On Sat, Mar 14, 2009 at 19:51, PJ <af.gour...@videotron.ca> 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:

// ....
$result1 = mysql_query($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

</Daniel P. Brown>
daniel.br...@parasane.net || danbr...@php.net
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