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Possible Duplicate:
Passing PHP variable into JavaScript

I want to obtain a PHP variable, to use it in an if-condition in JavaScript.

I was trying to use it like this: 

var phpLogin = <?php $_SESSION['login'] ?>;

But it's wrong in syntax terms. So, how can I use a PHP variable in JavaScript?

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2  
var phpLogin = <?php ECHO $_SESSION['login'] ?>; You forgot to echo varible. And check if key exists in $_SESSION before echo... – Glavić Nov 26 '11 at 17:27
and possible more in stackoverflow.com/… – Gordon♦ Nov 26 '11 at 17:42

marked as duplicate by Gordon♦, mario, Rob W, Esailija, Justin Ethier Nov 26 '11 at 18:54

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

5 Answers

up vote 2 down vote

Use json_encode and don't forget echo.

var phpLogin = <?php echo json_encode($_SESSION['login']) ?>;
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No semi colon in the php code? – Tim Nov 26 '11 at 17:28
@Tim It's not necessary in this case. – Artefacto Nov 26 '11 at 17:29
OP also did not indicate that the returned data was desired to be in JSON format – Tim Nov 26 '11 at 17:30
You dont need one because it only contains a single statement and then closes out of php... but you can add it in if you like. – prodigitalson Nov 26 '11 at 17:30
@Tim So? He has to use JSON, otherwise it'll only work if $_SESSION['login'] is a number. – Artefacto Nov 26 '11 at 17:31
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up vote 2 down vote

You forgot the echo.

<?php echo $_SESSION['login']; ?>

or just

<?=$_SESSION['login']?>
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No semi colon terminating the PHP code in the 2nd example – Tim Nov 26 '11 at 17:29
1  
Avoid using short tags. – dlondero Nov 26 '11 at 17:31
up vote 2 down vote

The problem in your code is that you forgot the echo:

var phpLogin = <?php echo $_SESSION['login'] ?>;
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up vote 1 down vote

Try: var phpLogin = "<?php =$_SESSION['login']; ?>";

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I tried, but, unfortunately, it returns it as a string. Like: "<?php =$_SESSION['login']; ?>", instead of a php variable. – Vitor Santos Nov 26 '11 at 18:12
up vote 1 down vote

It may look like overkill, but for safety reasons you should use json_encode when echo'ing directly to PHP:

var phpLogin = <?=json_encode($_SESSION['login'])?>;

Or, in its longer form:

var phpLogin = <?php echo json_encode($_SESSION['login']); ?>;

Or, to also accept empty sessions:

var phpLogin = <?php echo @json_encode($_SESSION['login']); ?>;

This will simply output NULL when there's no login session.

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