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 'music_artist'( I've got this), I want the id of the
artist they selected in the drop list to be inserted into the
table 'music_album' as $artist_id.
-----Original Message-----
From: Gurhan Ozen <[EMAIL PROTECTED]>
To: Barry Rumsey <[EMAIL PROTECTED]>, php-db list <php-
[EMAIL PROTECTED]>
Date: Wed, 17 Apr 2002 19:02:52 -0400
Subject: RE: [PHP-DB] drop list inserts
> 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/function.mysql-query.php
> for
> this..
> Second of all, in your INSERT INTO query you are trying to insert the
> value
> of a variable called $music_artist.id which doesn't exist anywhere. I
> think
> you meant to insert $artist_name instead???
>
> Gurhan
>
>
> -----Original Message-----
> From: Barry Rumsey [mailto:[EMAIL PROTECTED]]
> Sent: Wednesday, April 17, 2002 6:23 PM
> To: php-db list
> Subject: [PHP-DB] drop list inserts
>
>
> I have the following to pages( just testing them at the moment ):
> <? mysql_connect("host","xxxx","xxxx");
> mysql_select_db("music");
> echo "<form name='add_album' method='post'
> action='test-album-add.php'>";
> $getlist = mysql_query("SELECT * FROM music_artist ORDER BY
> artist_name ASC");
> echo " Artist Name : <select name=\"artist_name\">\n";
> while ($row = mysql_fetch_array($getlist)) {
> echo '<option
> value="'.$row["id"].'">'.$row["artist_name"]."</option>\n";
> }
> echo " </select>\n";
> echo "<br>Album Name : <input type='text' name='album'
> value='$album'>";
> echo "<input type='submit' name='Submit' value='Submit'>";
> echo "</form>";
> ?>
>
> and
>
> <?
> include("../mainfile.php");
> include("../header.php");
> OpenTable();
>
> mysql_connect( "host", "xxxx", "xxxx" );
> mysql_select_db( "xoops" );
>
> $query_id = mysql_query("INSERT INTO music_album VALUES
> (NULL, '$music_artist.id' ,'$album' ,NULL ,NULL)");
> echo " $artist_name and $album has been added to the
> database.";
> CloseTable();
> include("../footer.php");
> ?>
>
> What I am trying to do is insert the id of the artist they selected in
> the first page into a second table. At the moment all I get is 0
> inserted instead of the artist id from page 1.
>
> Could someone please point out what I'm doing wrong?
>
> Thanks in advance.
>
>
>
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