Re: Logout link in layout

If($this->Session->read('User.id'))
would only be appropriate in cakephp 1.3! (in the view)

On 24 Dez., 12:40, stefano <ssalvat...@gmail.com> wrote:
> if($Session->read("User.id")) {
> ...
>
> }
>
> On Thu, Dec 24, 2009 at 8:30 AM, Edinei Luis Cipriani
>
>
>
> <phpedi...@gmail.com> wrote:
> > If(isset($this->Session->read('User.id'))):
> >> echo $html->link('Logout', array('controller' => 'Users',
> >> 'action'=>'logout'));
>
> > endif;
>
> > happy christmas
>
> > Em 24/12/2009, às 07:46, emmexx escreveu:
>
> >> I'd like to put a logout link in my default layout. The link should be
> >> there only when a user has already logged in.
> >> I put the following code in my app/views/layouts/default.ctp:
>
> >> if ($this->Auth->user())
> >>       echo $html->link('Logout', array('controller' => 'Users',
> >> 'action'=>'logout'));
>
> >> but, of course, cake didn't like and screamed in horror:
>
> >> Notice (8): Undefined property:  View::$Auth [APP/views/elements/
> >> logout.ctp, line 3]
>
> >> Fatal error: Call to a member function user() on a non-object in /app/
> >> views/elements/logout.ctp on line 3
>
> >> Is there a way to accomplish that?
>
> >> Thank you and Merry Christmas!
>
> >>   maxx
>
> >> Check out the new CakePHP Questions sitehttp://cakeqs.organd help
> >> others with their CakePHP related questions.
>
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>
> > Check out the new CakePHP Questions sitehttp://cakeqs.organd help others with their CakePHP related questions.
>
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>
> --
> Stefano Salvatori M.http://stefano.salvatori.cl/

Check out the new CakePHP Questions site http://cakeqs.org and help others with their CakePHP related questions.

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