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

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JonStark

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

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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euromark (munich)

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

$session was already outdated in 1.3

$this->Session is the correct usage

On 23 Dez., 17:26, JonStark <[hidden email]> 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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Check out the new CakePHP Questions site http://ask.cakephp.org and help others with their CakePHP related questions.

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JonStark

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

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 <[hidden email]> wrote:

> $session was already outdated in 1.3
>
> $this->Session is the correct usage
>
> On 23 Dez., 17:26, JonStark <[hidden email]> wrote:
>
>
>
>
>
>
>
> > I want to do this on cake 2.0.4 (new to cake), but this does'nt work:
>
> > <?php
> > if ($session->read('Auth')) {
> > echo("logged");
> > }
>
> > else {
> > echo("not logged");
> > }
> > ?>
>
> > Thanks a lot for your help !

euromark (munich)

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 <[hidden email]> 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 <[hidden email]> wrote:
>
>
>
>
>
>
>
> > $session was already outdated in 1.3
>
> > $this->Session is the correct usage
>
> > On 23 Dez., 17:26, JonStark <[hidden email]> wrote:
>
> > > I want to do this on cake 2.0.4 (new to cake), but this does'nt work:
>
> > > <?php
> > > if ($session->read('Auth')) {
> > > echo("logged");
> > > }
>
> > > else {
> > > echo("not logged");
> > > }
> > > ?>
>
> > > Thanks a lot for your help !

euromark (munich)

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

more can be found here:
http://book.cakephp.org/2.0/en/core-libraries/helpers/session.html?highlight=session#SessionHelper

On 23 Dez., 20:01, euromark <[hidden email]> wrote:

> ok, that makes sense.
> no - the method call stays the same!
>
> btw:
> most cake developers would check on the id specifically:
>
> if ($this->Session->check('Auth.User.id')) {}
>
> On 23 Dez., 19:51, JonStark <[hidden email]> wrote:
>
>
>
>
>
>
>
> > I never used previous versions, I found this by searching a little on
> > the net.
>
> > So it's more like :
>
> > <?php
> > if ($this->Session('Auth')) {
> > echo("logged");
> > }
> > else {
> > echo("not logged");
> > }
> > ?>
>
> > ?
>
> > Thanks a lot for your help !
>
> > On 23 déc, 19:41, euromark <[hidden email]> wrote:
>
> > > $session was already outdated in 1.3
>
> > > $this->Session is the correct usage
>
> > > On 23 Dez., 17:26, JonStark <[hidden email]> wrote:
>
> > > > I want to do this on cake 2.0.4 (new to cake), but this does'nt work:
>
> > > > <?php
> > > > if ($session->read('Auth')) {
> > > > echo("logged");
> > > > }
>
> > > > else {
> > > > echo("not logged");
> > > > }
> > > > ?>
>
> > > > Thanks a lot for your help !

JonStark

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

Thanks a lot ! Now it works, and more importantly I understand why !
Thanks a lot for your time.

On 23 déc, 20:02, euromark <[hidden email]> wrote:

> more can be found here:http://book.cakephp.org/2.0/en/core-libraries/helpers/session.html?hi...
>
> On 23 Dez., 20:01, euromark <[hidden email]> wrote:
>
>
>
>
>
>
>
> > ok, that makes sense.
> > no - the method call stays the same!
>
> > btw:
> > most cake developers would check on the id specifically:
>
> > if ($this->Session->check('Auth.User.id')) {}
>
> > On 23 Dez., 19:51, JonStark <[hidden email]> wrote:
>
> > > I never used previous versions, I found this by searching a little on
> > > the net.
>
> > > So it's more like :
>
> > > <?php
> > > if ($this->Session('Auth')) {
> > > echo("logged");
> > > }
> > > else {
> > > echo("not logged");
> > > }
> > > ?>
>
> > > ?
>
> > > Thanks a lot for your help !
>
> > > On 23 déc, 19:41, euromark <[hidden email]> wrote:
>
> > > > $session was already outdated in 1.3
>
> > > > $this->Session is the correct usage
>
> > > > On 23 Dez., 17:26, JonStark <[hidden email]> wrote:
>
> > > > > I want to do this on cake 2.0.4 (new to cake), but this does'nt work:
>
> > > > > <?php
> > > > > if ($session->read('Auth')) {
> > > > > echo("logged");
> > > > > }
>
> > > > > else {
> > > > > echo("not logged");
> > > > > }
> > > > > ?>
>
> > > > > Thanks a lot for your help !

