How to prepare PHP variable for Javascript array() use?

Question

Let's say that we have input field or textarea where user can put anything. This means that user can put there:

text
white spaces
multilines
HTML tags
single and double quotes
slashes
and whatnot...

My current code does this: <?php $data = addslashes($content_of_input); ?>

and soon after that...

<?php
    $php_generate_javascriptArray .='
        javascriptArray[0] ="'.$data.'";
    ';
?>
<script>
    javascriptArray = [];
    <?php echo $php_generate_javascriptArray; ?>
</script>

Adding slashes unfortunately isn't enough - Javascript breaks when user puts for instance multiple lines or HTML links into that. Is there any way to prevent that and still ALLOW Javascript array to contain LINK, MULTIPLE LINES, HTML TAGS? I'm looking for some universal filters.

Quentin · Accepted Answer · 2012-09-07 11:30:44Z

5

json_encode will convert a PHP data structure (or string, or number) to the appropriate JavaScript data structure while making it safe for injecting into a script element in an HTML document (by escaping slashes).

<script>
    var data = <?php echo json_encode($data); ?> ;
</script>

edited Sep 7, 2012 at 11:30

answered Sep 7, 2012 at 11:24

Quentin

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5 Comments

Atadj Over a year ago

For some reason - in the above example - adding json_encode() instead of addslashes() breaks Javascript altogether even when using plain text. Perhaps because I'm using double quotes?

Quentin Over a year ago

@Paul — Yes, json_encode will output a literal (string, array, object, etc). You need to output it directly and not wrap it in quotes (which it might contain) to make it a string.

Atadj Over a year ago

I need to use PHP for() to generate very, very long multi-dimensional Javascript array(). The above example is my only choice. In my question I simplified it. Is there any way to do it using my above example?

Quentin Over a year ago

Use for() to generate a very long multi-dimensional PHP array. Then json_encode it.

Atadj Over a year ago

EDIT: Okay, I tested and it so far it is okay.

FIG-GHD742 · Accepted Answer · 2012-09-07 11:28:02Z

1

Use PHP function urlencode(...) or base64_encode(...) of need of more advanced protection.

I normal use urlencode and on Javascript side use unescape for decode the URL format data.

answered Sep 7, 2012 at 11:28

FIG-GHD742

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