I am trying to write a script that checks a table for a value in MySQL. It opens the table and queries it fine but if my query returns nothing i have no way of testing to see if it returned anything. I have tried testing for "false" and "" and ! $result in a if statement but i don't know what to do. How do I check to see if i got anything or not without getting this: "Notice: Trying to get property of non-object ". This is my most current version of the code: $id = session_id(); function logOut($session__id) { $link = mysql_connect( "localhost", "shopper", "shopper" ); if ( ! $link ) { die("Couldn't connect to MySQL"); } mysql_select_db( "shopingsystem" ) or die("Couldn't open Datatabase"); $query1 = "select * from userInfo WHERE session_id=\"$session__id\""; $result = mysql_query( $query1, $link ); // query for username of person with this session id if ( $result ) // Here i test for query failure but it does no good for empty but successful results { if (! $result) // Here is where i want to test for empty result but it does no good! // I have tried ($result == false), ($result == "") and (! $result) // None do any good they all let an empty result pass. { print "Query failed for $session__id <br>"; } print "Query successful for $session__id <br>"; $temp = mysql_fetch_object( $result ); $un = $temp->username; // Here is where i get "Notice: Trying to get property of non-object" error! $query2 = "UPDATE userinfo SET session_id='NULL', is_logged_on='F', last_logon='NULL' WHERE username='$un'"; $result = mysql_query( $query2, $link ); if ( $result ) print "Logout successful for $un"; session_unset(); session_destroy(); } else print "Query Failed for $session__id"; mysql_close( $link ); } Can you help me? "Knowledge is power, Those who have it must share with those who don't" -Walter Wojcik --------------------------------- Do you Yahoo!? vote.yahoo.com - Register online to vote today!