Received: from qmail-a.directi.com ([203.199.89.139]) by postgresql.org (8.11.3/8.11.4) with SMTP id f937f5n96866 for ; Wed, 3 Oct 2001 03:41:07 -0400 (EDT) (envelope-from azaidi@directi.com) Received: (qmail 15744 invoked by uid 0); 3 Oct 2001 07:39:36 -0000 Received: from unknown (HELO default) (203.197.51.84) by 203.199.89.139 with SMTP; 3 Oct 2001 07:39:36 -0000 Message-ID: <006101c14bde$b6f1b000$4301a8c0@directi.com> From: "Arsalan Zaidi" To: "PostgreSQL" Subject: select in a LIKE? Date: Wed, 3 Oct 2001 13:10:43 +0530 MIME-Version: 1.0 Content-Type: text/plain; charset="iso-8859-1" Content-Transfer-Encoding: 7bit X-Priority: 3 X-MSMail-Priority: Normal X-Mailer: Microsoft Outlook Express 5.00.2615.200 X-MimeOLE: Produced By Microsoft MimeOLE V5.00.2615.200 X-Archive-Number: 200110/5 X-Sequence-Number: 1973 Hi Can I construct a LIKE out of a select query? Something like select * from users where name like '%(select name from blah)' Currently, this won't work. --Arsalan ------------------------------------------------- WAR IS PEACE FREEDOM IS SLAVERY IGNORANCE IS STRENGTH --1984