Re: Cake 2.0.4 : how to display "logout" if logged, else display "login"

ok, that makes sense.
no - the method call stays the same!

btw:
most cake developers would check on the id specifically:

if ($this->Session->check('Auth.User.id')) {}

On 23 Dez., 19:51, JonStark <jean...@gmail.com> wrote:
> I never used previous versions, I found this by searching a little on
> the net.
>
> So it's more like :
>
>       <?php
>          if ($this->Session('Auth')) {
>                 echo("logged");
>                 }
>          else {
>                 echo("not logged");
>                 }
>         ?>
>
> ?
>
> Thanks a lot for your help !
>
> On 23 déc, 19:41, euromark <dereurom...@googlemail.com> wrote:
>
>
>
>
>
>
>
> > $session was already outdated in 1.3
>
> > $this->Session is the correct usage
>
> > On 23 Dez., 17:26, JonStark <jean...@gmail.com> wrote:
>
> > > I want to do this on cake 2.0.4 (new to cake), but this does'nt work:
>
> > >         <?php
> > >          if ($session->read('Auth')) {
> > >                 echo("logged");
> > >                 }
>
> > >          else {
> > >                 echo("not logged");
> > >                 }
> > >         ?>
>
> > > Thanks a lot for your help !

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