PHP variable in javascript "Invalid or unexpected token"

Question

I'm having a problem trying to use a PHP variable within JavaScript. I keep getting the following message.

"Invalid or unexpected token. Message: Undefined variable: example."

I'm unsure why example is being undefined, as it is defined within the php code. Here is my code:

<?php
    $example = '2';
?>
<script type="text/javascript">
    var php_var = "<?php echo json_encode($example); ?>";
</script>

Does anyone have any suggestions? I have also tried the following javascript that results in the same problem:

<script type="text/javascript">
    var php_var = "<?php echo $example; ?>";
</script>

you missed semi colon after $example = '2' that might be problem. — Ranjit Shinde
– Ranjit Shinde, Commented Jan 10, 2017 at 11:43
@SoniyaReddy I think you missed what he is trying to do. $example should be output as var php_var = "2"; — AndFisher
– AndFisher, Commented Jan 10, 2017 at 11:46
I deleted my answer because I realized I had read (skimmed) too fast. — TecBrat
– TecBrat, Commented Jan 10, 2017 at 11:52
json_encode will add double quotes, making var php_var = ""2""; Having said that var php_var = "<?php echo $example; ?>"; worked as expected for me. — AndFisher
– AndFisher, Commented Jan 10, 2017 at 11:53

Monty · Accepted Answer · 2017-01-10 11:55:48Z

1

Firstly, your original code has a syntax error: $example = '2' needs a semicolon. Secondly, the next piece of code is just assigning the string <?php echo $example; ?> to the JavaScript variable php_var where the $example PHP variable is first substituted. The $example variable should be initiated properly first, however, for this to work.

As a separate note: JS cannot execute PHP directly -- only a PHP server can do so. What you're most likely trying to do is this:

<?php
    $example = '2';
?>
<script type="text/javascript">
    var php_var = '<?php echo $example ;?>';
</script>

edited Jan 10, 2017 at 11:55

answered Jan 10, 2017 at 11:50

Monty

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3 Comments

Jake Over a year ago

Thank you for the help! I added this in my footer.php and it worked. For some strange reason, $example when created on my body page is not going to the footer.

nclsvh Over a year ago

I'm confused as to what is different in your answer and the question? I cannot see any difference except ' being changed to ". But I doubt that is the actual solution... Both single and double quotes don't work for me. Why could that be? UPDATE: they actually do print the php var, but because of the error, the rest of the javascript is broken.

Nguai al Over a year ago

this is assigning php_var to a literal string. This didn't work for me. i am using chrome.

user3569377 · Accepted Answer · 2017-01-10 11:47:20Z

1

This should work, use single quotes

<?php
  $example = '2';
?>
  <script type="text/javascript">
      var php_var = '<?php echo  $example; ?>';
  </script>

answered Jan 10, 2017 at 11:47

user3569377

213 bronze badges

Comments

Abhay Maurya · Accepted Answer · 2017-01-10 11:57:47Z

0

Tried and working both variant:

<?php
    $example = '2';
?>
<script type="text/javascript">
    var php_var = <?php echo json_encode($example); ?>;
    console.log(php_var);
</script>

<script type="text/javascript">
    var php_va = "<?php echo $example; ?>";
    console.log(php_va);
</script>

answered Jan 10, 2017 at 11:57

Abhay Maurya

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