It seems that error is from an ill-formed MySql statement. Try this, $descripQuery = "SELECT descrip from project_descrip where rowid=$foo"; echo $descripQuery ; // put in this echo after the query. $descripResult = mysql_query($descripQuery); echo mysql_error(); file://put this echo after query execution
Now, run it and see on screen what query is actually executed. "Jennifer" <[EMAIL PROTECTED]> wrote in message 003401c2bf2b$0a042d20$8101a8c0@fvcrx01">news:003401c2bf2b$0a042d20$8101a8c0@fvcrx01... > 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 > > > > 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)){ > > $foo = $row->descrip; > > echo("<font face=verdana size=1>"); > echo("<a href='projects_descrip.php?rowid=$foobar'>" . "$foo</a>" . " | "); > echo("</font>"); > > } > > > ?> > > thanks in advance, > jennifer > -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
