Re: How can I debug an ajax login call?

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 <heidi.anselstet...@consultingteam.de> 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?

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