CakePHP - can't get session variable in view

Question

In ProjectsController.php I'm setting a session variable as I'd like that info to be accessible in ALL controllers, models and views:

$this->Session->write('Project.title', $this->Project->title);

Now, when I try access it from Projects view, like this:

        <p>Project: <strong>
        <?php if (isset($session->read('Project.title'))): 
            $session->read('Project.title');
        ?>
        <?php else: ?>
            Not selected
        <?php endif; ?>
        </strong></p>

I'm getting the following error:

Fatal error: Can't use method return value in write context

Which refers to the second line of above code.

I've been through CakePHP documentation and also searched SO, what am I doing wrong here?

Thanks!

EDIT:

I've also tried using:

$this->Session->read('Project.title')

resulting in the same error message.

mark · Accepted Answer · 2012-12-16 14:27:09Z

6

You should do what PHP tells you once in a while :) The error message is pretty clear. You cannot use isset() and empty() this way. They only work with variables direcly, not methods. So use

<?php if ($this->Session->check('Project.title')) {
    echo $this->Session->read('Project.title');
} ?>

as documented in the cookbook

you could also do

<?php
$title = $this->Session->read('Project.title');
if ($title) {
    echo $title;
} ?>

or even

<?php if ($title = $this->Session->read('Project.title')) {
    echo $title;
} ?>

the last one is not cakephp coding convention, though.

answered Dec 16, 2012 at 14:27

mark

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2 Comments

CodeVirtuoso Over a year ago

Yes Mark, I should definitely listen to it:) This resolved it, thanks.

Imdad Over a year ago

For a CakePHP 3.x solution refer to stackoverflow.com/a/32668870

Aditya M P · Accepted Answer · 2012-12-16 14:27:18Z

0

isset() cannot deal with functions like that. Store the return value of that $session object's read() function, then test that variable inside the if() condition:

$title = $session->read('Project.title');
echo ($title) ? $title : '';

answered Dec 16, 2012 at 14:27

Aditya M P

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