Use php var in javascript file fault

Question

I have a question about using a php variable in my javascript file. This is my index.php file:

<body>
    <div class="container">
        <div class="row">
            <div class="span12">
                <?php
                    if(isset($_GET['appid'])) {
                        // GET appid
                        $appid = $_GET['appid'];

                        $json_url  ='http://api.url.com/api/gateway/call/1.4/getApp?appid=' . $appid;

                        $ch = curl_init($json_url);
                        curl_setopt($ch, CURLOPT_RETURNTRANSFER, true);

                        $str = curl_exec($ch);
                        curl_close($ch);

                        $data = json_decode($str);
                        $array = json_encode($data);
                    }  

                    else{

                    }    
                ?>
                <p id="errors" class="text-error"></p>
            </div>
        </div>

        <hr>

    </div> <!-- /container -->

    <script src="//ajax.googleapis.com/ajax/libs/jquery/1.9.1/jquery.min.js"></script>
    <script>window.jQuery || document.write('<script src="js/vendor/jquery-1.9.1.min.js"><\/script>')</script>

    <script src="js/vendor/bootstrap.min.js"></script>

    <script src="js/plugins.js"></script>
    <script src="js/main.js.php"></script>
</body>

As you can see I check if an appid is sent. When I received it I load data from an api. In the bottom of the page I have a link to my javascript file. (.php because I want to use the php var $ array in my js file)

My main.js.php:

$(document).ready(function() {
    var data = <?php echo $array;?>;
    alert(data);
});

But I got always error in console:

Uncaught SyntaxError: Unexpected token <

Does anyone know what I do wrong?

is it the full source of `main.js.php`? the code you pasted here you sets `$array` in your `index.php` and tries use it in your `main.js.php` ? — t.niese, May 23 '13 at 08:03
The `$array` variable from your `index` file is'nt available in your `main.js.php` file. — adeneo, May 23 '13 at 08:03

score 2 · Accepted Answer · edited May 23 '17 at 11:53

You are creating the array in a completely different file! The two variables are not in the same scope. What's more, the Javascript file is apparently not interpreted as PHP (and neither should it). So:

Javascript complains about the <? tag, which it should never see.
Even if you solved that, it won't work since there's no PHP $array variable in main.js.php.

Start by understanding how Javascript and PHP are interpreted, see Reference: Why does the PHP code in my Javascript not work?.

PLPeeters · Answer 2 · 2013-05-23T09:02:45.023

Your variable is in a different scope since you're not using the PHP include function. Here's the easiest way I know to achieve what you want:

Rename your JavaScript file to main.js

Since you cannot use PHP in a .js file, declare your JavaScript variable before you call your script, like this:

<script type="text/javascript">
    var array = '<?php echo $array ?>';
</script>
<script src="js/main.js"></script>

Then, in your main.js file, just replace the code you posted by this:

$(document).ready(function() {
    var data = array;
    alert(data);
});

score 1 · Answer 3 · answered May 23 '13 at 07:59

1

Try this,

$(document).ready(function() {
    var data = '<?php echo isset($array) ? $array : 
                    json_encode(array("nothing in array data"));?>';
    // if $array not set then it should return {"0":"nothing in array data"}
    alert(data);
});

answered May 23 '13 at 07:59

Rohan Kumar

40,431
11
76
106

I still get the error: Uncaught SyntaxError: Unexpected token – nielsv May 23 '13 at 08:02
Remove the `curl code` from your page then check if error comes then there no error in `javascript`. – Rohan Kumar May 23 '13 at 08:19

score -2 · Answer 4 · answered May 23 '13 at 08:11

-2

you forgot to Quote Value of var data use this

var data = '<?php echo $array;?>';

answered May 23 '13 at 08:11

Shushant

1,625
1
13
23

Use php var in javascript file fault

4 Answers4