Re: Join - problem

not having the time to look at your code I am not sure but do this to see if its mysql causeing the errors $result = mysql_query($query) or die(mysql_error()); ----- Original Message ----- From: "Henning" <henningolsen36@netscape.net> To: <php-general@lists.php.net> Sent: Wednesday, September 18, 2002 5:12 PM Subject: [PHP] Join - problem > 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"); > $query = ("SELECT fornavn, efternavn, gade, liste.postnummer, > postnummer.postbynavn FROM liste > left join postnummer on liste.postnummer = postnummer.postnummer > WHERE MATCH (fornavn, efternavn) AGAINST ('$search');"); > $result = mysql_query($query); > while(list( $fornavn, $efternavn, $gade, $postnummer, $postbynavn) = > mysql_fetch_row($result)) > print("<TR><TD>$fornavn</TD><TD>$efternavn</TD><TD>$gade</TD><TD>$postnummer </TD><TD>$postbynavn</TD></TR>\n"); > > > 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! > > Henning > > > -- > PHP General Mailing List (http://www.php.net/) > To unsubscribe, visit: http://www.php.net/unsub.php