> hi, i'm hoping someone can help me out here - im trying to pass a variable > from a string in the href tag, however my sql breaks when i put a var in > the statment. > when i replace $foo with a number, the page works fine. can someone please > tell me what im doing wrong? > > here is the error i get: > > Warning: Supplied argument is not a valid MySQL result resource in > /home/villany2k1/www.villany2k1.com/htdocs/projects/projects_descrip.php > on line 11
That means your query failed. Use mysql_error() to find out why. > here is my code: > > <?php > $hostName="localhost"; > $userName="xxxx"; > $password="xxxx"; > $database= "projects"; > $tableName="project_descrip"; > > $descripQuery = "SELECT descrip from project_descrip where rowid=$foo"; > mysql_connect($hostName, $userName, $password); > $descripResult = mysql_db_query($database,$descripQuery); > while($row = mysql_fetch_object($descripResult)){ You shouldn't be using mysql_db_query() for one, it's depreciated. Use mysql_select_db() along with mysql_query(). If rowid is an integer, then you don't need quotes around $foo, otherwise, if it's a string, then you do. where rowid='$foo' ---John W. Holmes... PHP Architect - A monthly magazine for PHP Professionals. Get your copy today. http://www.phparch.com/ -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php