Sorry, just say the message.
> Your code looks well. But is the variable $db the name of your database or
> your link-identifier. When it is the name of your database i'm not really
> surpised your code wouldn't work. mysql_query requires as second argument a
> link identifier.
Read what Rolf says. Very wise words here. :-)
> So with all that said here is what i have done that doesn't work,
>
> $result = mysql_query("SELECT artist_id FROM songs",$db) or
> die(mysql_error());
>
> if ($row = mysql_fetch_row($result)){
> do {
>
> $artist_name = $row["artist_id"];
> $result_1 = mysql_query("SELECT artist_id,artist_name FROM
> artists WHERE artist_name = '$artist_name'",$db);
> $row_1 = mysql_fetch_array($result_1);
Why don't you make one query with a union between the 2 tables and then pass
through the results with a loop and maybe an if inside it? Ypou are making to
many connections to the DB.
P.D.: Any way, your problem will be solved with Rolfs advice.
--
select 'mmarques' || '@' || 'unl.edu.ar' AS email;
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Martín Marqués | [EMAIL PROTECTED]
Programador, Administrador, DBA | Centro de Telemática
Universidad Nacional
del Litoral
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