PHP if statement not displaying HTML

Question

So I'm checking if a user is logged in, and if so I'll remove the signup forms. So I thought I'd do it with if else statements. So here's my code

<?php if($_SESSION['loggedin'] === false): ?>
<h1>Sign Up</h1>
<?php elseif($_SESSION['loggedin'] === true):?>
<h1>Welcome</h1>
<?php endif;?>

For some reason all I get is a blank page. I'm using

error_reporting(E_ALL);
ini_set('display_errors', '1');

To display errors, but I'm not getting anything. On another page when doing a var_dump on $_SESSION['loggedin' I get bool(true), so I know that I'm logged in and expect Welcome. I feel like its a syntax error. Any ideas?

session_start(); have you put this on the top just after <?php — mehulmpt
– mehulmpt, Commented Jun 26, 2014 at 14:46
Check your apache (or whatever webserver) error logs, maybe you're having an error you can't see with display_errors? — David Wilkins
– David Wilkins, Commented Jun 26, 2014 at 14:47
I am using session_start(), its in another <?php tag right at the top @MehulMohan — Idris
– Idris, Commented Jun 26, 2014 at 14:47
Change your else line to: <?php else:?> to troubleshoot. Looks like loggedin is neither false nor true. — bloodyKnuckles
– bloodyKnuckles, Commented Jun 26, 2014 at 14:49

stuartbates · Accepted Answer · 2014-06-26 14:51:49Z

2

If you tail your server logs (e.g. tail -F logs/php_error.log) you'll probably see something like:

PHP Notice: Undefined variable: _SESSION

Try changing the code to:

if (isset( $_SESSION['loggedin'] ) && $_SESSION['loggedin'] == true )
{
  echo 'Welcome';
} else {
  echo 'Sign up';
}

Personally I'd also recommend changing your sign out logic to use unset($_SESSION) that way you only have to check for:

if (isset( $_SESSION['loggedin'] ) ){
  echo 'Welcome';
} else {
  echo 'Sign up';
}

answered Jun 26, 2014 at 14:51

stuartbates

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3 Comments

Pitchinnate Over a year ago

What happens if there is session info that should persist despite login state? Not saying there is in this case but there could be.

stuartbates Over a year ago

I have sites which require that - in that case I name space everything inside the session e.g. $_SESSION['user']['id'], $_SESSION['user']['name'] etc... Then during logout you can just call unset($_SESSION['user']) and check using if (isset($_SESSION['user']))

Pitchinnate Over a year ago

Correct, I was just putting out a word of caution. In this case they could just use unset($_SESSION['loggedin']);

Reactgular · Accepted Answer · 2014-06-26 14:48:48Z

2

Try this:

<?php if(isset($_SESSION['loggedin']) && $_SESSION['loggedin'] === true): ?>
<h1>Welcome</h1>
<?php else:?>
<h1>Sign Up</h1>
<?php endif;?>

answered Jun 26, 2014 at 14:48

Reactgular

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1 Comment

Joseph Over a year ago

I agree. Geting rid of the elseif statement will help, plus the isset() check.

Mr. Llama · Accepted Answer · 2014-06-26 14:48:54Z

1

It's possible that neither code path gets executed if $_SESSION['loggedin'] isn't a boolean because the === comparison checks type and value. (See documentation for details)

A more safe and sane approach would be:

<?php if($_SESSION['loggedin'] === true): ?>
<h1>Welcome</h1>
<?php else:?>
<h1>Sign Up</h1>
<?php endif;?>

answered Jun 26, 2014 at 14:48

Mr. Llama

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Comments

Pitchinnate · Accepted Answer · 2014-06-26 14:49:25Z

1

Try adding an isset() to your if test.

<?php if(!isset($_SESSION['loggedin']) || $_SESSION['loggedin'] === false): ?>
<h1>Sign Up</h1>
<?php else: ?>
<h1>Welcome</h1>
<?php endif;?>

answered Jun 26, 2014 at 14:49

Pitchinnate

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Comments

mehulmpt · Accepted Answer · 2014-06-26 14:47:36Z

0

Try this way:

<?php if(!$_SESSION['loggedin']) { ?>
<h1>Sign Up</h1>
<?php } elseif($_SESSION['loggedin']) { ?>
<h1>Welcome</h1>
<?php } ?>

answered Jun 26, 2014 at 14:47

mehulmpt

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Comments

ameenulla0007 · Accepted Answer · 2014-06-26 18:46:52Z

0

it is not resulting due to Undefined variable: _SESSION error. so you can utilize any ignorant (@ or &) Ex. @$_SESSION['loggedin'] OR &$_SESSION['loggedin']

answered Jun 26, 2014 at 18:46

ameenulla0007

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