RE: [PHP-DB] NEWBIE - Needs Assistance with Joins

actually artist_id = 'aid' will not work, since you are comparing artist_id with the string "aid" and not with the artists id in the second table... from a sql point of view the statement SELECT artist_name, album_title FROM artists, album_titles WHERE artist_id = aid is correct (if your names are UNIQUE and correct and if you only want 1:1 relations). For starters you could write the statement as one of the following: SELECT artists.artist_name, album_titles.album_title FROM artists LEFT JOIN album_titles ON artists.artist_id = album_titles.aid (might be RIGHT JOIN, I'm never sure...) or if everything is unique: SELECT artist_name, album_title FROM artists LEFT JOIN album_titles ON artist_id = aid Greets, Remo > -----Original Message----- > From: Jason Wong [mailto:phplist@gremlins.com.hk] > Sent: Monday, September 17, 2001 10:10 PM > To: php-db@lists.php.net; Cecily Walker Kidd > Subject: Re: [PHP-DB] NEWBIE - Needs Assistance with Joins > > > > ----- Original Message ----- > From: Cecily Walker Kidd <babeephat@hotmail.com> > To: <php-db@lists.php.net> > Sent: Monday, September 17, 2001 5:42 AM > Subject: [PHP-DB] NEWBIE - Needs Assistance with Joins > > > > Hello, > > > > I have two tables, one that contains an item id, album title name, and > > artist ID number. The second table is a list of artists, with an > > auto-increment artist ID. > > > > I want to join the two tables and have them output to a single PHP page. > > I was following along with the tutorial at > > http://www.webmasterbase.com/article/228/, and tried to > > modify it for my > > own needs. When I do this, I get a parse error on the line that starts > > with SELECT. Here's the code: > > > > $Link = mysql_connect ($Host, $User, $Password); > > > > $CDList =mysql_query( > > "SELECT artist_name, album_title ". > > "FROM artists, album_titles WHERE artist_id = aid"); > > > > > > What am I doing wrong? > > > > Thanks in advance. > > Try: > > -------------------------------------------------------- > $CDList = mysql_query("SELECT artist_name, album_title > FROM artists, album_titles > WHERE artist_id = 'aid'" > ); > -------------------------------------------------------- > > > hth > -- > Jason Wong > Gremlins Associates > www.gremlins.com.hk > Tel: +852-2573-5033 > Fax: +852-2573-5851 > > > > > -- > PHP Database Mailing List (http://www.php.net/) > To unsubscribe, e-mail: php-db-unsubscribe@lists.php.net > For additional commands, e-mail: php-db-help@lists.php.net > To contact the list administrators, e-mail: php-list-admin@lists.php.net > >