kts Posted February 1, 2008 Share Posted February 1, 2008 SELECT * FROM V_Providers WHERE Pname LIKE '%b' AND SID = 3 is what i get when I echo and the code is $this->query = "SELECT * FROM " . $V_Providers . " WHERE ".$searchtype." LIKE '%".$state."' AND SID = ".$serviceid.""; the dilema is i paste the echo into phpmyadmin and it works, but when executed in php it does not show any results i believe it would have to be syntax, anyone? Quote Link to comment Share on other sites More sharing options...
Barand Posted February 1, 2008 Share Posted February 1, 2008 does echo mysql_error(); give any clues? Quote Link to comment Share on other sites More sharing options...
kts Posted February 1, 2008 Author Share Posted February 1, 2008 Unfortunately there is no error...... im thinking it has to be something with one of my if statements thats making it bypass improperly. Just wondering if it was a common error, clearly its a booboo on my behalf haha thanks though. Quote Link to comment Share on other sites More sharing options...
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