change your query to this: select count(distinct itemid) from business where name like 'word1 word2 word3%' or description like 'word1 word2 word3%'; Peter On 4 Nov 2002, Chris Barnes wrote: > Hi, > I've got a dilly of a problem. I'm probably doing something wrong but I > don't know what. I'm trying to use the LIKE statement in a query where > more than one word is used in with LIKE..e.g. > > select count(distinct itemid) from business where name or description > like 'word1 word2 word3%' > > The problem I'm having is probably obvious to you but I don't know why > this returns no matches but if i specify only 1 word in the LIKE > statement then it returns a match. > > Am i not able to specify more than 1 word with LIKE or am I just doing > it wrong? > > It has been designed to take input from a web form by the variable > $search_string and then the query string is constructed from that e.g. > > $query = "select count(distinct itemid) from business where name or > description like'" . $search_string . "'"; > > > Any help or suggestions greatly appreciated. > --------------------------------------------------------------------------- Peter Beckman Systems Engineer, Fairfax Cable Access Corporation beckman@purplecow.com http://www.purplecow.com/ --------------------------------------------------------------------------- -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
