How to use a PHP variable in JavaScript?

Question

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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?

var phpLogin = ; 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 http://stackoverflow.com/search?q=[php]+pass+php+variable+to+javascript — Gordon, Nov 26 '11 at 17:42

score 3 · Answer 1 · answered Nov 26 '11 at 17:27

3

Use json_encode and don't forget echo.

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

answered Nov 26 '11 at 17:27

Artefacto

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
@Tim, that's the easiest way to escape string. – Michael Krelin - hacker Nov 26 '11 at 17:31
@Tim: Why would he not want it json encoded. IF hes going to be using it in js it saves him the trouble of trying to manually escape it himself. – prodigitalson Nov 26 '11 at 17:31
My objective is to obtain this variable, to compare it with the content of a XML reply. So, the answer will be returned in XML. Either way, it keeps me sending the following error: `unexpected token '<'`. It happens on every solution presented by you.. Any clue about what's happening? – Vitor Santos Nov 26 '11 at 18:06

score 2 · Answer 2 · answered Nov 26 '11 at 17:27

2

You forgot the echo.

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

or just

<?=$_SESSION['login']?>

answered Nov 26 '11 at 17:27

Simone

No semi colon terminating the PHP code in the 2nd example – Tim Nov 26 '11 at 17:29

score 2 · Answer 3 · answered Nov 26 '11 at 17:28

2

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

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

answered Nov 26 '11 at 17:28

Aurelio De Rosa

score 1 · Answer 4 · answered Nov 26 '11 at 17:28

1

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

answered Nov 26 '11 at 17:28

Tim

I tried, but, unfortunately, it returns it as a string. Like: `""`, instead of a php variable. – Vitor Santos Nov 26 '11 at 18:12

score 1 · Answer 5 · answered Nov 26 '11 at 17:34

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.

5 Answers5