On Jan 11, 3:37 pm, Ryan Schmidt <[email protected]> wrote:
> On Jan 11, 2011, at 07:10, javier0051 wrote:
> > Code | Context
>
> > $sql = "SELECT COUNT(*) AS `count` FROM `articles` AS `Article`
> > LEFT
> > JOIN `categories` AS `Category` ON (`Article`.`category_id` =
> > `Category`.`id`) LEFT JOIN `images` AS `Image` ON
> > (`Image`.`foreign_id` = `Article`.`id` AND `Image`.`thumb` = 1 AND
> > `Image`.`model` = 'Article') LEFT JOIN `galleries` AS `Gallery` ON
> > (`Gallery`.`article_id` = `Article`.`id`) WHERE conditions = ('43')
> > "
> > $error = "1054: Unknown column 'conditions' in 'where clause'"
>
> At the risk of stating the obvious, it sounds like there is no column called
> "conditions".
a missing field named conditions shouts app-code error to me. Given it
comes from a paginate call, I guess this is in the code:
->paginate(array('conditions' => array(43)));
which is simply wrong.
javier I'd suggest doing the blog tutorial before continuing - it'll
answer these sort of problems for you by example.
AD
ps. there's a spanish google group cakephp-esp
Check out the new CakePHP Questions site http://cakeqs.org and help others with
their CakePHP related questions.
You received this message because you are subscribed to the Google Groups
"CakePHP" group.
To post to this group, send email to [email protected]
To unsubscribe from this group, send email to
[email protected] For more options, visit this group at
http://groups.google.com/group/cake-php?hl=en
