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
if you want to search for multiple words, u have to use multiple like
operators:
select count(distinct itemid) from business where name like 'word1' or
name like 'word2' or name like 'word3';
or the IN statement with wildcards:
select count(distinct itemid) from business where name IN
Yeah I really need to search for multiple words. Can anyone confirm if
the IN statement will work for me in this situation?
On Mon, 2002-11-04 at 09:31, [EMAIL PROTECTED] wrote:
if you want to search for multiple words, u have to use multiple like
operators:
select count(distinct itemid) from
You can't use wildcards with IN, only with LIKE or regular expressions.
---John Holmes...
-Original Message-
From: [EMAIL PROTECTED] [mailto:Amit_Wadhwa;Dell.com]
Sent: Sunday, November 03, 2002 5:31 PM
To: [EMAIL PROTECTED]
Subject: RE: [PHP-DB] LIKE statement or IN statement
Chris Barnes wrote:
Yeah I really need to search for multiple words. Can anyone confirm if
the IN statement will work for me in this situation?
Chris --
Why not just try it you self and let's us know.
Also check to MySQL doc at http://mysql.org
David
On Mon, 2002-11-04 at 09:31, [EMAIL
expressions.
---John Holmes...
-Original Message-
From: [EMAIL PROTECTED] [mailto:Amit_Wadhwa;Dell.com]
Sent: Sunday, November 03, 2002 5:31 PM
To: [EMAIL PROTECTED]
Subject: RE: [PHP-DB] LIKE statement or IN statement?
if you want to search for multiple words, u have to use multiple
ok so you would have to use :
--select count(distinct itemid) from business where name like 'word1'
or
name like 'word2' or name like 'word3';
no other go.
If you're not going to use wildcards, then you can use IN. The whole
idea of using LIKE is that you can use _ and % as wildcards when
