LIKE (URL + '%'); > > But this doesn't return any results. I would like the following as output: > 'http://www.microsoft.com/kb/' Microsoft Knowledgebase
Hi! How about the following? SELECT URL, Name FROM websites WHERE 'http://www.microsoft.com/kb/knowledgeb.asp?id=3&strse=12' LIKE CONCAT(URL, '%'); - Ville -- MySQL General Mailing List For list archives: http://lists.mysql.com/mysql To unsubscribe: http://lists.mysql.com/[EMAIL PROTECTED]
