Read this for AJAX debugging: http://bakery.cakephp.org/articles/HeathNail/2010/08/20/debugging-with-firephp
best regards Roland On 2 Dez., 11:29, heohni <[email protected]> wrote: > Hi, > > I cannot login into my users area and I am using a ajax form: > > function login() { > > Configure::write('debug', 2); > $this->layout = 'ajax'; > > $this->RequestHandler->isAjax()) { > if($this->Auth->user()){................ <==== > > This is always wrong and I don't see a way to debug this...? > How can I make the sql query visible? > I am using a ajax form as I said... > Via Jquery I catch the submit and send a post to > $.post('/users/login', $(this).serializeArray(), afterValidate, > 'json'); > > I could send the sql statement debug info to the json output if I > would know how to make the query visible? > Any ideas? -- Our newest site for the community: CakePHP Video Tutorials http://tv.cakephp.org Check out the new CakePHP Questions site http://ask.cakephp.org and help others with their CakePHP related questions. To unsubscribe from this group, send email to [email protected] For more options, visit this group at http://groups.google.com/group/cake-php
