Unable to parse JSON object in JS from PHP json_encode

Question

I have an array in PHP, which I pack to JSON object with _json_encode(..)_ . Then I send it to JS function as parameter. When I want to parse the object in Javascript with eval(..) nothing happens (there is an error behind the curtains I guess). What could be wrong?
Code:

<script type="text/javascript">
    function testFun(inArr) {
      var obj=eval('('+inArr+')');
      alert(obj.m); //alert(obj) also doesnt work
    }
</script>  


//PHP
$spola_array = array('m' => 1, 'z' => 2);
$json_obj=json_encode($spola_array);
echo '<script type="text/javascript">testFun('.$json_obj.');</script>';

Corbin · Accepted Answer · 2012-05-27 21:42:30Z

5

It's already parsed since you're outputting it as an object literal and not a string. That will look like:

<script type="text/javascript">testFun({m: 1, z: 2});</script>

So in your function, it's just:

alert(inArr.m) //1

You would only need to parse it if it were a string:

<script type="text/javascript">testFun('{m: 1, z: 2}');</script>

answered May 27, 2012 at 21:42

Corbin

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5 Comments

c0dehunter Over a year ago

BIG thanks for this one. Couldn't figure it out in past hours.

Corbin Over a year ago

No problem. Sometimes it can help to debug things like this with console.log by the way. It typically provides a better formatted output than alerts.

c0dehunter Over a year ago

Where is this console.log? Edit: do you mean Firebug? I've never used it before and maybe I should start :)

Corbin Over a year ago

It's just a function the dev tools of Chrome and IE expose, and a function exposed by Firebug in Firefox. You just use it like console.log("test!"); or console.log(someVar); It's not meant to be used in production code. It's just a slightly more useful way of doing alert(someVar). It typically includes a little bit of information about the type of variable, and in the case of object literals, it will show that they are objects and show which keys are defined.

c0dehunter Over a year ago

Thanks, great piece of advice.

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