Don't include the ; in your query, for one thing.
---John Holmes...
----- Original Message -----
From: "John Wards" <[EMAIL PROTECTED]>
To: "Henning" <[EMAIL PROTECTED]>
Cc: <[EMAIL PROTECTED]>
Sent: Wednesday, September 18, 2002 11:26 AM
Subject: Re: [PHP] 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" <[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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>
>
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