> 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 

> 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.

David Robley
Temporary Kiwi!

Quod subigo farinam

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