y Rumsey <[EMAIL PROTECTED]>
Date: Thu, 18 Apr 2002 12:17:34 +0200
Subject: Re: [PHP-DB] drop list inserts
> Hi Barry,
>
> Barry Rumsey wrote:
> > $query_id = mysql_query("INSERT INTO music_album VALUES
> > (NULL, '$music_artist.id' ,'$album' ,NU
rry Rumsey [mailto:[EMAIL PROTECTED]]
Sent: Wednesday, April 17, 2002 7:35 PM
To: php-db list
Subject: RE: [PHP-DB] drop list inserts
The $music_artist.id is the id from the first page. ( database = music ,
table = music_artist id ). This is what I need, a drop down list of the
artists in table &#x
Hi Barry,
Barry Rumsey wrote:
> $query_id = mysql_query("INSERT INTO music_album VALUES
> (NULL, '$music_artist.id' ,'$album' ,NULL ,NULL)");
> echo " $artist_name and $album has been added to the
>
> Could someone please point out what I'm doing wrong?
>
is the value of $music_
into the
table 'music_album' as $artist_id.
-Original Message-
From: Gurhan Ozen <[EMAIL PROTECTED]>
To: Barry Rumsey <[EMAIL PROTECTED]>, php-db list
Date: Wed, 17 Apr 2002 19:02:52 -0400
Subject: RE: [PHP-DB] drop list inserts
> Hi Barry,
> First of all,
> $quer
Hi Barry,
First of all,
$query_id = mysql_query("INSERT INTO...); is wrong. That line will just
assign the resultset of the whatever mysql_query() function returns to the
variable $query_id .. Get rid of $query_id and just have mysql_query(INSERt
INTO ); See: http://www.php.net/manual/en/func
