Access PHP var from external javascript file

PhpJavascriptVariablesExternal

Php Problem Overview


I can access a PHP var with Javascript like this:

<?php
    $fruit = "apple";
    $color = "red";
?>

<script type="text/javascript">
    alert("fruit: " + "<?php echo $fruit; ?>"); // or shortcut "<?= $fruit ?>"
</script>

But what if I want to use an external JS file:

<script type="text/javascript" src="externaljs.js"></script>

externaljs.js:

alert("color: " + "<?php echo $color; ?>");

Php Solutions


Solution 1 - Php

You don't really access it, you insert it into the javascript code when you serve the page.

However if your other javascript isn't from an external source you can do something like:

<?php
    $color = "Red";
?>
<script type="text/javascript">var color = "<?= $color ?>";</script>
<script type="text/javascript" src="file.js"></script>

and then in the file.js use color like so:

alert("color: " + color);

Solution 2 - Php

You can also access data from php script in Javascript (I'll use jQuery here) like this

Create input hidden field within you php file like this

<input type="hidden" id="myPhpValue" value="<?php echo $myPhpValue ?>" />

in your javascript file:

var myPhpValue = $("#myPhpValue").val();
//From here you can the whaterver you like with you js Value
if(myPhpValue != ''){
//Do something here
}

This will do the job as well :)

Solution 3 - Php

What I've seen done is let .js files run through the php interpreter. Which I can not recommend.

What I do recommend is fetching the values through AJAX and have the PHP file return the value to the JS file. Which is a much cleaner method.

Solution 4 - Php

First of all you have to understand that no program can actually have access to the other program's variable.

When you realize that, the rest is simple.
You can either set up a js variable in the main file and then include your external js, or make this external js dynamic, generated by PHP as well

Solution 5 - Php

What you likely want, is called Asynchronous JavaScript and XML (AJAX): http://www.w3schools.com/ajax/default.aspa

Basically, imagine being able to send messages from the clients JavaScript to your PHP scripts on the server. In the example you gave (externaljs.js), you would have the script ask the server what $color is, via HTTP. You can also point the script tag at a PHP script that generates the JavaScript you want. It depends on what you need to do.

It helps to have some understanding of taint checking, data verification, and security ;)

Solution 6 - Php

As the others are saying, javascript doesn't have access to php variables. However, it does have access to the DOM. So, you can use php to add attributes to some page element. And then you can access those attributes with javascript.

e.g. <div id='apple' class='red'> is completely available to javascript

Solution 7 - Php

Don solution is good, furthermore if you want to use a php array in an external javascipt this can help you:

PHP:

<?php
    $my_php_array = [];
?>     

HTML:

<script type="text/javascript"> var my_js_array = <?php echo json_encode($my_php_array);?> ; </script>
<script src = "../public/js/my-external-js.js"></script>

Javasript: (You can now use the array like a normal Javascript array)

 my_js_array[0] 
 my_js_array.length

Solution 8 - Php

externaljs.js is a static file. Of course it can't access PHP data. The only way to pass PHP data to a js file would be to physically alter the file by writing to it in your PHP script, although this is a messy solution at best.

Edit in response to Ólafur Waage's answer: I guess writing to the js file isn't the only way. Passing the js through the PHP interpreter never crossed my mind (for good reason).

Solution 9 - Php

<script type="text/javascript" src="externaljs.js"></script>

You could change it to

<script type="text/javascript" src="externaljs.php"></script>

And the PHP script could just write JavaScript like that :

<?php
$fruit = "apple";
echo 'var fruit = '.json_encode($fruit);
...

Though using AJAX like said Sepehr Lajevardi would be much cleaner

Solution 10 - Php

2017-2018 and above solution:

Since nobody bringed it up yet and I guess no one thought of combining the functions base64_encode and json_encode yet, you could even send PHP Array variables like that:

> index.php

<?php
      $string = "hello";
      $array = ['hi', 'how', 'are', 'you'];
      $array = base64_encode(json_encode($array));

Then you could just load your desired js file with the parameter for a query string like this:

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

Then js/main.php will look like this for example. You can test your variables this way:

> js/main.php

    <?php
    if ($_GET['string']) {
        $a = $_GET['string'];
    }
    if ($_GET['array']) {
        $b = $_GET['array'];
    }
    $b = json_decode(base64_decode($b));

    echo 'alert("String $a: + '.$a.'");';
    echo 'alert("First key of Array $array: + '.$b[0].'");';
    exit();
    ?>

The following will then output when you open your index.php. So you see, you don't open js/main.php and you still got the javascript functionality from it.

enter image description here

Solution 11 - Php

You cant do that and dont try to as this is not a recommended approach, However you can pass php variables as a function parameters to function written in external js

Solution 12 - Php

You can include() them just as you would anything else:

<?php
    $fruit = "apple";
    $color = "red";
?>

<script type="text/javascript">
    <?php include('/path/to/your/externaljs.js'); ?>
</script>

This will basically render the external file as inline js. The main disadvantage here is that you lose the potential performance benefit of browser caching. On the other hand, it's much easier than re-declaring your php variables in javascript.

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QuestionFFishView Question on Stackoverflow
Solution 1 - PhpDonView Answer on Stackoverflow
Solution 2 - PhpAtanas KovachevView Answer on Stackoverflow
Solution 3 - PhpÓlafur WaageView Answer on Stackoverflow
Solution 4 - PhpYour Common SenseView Answer on Stackoverflow
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