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:[EMAIL PROTECTED]]
> Sent: Monday, September 17, 2001 10:10 PM
> To: [EMAIL PROTECTED]; Cecily Walker Kidd
> Subject: Re: [PHP-DB] NEWBIE - Needs Assistance with Joins
>
>
>
> ----- Original Message -----
> From: Cecily Walker Kidd <[EMAIL PROTECTED]>
> To: <[EMAIL PROTECTED]>
> 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/)
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