Re: left join does not work, why?

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On Sun, Mar 15, 2009 at 7:02 PM, PJ <af.gourmet@xxxxxxxxxxxx> wrote:

> Daniel Brown wrote:
> > On Sat, Mar 14, 2009 at 19:51, PJ <af.gourmet@xxxxxxxxxxxx> wrote:
> >
> >> $sql1 = "SELECT b.id, b.title, b.sub_title, b.descr, b.comment,
> >> b.bk_cover, b.copyright, b.ISBN, b.language, b.sellers, c.publisher,
> <---
> >> CONCAT_WS(' ', first_name, last_name) AS Author
> >> FROM book AS b
> >> LEFT JOIN book_author AS ab ON b.id = ab.bookID
> >> LEFT JOIN author AS a ON ab.authID=a.id
> >> LEFT JOIN book_publisher as abc ON b.id = abc.bookID // <---
> >> LEFT JOIN publishers AS c ON abc.publishers_id = c.id // <---
> >> ORDER BY title ASC ";
> >> $result1 = mysql_query($sql1, $db);
> >> $bookCount = mysql_num_rows($result1);
> >>
> >
> >     For this and future problems of a similar nature, the quickest and
> > easiest way to start debugging is to change the mysql_query() line to
> > output the error.  Like so:
> >
> > <?php
> > // ....
> > $result1 = mysql_query($sql1,$db) or die("SQL:
> ".$sql1."\n".mysql_error());
> > // ....
> > ?>
> >
> >     If the query fails on $sql1, PHP runs the 'or' as a failure
> > fallback and dies by printing the SQL query given and MySQL's error
> > response.
> >
> Thank you for that. I had not used that as I thought that
> error_reporting(E_ALL);
> ini_set('display_errors', 1);
> would be enough.   =-O
>
> The error reported now is to check the syntax near the commented out lines.
> So, what is going on here? I understood that when the lines are
> commented out they are not lprocessed.
> When I deleted the commented out lines the errors went away.


:P

>
> This does not make sense and obviously could be very misleading in
> debugging.
>
>
> --
> 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
>
>
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>

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