Passing JSON to Javascript from PHP using direct output

Question

Without using AJAX, how to correctly write a JSON string to JS from PHP. Example below, some case will break this code. I'm trying to find a perfect way to display json correctly, for both case "a" and "b"

Write the output JS to browser console, you'll see it breaks.

<?php
$var = array(
        'nokey',
        5 => 'xyz',
        'key' => '456',
        //bug
        "apostrophe ' xx" => 'quotes " xx',
        '\'' => "\"",
        'malicious backslash \ ' => 'double \\',
        "line break \n"
    );
$var = json_encode($var);
?>

<script>
    //bug
    var a = JSON.parse('<?php echo $var ?>');
    var b = JSON.parse("<?php echo $var ?>");
</script>

http://sandbox.onlinephpfunctions.com/code/5c860f978ddd6d196b15c58f55db20de34bcf72c

use JSON_HEX_APOS and JSON_HEX_QUOT php.net/manual/en/json.constants.php — bassxzero
– bassxzero, Commented Nov 27, 2017 at 18:46
You don't need JSON.parse(); just embed it as raw JS code. — SLaks
– SLaks, Commented Nov 27, 2017 at 18:47
What is the problem by the way ? Try the outputed JSON into this tool : jsoneditoronline.org — Vincent Decaux
– Vincent Decaux, Commented Nov 27, 2017 at 18:47

Ice76 · Accepted Answer · 2017-11-27 18:48:10Z

2

You dont need to parse a JSON object when it is already JSON.

remove JSON.parse()

<?php
$var = array(
        'nokey',
        5 => 'xyz',
        'key' => '456',
        //bug
        "apostrophe ' xx" => 'quotes " xx',
        '\'' => "\"",
        'malicious backslash \ ' => 'double \\',
        "line break \n"
    );
$var = json_encode($var);
?>

<script>
    //bug
    var a = <?php echo $var ?>;
    var b = <?php echo $var ?>;
</script>

answered Nov 27, 2017 at 18:48

Ice76

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2 Comments

Juan Over a year ago

In case the code is processed as xhtml I have seen as a common practice to surround code in the script tags as commented CDATA (//<![CDATA[ ... //]]>

Thanh Trung Over a year ago

This is correct. It covers all the cases without fail

samanime · Accepted Answer · 2017-11-27 18:47:00Z

1

You need to also addslashes().

$var = addslashes(json_encode($var));

Then, once outputted, it'll be output with it's quotes slashed, allowing it to be parsed properly.

It doesn't matter if you are using single-quotes ('') or double-quotes (""), you can slash all the quotes and it'll be just fine.

answered Nov 27, 2017 at 18:47

samanime

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1 Comment

Thanh Trung Over a year ago

This is correct too. It works if I want a valid JSON string

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