I ran into the same problem on a script yesterday.. It turns out that one of the vars in my mysql_connect() command wasn't assigned.. It gave me no indication of this so I hadn't thought to look there.. I just happened to be reading back through the whole page to see what I could find and noticed that I hadn't set the var at the top..
Something else that I noticed in your code is that there doesn't seem to be a mysql_select_db() call.. Unless that is somewhere else in the code and just not shown here, you'll need it I believe.. Just a couple thoughts, Bob Weaver "Luke Kearney" <[EMAIL PROTECTED]> wrote in message [EMAIL PROTECTED]">news:[EMAIL PROTECTED]... > 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. > > Thanks > > Luke K > > -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
