In article <[EMAIL PROTECTED]>, [EMAIL PROTECTED] says... > Hello All, > I am struggling to understand why my query throws an error the code as below > > $link = mysql_connect ( "db.localhost.org", $user , $pass ); > if ( ! $link ) > die ( "Couldn't Connect to Mysql Server" ); > print "Sucessfully connect to server <p>"; > $query = "select * from users " > ."where screen_name='$screen_name' " > ." and passwd=password('$passwd')"; > $result = mysql_query($query, $link); > if (mysql_num_rows($result) >0 ) > > When executed you get a little message back like this one > > Warning: Supplied argument is not a valid MySQL result resource in > /usr/local/share/doc/apache/trial/authmain.php on line 20 > > What does this generally mean as I have noted that others use this syntax so > what is my issue ? > > Any help or pointers to references would be most appreciated.
This generally means there is a problem with your SQL syntax; add to your mysql_query line 'or exit(mysql_error()' and see waht you get. mysql_error should return a descriptive error message from mysql. -- David Robley Temporary Kiwi! Quod subigo farinam -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
