Hi, I fixed it, I have not realize that my database was not connected to a user.
so when i logged as prog_tony dealer was not a member of prog_tony so it could not be found. thanks for the help "Tony" <[EMAIL PROTECTED]> wrote in message news:[EMAIL PROTECTED] > hello > > I'm new with php just learning and i have just a problem with the following > code > > $dbconnect = mysql_connect("localhost", "prog_tony","PASSWORD"); > mysql_select_db("prog_dealer", $dbconnect); > $stop = 0; > $counter = 1; > while(!$stop){ > $query = "SELECT name FROM category WHERE id = $counter"; > $dbdo = mysql_query($query,$dbconnect); > while($row= mysql_fetch_array($dbdo)){ > if($row[0]){ > $counter++; > }else{ > $stop=1; > } > $row[0] = ""; > } > > } > > > > I'm getting an error with mysql_fetch_array() which is line 14 because I > didn't show the other lines since it is not relevant. > here is the error > Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result > resource in /home/prog/public_html/php/cateadded.php on line 14 > > Thank you -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
