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?

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. — putvande
– putvande, Commented Jan 15, 2014 at 13:51
FYI: That is not JSON. JSON does not have variables in it and it does not call methods. — epascarello
– epascarello, Commented Jan 15, 2014 at 13:51
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. — BenM
– BenM, Commented Jan 15, 2014 at 13:52
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. — Sirko
– Sirko, Commented Jan 15, 2014 at 13:52
JSON is a notation, not an object. What you have in JS above is a JS object. — crush
– crush, Commented Jan 15, 2014 at 13:54

hanzi · Accepted Answer · 2014-01-15 13:52:46Z

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.)

answered Jan 15, 2014 at 13:52

hanzi

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1 Comment

crush Over a year ago

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

Serge Seredenko · Accepted Answer · 2014-01-15 14:13:02Z

0

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.

answered Jan 15, 2014 at 14:13

Serge Seredenko

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