Variable conversion problem

From: Date: Fri, 21 Sep 2001 01:45:23 +0000
Subject: Variable conversion problem
Groups: php.general 
Request: Send a blank email to php-general+get-68095@lists.php.net to get a copy of this message
Another PHP problem has kept me up all night, scouring through my database books and trying everything I could think of. I am still new to the MySQL and PHP field, but never the lass I will not give up. $cat is a variable that is passed to the query from the previous page. The SELECT statement works faultlessly, except for the point at which no choice is made from the drop down on the page prior to this one. I have set a value of "x" to the drop down for a non selection, but am having problems converting that "x" value to a "%" value (the MySql wildcard symbol) withing my SELECT statement. I am assuming that I will need an IF THEN before the SELECT, but I have tried several variation to no avail. $result = mysql_query("SELECT CompetitorName FROM competitor, competitorproducts, productcategory WHERE competitorproducts.CompID=competitor.ID AND productcategory.ID=\"$cat\" AND competitorproducts.CatID=productcategory.ID" ); Any help would be mostly appreciated. Thanks. Neil Silvester Webmaster / Systems Administrator Heat and Control Inc.

« previous php.general (#68095) next »