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' ,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_artist.id correct? Also, your select has the name of music_artist and the value of the respective id, isn't that what you want? Regards, Christian -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
RE: [PHP-DB] drop list inserts
Barry,
All I saw was $artist_name variable that will be passed thru the form.. Can
you do a
echo $music_artist.id to see if it has a value ..
And you still need to get rid of $query_id = in the line that you are
trying to do insert.
Gurhan
-Original Message-
From: Barry 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 '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,,);
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].''.$row[artist_name]./option\n;
}
echo /select\n;
echo brAlbum Name : input type='text' name='album'
value='$album';
echo input type='submit' name='Submit' value='Submit';
echo /form;
?
and
?
include(../mainfile.php);
include(../header.php);
OpenTable();
mysql_connect( host, , );
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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Re: [PHP-DB] drop list inserts
Yes I see what I have done wrong. I named it music_artist instead of nameing it music_artist.id Sorry about that but thanks for the help. I think I'll need more help as I'm still learning. -Original Message- From: Christian Schneider [EMAIL PROTECTED] To: Barry 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' ,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_artist.id correct? Also, your select has the name of music_artist and the value of the respective id, isn't that what you want? Regards, Christian -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
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,,);
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].''.$row[artist_name]./option\n;
}
echo /select\n;
echo brAlbum Name : input type='text' name='album' value='$album';
echo input type='submit' name='Submit' value='Submit';
echo /form;
?
and
?
include(../mainfile.php);
include(../header.php);
OpenTable();
mysql_connect( host, , );
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.
--
PHP Database Mailing List (http://www.php.net/)
To unsubscribe, visit: http://www.php.net/unsub.php
--
PHP Database Mailing List (http://www.php.net/)
To unsubscribe, visit: http://www.php.net/unsub.php
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 '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,,);
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].''.$row[artist_name]./option\n;
}
echo /select\n;
echo brAlbum Name : input type='text' name='album'
value='$album';
echo input type='submit' name='Submit' value='Submit';
echo /form;
?
and
?
include(../mainfile.php);
include(../header.php);
OpenTable();
mysql_connect( host, , );
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.
--
PHP Database Mailing List (http://www.php.net/)
To unsubscribe, visit: http://www.php.net/unsub.php
--
PHP Database Mailing List (http://www.php.net/)
To unsubscribe, visit: http://www.php.net/unsub.php
--
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To unsubscribe, visit: http://www.php.net/unsub.php
