Forum Moderators: coopster

Message Too Old, No Replies

Querying with LIKE

I've tried everything!

         

inveni0

10:16 pm on Nov 30, 2006 (gmt 0)

10+ Year Member



How do I set up the wild cards in this:

$query_Members = sprintf("SELECT * FROM Members WHERE Name LIKE '%s' ORDER BY Name ASC", $colname_Members);

I have tried:

$query_Members = sprintf("SELECT * FROM Members WHERE Name LIKE '%$colname_Members%' ORDER BY Name ASC");

But it doesn't work. I get a "Query Empty" error, even though I can echo the value of $colname_Members on the same page as the error.

Thanks.

justageek

10:26 pm on Nov 30, 2006 (gmt 0)

WebmasterWorld Senior Member 10+ Year Member



I always create my vars first just remove any problems like you are seeing now. It looks like the db is trying to eval the variable and it is blank since it is not a db var. Try this:

$query = "SELECT * FROM Members WHERE Name LIKE '%$colname_Members%' ORDER BY Name ASC";

If echoing out the $query var looks fine then use it for the db query like this:

$result = mysql_query($query);

JAG

justageek

10:28 pm on Nov 30, 2006 (gmt 0)

WebmasterWorld Senior Member 10+ Year Member



I should also add that you always want to wrap your variables with a mysql_real_escape_string().

JAG

inveni0

10:41 pm on Nov 30, 2006 (gmt 0)

10+ Year Member



I got it worked out. Thanks, though!