Re: Join - problem

Oh dear - I forgot the rights in the new table: select command denied to user: 'henning@localhost' for table 'postnummer' Thanks for helping, John! Henning John Wards wrote:

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