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I have some JSON which has been passed from a servlet and is stored in xmlhttp.responseText. I want to decompose this JSON so that i can have values of data, size, style, name, etc. Also I wish to have the widget value to be in separate variable.

Here is the JSON:

{
  "widget vlaue=2": {
    "debug": "on",
    "window": {
      "title": "Sample Konfabulator Widget",
      "name": "main_window",
      "width": 500,
      "height": 500
    },
  },
  "image": { 
    "src": "Images/Sun.png",
    "name": "sun1",
    "hOffset": 250,
    "vOffset": 250,
    "alignment": "center"
  },
  "text": {
    "data": "Click Here",
    "size": 36,
    "style": "bold",
    "name": "text1",
    "hOffset": 250,
    "vOffset": 100,
    "alignment": "center",
    "onMouseUp": "sun1.opacity = (sun1.opacity / 100) * 90;"
  }
}

I have tried this:

obj = JSON.parse(xmlhttp.responseText);

but this failed. I could not find anything related to it online. Can anyone please help me with this?

share|improve this question
2  
What do you mean by it "failed"? Did you get any errors from the browser? – Vivin Paliath Nov 8 '11 at 1:34
is it something to do with vlaue=2? – dibs Nov 8 '11 at 1:39
the output is "undefined" for this – typedefcoder2 Nov 8 '11 at 1:40
1  
Lol, when somebody says "this might sound really dumb, but..." that normally is not directed at the person they are speaking to. It usually means "this might sound obvious, but why don't you check anyway". There is no disrespect towards the person it is being said to. – Strelok Nov 8 '11 at 2:15
1  
@typedef1 dude, he wasn't calling you dumb, relax. Fact is we make silly mistakes or typo sometimes, Strelok was just reminding you that is one thing that you can check. Its just a gentle reminder/hint, I would not be offended. – Gapton Nov 8 '11 at 3:02
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2 Answers

up vote 4 down vote accepted

From json.org :

To convert a JSON text into an object, you can use the eval() function. eval() invokes the JavaScript compiler. Since JSON is a proper subset of JavaScript, the compiler will correctly parse the text and produce an object structure. The text must be wrapped in parens to avoid tripping on an ambiguity in JavaScript's syntax.

var myObject = eval('(' + myJSONtext + ')');

However JSON.parse is still recommended:

The eval function is very fast. However, it can compile and execute any JavaScript program, so there can be security issues. The use of eval is indicated when the source is trusted and competent. It is much safer to use a JSON parser. In web applications over XMLHttpRequest, communication is permitted only to the same origin that provide that page, so it is trusted. But it might not be competent. If the server is not rigorous in its JSON encoding, or if it does not scrupulously validate all of its inputs, then it could deliver invalid JSON text that could be carrying dangerous script. The eval function would execute the script, unleashing its malice.

To defend against this, a JSON parser should be used. A JSON parser will recognize only JSON text, rejecting all scripts. In browsers that provide native JSON support, JSON parsers are also much faster than eval. It is expected that native JSON support will be included in the next ECMAScript standard.

var myObject = JSON.parse(myJSONtext, reviver);

Perhaps there is something wrong with your JSON, visit jsonlint.com for a free web-base JSON validator.

share|improve this answer
My JSON has been verifies from JSONLint. I have been doing that before i cam to Javascript. – typedefcoder2 Nov 8 '11 at 1:49
You said it was from a servlet, are you 100% sure that the string was not auto-escaped or modified? I was once using PHP to pass a JSON string and as you may know, the double quotes messes everything up because it wasn't properly escaped in PHP. Since every language has different escape characters, the expected output from the servlet might have been changed. Try printing out the JSON string on the server side to see if it is EXACTLY as you have posted above. – Gapton Nov 8 '11 at 1:55
Well the one above is a similar kind of example. I can not post the exact JSON due to confidentiality. though my JSON has been verifies from JSONLint. – typedefcoder2 Nov 8 '11 at 1:59
In your first post, is the JSON copy-and-pasted from the OUTPUT of the servlet, or from some sourcecode that you wrote? I really suspect that the JSON has been unknowingly modified during execution. – Gapton Nov 8 '11 at 2:09
up vote 4 down vote

Your JSON, that you provided is INVALID. Line 9 contains an extra , that shouldn't be there.

Here is a JSFiddle that works without the ,.

http://jsfiddle.net/ApDsP/

share|improve this answer
Good eye, indeed I believe that to be the error. Now I learned that I should not 100% rely on JSON validator since not all validators check for trailing comma. (I guess this is because not all JSON libraries forbid it either) – Gapton Nov 8 '11 at 3:06
I don't know what JSON the OP was pasting into JSlint, but the JSON in the question doesn't validate in JSlint either. – Strelok Nov 8 '11 at 3:14
Actually, I DID copy and pasted his JSON into JSONLint too, and first time I did that, it worked. Now I just tried again 5 seconds ago, and it FAILED (of course, due to the comma in line 9). I think the OP or someone edited the JSON he posted after my first attempt of validating it in JSONLint. – Gapton Nov 8 '11 at 5:21

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