Re: difficult select problem

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On Mon, Apr 6, 2009 at 12:32 PM, PJ <af.gourmet@xxxxxxxxxxxx> wrote:

> I've searched the web, the tutorials, etc. with no luck and have asked
> on MySql list with no luck and have even posted here with no replies.
> So, let's try again:
>
> I am trying to limit the search for books to only those that start with
> "A" (does not have to be case sensitive); but within that selection
> there may be a second author whose name may start with any letter of the
> alphabet.
>
> So, the real problem is to find if there is a second author in the
> selection of books and to display that name.
> My query shows all the books that begin with A:
>
> $SQL = "SELECT b.*, c.publisher, a.first_name, a.last_name
>        FROM book AS b
>        LEFT JOIN book_publisher as bp ON b.id = bp.bookID
>        LEFT JOIN publishers AS c ON bp.publishers_id = c.id
>        LEFT JOIN book_author AS ba ON b.id = ba.bookID
>        LEFT JOIN author AS a ON ba.authID = a.id
>        WHERE LEFT(last_name, 1 ) = '$Auth' ";
>
> Within the results there are some books that have 2 authors and now I have
> to find if
> there is a second author in the results and then be able to echo the second
> author.
> So far, I have not been able to figure out how to go about that.
> Do I need to do another query to find the second author or can it somehow
> be incorporated into the original query? Or can it be done with a UNION ?
> Please help.
>
>
>
> --
> unheralded genius: "A clean desk is the sign of a dull mind. "
> -------------------------------------------------------------
> Phil Jourdan --- pj@xxxxxxxxxxxxx
>   http://www.ptahhotep.com
>   http://www.chiccantine.com/andypantry.php
>
>
> --
> PHP General Mailing List (http://www.php.net/)
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>
>
$SQL = "SELECT b.*, c.publisher, a.first_name, a.last_name
       FROM book AS b
       LEFT JOIN book_publisher as bp ON b.id = bp.bookID
       LEFT JOIN publishers AS c ON bp.publishers_id = c.id
       LEFT JOIN book_author AS ba ON b.id = ba.bookID
       LEFT JOIN author AS a ON ba.authID = a.id
       WHERE last_name like '$Auth%' ";


The above assumes that $Auth has a value of A and should pull all the
authors whose last name starts with the letter A

-- 

Bastien

Cat, the other other white meat

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