JS variable inside JSON string from PHP

Question

I am generating a JSON string from a PHP array to echo a JS object.
This is what I want to get in js:

var myVar = 123; 
//php output: 
var obj = {a:1, b:[1,2], c: myVar, d:Date.UTC(2014, 0, 07)} 

This is what I have:

<?php 
$array = array('a'=>1, 'b'=>array(1,2), 'c'=>???, 'd'=>???); 
echo json_encode($array);
?> 

The question is: What I put in PHP instead of question marks so that it won't be converted to string?

• putvande: so that it won't be converted to string? What type of variable is it? If you do json_encode it will always be parsed as a string I think.
• epascarello: FYI: That is not JSON. JSON does not have variables in it and it does not call methods.
• BenM: JSON should be uncoupled from your application. You shouldn't be placing variable pointers inside of it... It's a data store, not a collection of variables.
• Sirko: You can't do this (at least not with jsoon_encode(). Have a look at json.org definition for values. You could just code your own encoder, but that would not be according to the standard.
• crush: JSON is a notation, not an object. What you have in JS above is a JS object.

Answer 1

JSON doesn't support variables or special Date objects. You can only use scalar values (strings, numbers, booleans), arrays and objects (associative arrays).

A way to get what you want would be to return a .js file and have the browser execute that (by including it as a script) instead of transferring simple JSON data. Otherwise you could only define "special" strings that are handled by the receiving side. (For example, array ["var", "myVar"] could be parsed accordingly.)

• crush: There is a commonly accepted date format for use with JSON, but it's type is still just a string.

Answer 2

You could actually do something like that:

<?php 
 $array = array('a'=>1, 'b'=>array(1,2),
 'c'=>'@#myVar#@',
 'd'=>'@#Date.UTC(2014, 0, 07)#@'
 );
 $json = json_encode($array);
 echo preg_replace('/\"\@\#(.*?)\#\@\"/', '${1}', $json);
?>

But in js JSON.parse won't work, so:

eval("var x = " + json_from_php);

Not such a good idea, but if you need it, it'll work. Just remember not to use this with any "json" which are generated not by your server.