JonStark

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

Just a last question :

I can echo this :

<?php

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

echo("

Logged in

")

} else {
echo ("Not logged in");
}

?>

but this doesn't work :

<?php

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

echo("

<div class="well" style="padding: 16px 19px;">
<a href="<?php echo $this->Html->url('/items/add'); ?>" class="btn
primary small">Add an activity</a>
</div>

")

} else {
echo ("");
}

?>

I'm sorry if this code looks supid, I'm completely new to cakephp, and
more generally php !

On 23 déc, 20:42, JonStark <[hidden email]> wrote:

> Thanks a lot ! Now it works, and more importantly I understand why !
> Thanks a lot for your time.
>
> On 23 déc, 20:02, euromark <[hidden email]> wrote:
>
>
>
>
>
>
>
> > more can be found here:http://book.cakephp.org/2.0/en/core-libraries/helpers/session.html?hi...
>
> > On 23 Dez., 20:01, euromark <[hidden email]> wrote:
>
> > > ok, that makes sense.
> > > no - the method call stays the same!
>
> > > btw:
> > > most cake developers would check on the id specifically:
>
> > > if ($this->Session->check('Auth.User.id')) {}
>
> > > On 23 Dez., 19:51, JonStark <[hidden email]> wrote:
>
> > > > I never used previous versions, I found this by searching a little on
> > > > the net.
>
> > > > So it's more like :
>
> > > > <?php
> > > > if ($this->Session('Auth')) {
> > > > echo("logged");
> > > > }
> > > > else {
> > > > echo("not logged");
> > > > }
> > > > ?>
>
> > > > ?
>
> > > > Thanks a lot for your help !
>
> > > > On 23 déc, 19:41, euromark <[hidden email]> wrote:
>
> > > > > $session was already outdated in 1.3
>
> > > > > $this->Session is the correct usage
>
> > > > > On 23 Dez., 17:26, JonStark <[hidden email]> wrote:
>
> > > > > > I want to do this on cake 2.0.4 (new to cake), but this does'nt work:
>
> > > > > > <?php
> > > > > > if ($session->read('Auth')) {
> > > > > > echo("logged");
> > > > > > }
>
> > > > > > else {
> > > > > > echo("not logged");
> > > > > > }
> > > > > > ?>
>
> > > > > > Thanks a lot for your help !

Geoff Douglas

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

Under the logged in condition... you are using double quotes twice... could cause a problem.

euromark (munich)

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

it does cause the problem :)
usually a good IDEA would automatically display a parse error

On 24 Dez., 02:52, Geoff Douglas <[hidden email]> wrote:
> Under the logged in condition... you are using double quotes twice... could
> cause a problem.

--
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.

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JonStark

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

Indeed, stupid mistake... Thanks again ! I need to be more careful.

This works :

<?php
$addlink = $this->Html->url('/items/add');

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

<div class='well' style='padding: 16px 19px;'>
<a href='$addlink' class='btn
primary small'>Add an activity</a>
</div>

");
} else
{
echo ("");
}

?>

On 24 déc, 11:25, euromark <[hidden email]> wrote:

> it does cause the problem :)
> usually a good IDEA would automatically display a parse error
>
> On 24 Dez., 02:52, Geoff Douglas <[hidden email]> wrote:
>
>
>
>
>
>
>
> > Under the logged in condition... you are using double quotes twice... could
> > cause a problem.