In article <[EMAIL PROTECTED]>, [EMAIL PROTECTED] says... > Hello > I'm using PHP4 and MySQL on Apache webserver. > I have a table valled "liste" with names of some people, and a table > called "postnummer" with zip-codes and citynames. > My select should join the adress and zip from one table with the > cityname from the other table. > But my join-line does not work. =:( > > My code results in this line: > > Warning: Supplied argument is not a valid MySQL result resource in > /var/www/html/find/resultat.php on line 27 > > The code is: > mysql_select_db("adresser"); > $query = ("SELECT fornavn, efternavn, gade, postnummer.postnummer, > postnummer.postbynavn FROM liste > left outer join postnummer on liste.postnummer = postnummer.postnummer > WHERE MATCH (fornavn, efternavn) AGAINST ('$search');"); > $result = mysql_query($query);
Add here echo mysql_error(); which will return you an error string from mysql which will probably be an error in your query syntax. It might also be useful to echo $query to ensure that it contains what you expect it does (possible register_globals problem?) > while(list($cpr, $fornavn, $efternavn, $gade, $postnummer, $postbynavn) = > mysql_fetch_row($result)) > > What is going on? > The select is ok if I do not join, but then I'm not able to get the > field "postbynavn" from the other table. > > PLEASE HELP! -- David Robley Temporary Kiwi! Quod subigo farinam -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
