PJ wrote:
Ian wrote:
On 31 Mar 2009 at 9:08, PJ wrote:
I must be doing something wrong. Can't figure it out even though I've
been searching the manuals & tutorials, it still does not work. Here is
the exact code that I have tried; the first version is commented out and
obviously does not work either (the spelling, the table names and column
names are correct):
$books = array();
/*$SQL = "SELECT * FROM book b
JOIN book_author c
ON b.id = c.bookID
JOIN author a ON a.id = c.authID
WHERE LEFT(a.last_name,1) = $Auth
ORDER BY $sort $dir
LIMIT $offset, $records_per_page"; */
$SQL = "SELECT * FROM book b
INNER JOIN book_author c
ON b.id = c.bookID
WHERE c.authID = (SELECT id FROM author a WHERE LEFT(a.last_name, 1
) = $Auth )
Hi,
I think this should be '$Auth' as MySQL will be expecting a string.
Hmmm.... That was it. I'm not clear on this... what was MySQL getting?
if not a string...
sorry for my ignorance...
You have to quote strings in sql, only numbers (and booleans) don't have
quotes.
ie
$sql = ".. where left(a.last_name, 1) = '" .
mysql_real_escape_string($Auth) . "'";
single quotes around the value and it's properly escaped.
--
Postgresql & php tutorials
http://www.designmagick.com/
--
PHP General Mailing List (http://www.php.net/)
To unsubscribe, visit: http://www.php.net/unsub.php
