Hi Thats a php issue. if you put variables in single quotes, it won't resolve the variable. You'll need to put it in double quotes for php to replace the variable with the name. :)
On Mar 12, 2:11 am, adeveloper guy <[email protected]> wrote: > hi, > > i have a simple form with a text box. user types some value in the > text box and i want that to be searched from the db > > i want to know the correct way of putting that typed in value and use > in LIKE command in my query > > this is what i have done ... > > var $paginate = array( > 'conditions' => array('Book.name like' => '%$this->data[Books] > [name]%'), > 'limit' => 2, > 'order' => array( > 'Book.name' => 'asc' > ) > ); > > the problem is, i m not getting $this->data[Books][name] being > replaced by the actual value .. it is taking this whole string as is > and hence not working > > need guidance. > > thanks in advance > > Dev 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
