Passing PHP JSON to [removed] echo json_encode vs echo json declaration

Question

I\'m trying to create a common constants file to share between php and javascript, using JSON to store the constants. But I\'m wondering why pass the JSON from PHP to javasc

Answer 1

In your case $json_obj is already a string. So it is not necessary. But if you have an array you want to pass to javascript json_encode will help you with this.

Answer 2

Usually this is what I do, the safest way I've found:

// holds variables from PHP
var stuff = {};
try {
    // stuff will always be an object
    stuff = JSON.parse('<?php echo empty($stuff) ? '{}' : json_encode($stuff) ?>');
} catch (e) {
    if (e instanceof SyntaxError)
    {
        // report syntax error
        console.error("Cannot parse JSON", e);
    }
}
// show resulting object in console
console.log("stuff:", stuff);

Answer 3

Passing PHP JSON to Javascript and reading
var stuff = <?php print json_encode($datajson); ?>; var arr = new Array(); arr= JSON.parse(stuff); document.write((arr[0].cust_code );

Answer 4

json_encode is a function that converts a PHP array into a JSON string, and nothing more. Since your $json_obj variable is already a JSON string, no further conversion is needed and you can simply echo it out.

To get to your $json_obj string from an array your code would have looked like this

$json_array = array(
    "const1" => "val",
    "const2" => "val2"
);

$json_obj = json_encode($json_array);

Answer 5

Even though it may seem like overkill for your particular problem, I would go for the json_encode/parse option still. Why? you ask. Well, think of it as avoiding duplication. If you encode/parse you can keep the constants in an object easily readable by you PHP-code. And the same for your JS code.

It simply eliminates the need to fiddle with it.

Answer 6

The argument to json_encode() should be a PHP data structure, not a string that's already in JSON format. You use this when you want to pass a PHP object to Javascript.