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" <[EMAIL PROTECTED]>
To: <[EMAIL PROTECTED]>
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
>
>
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