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. -- 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 -- 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
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 -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php