JSONP parsing in javascript/node.js

Question

If I have an string containing a JSONP response, for example"jsonp([1,2,3])", and I want to retrieve the 3rd parameter 3, how could I write a function that do that for me? I want to avoid using eval. My code (below) works fine on the debug line, but return undefined for some reason.

  function unwrap(jsonp) {
    function unwrapper(param) {
      console.log(param[2]); // This works!
      return param[2];
    }
    var f = new Function("jsonp", jsonp);
    return f(unwrapper);
  }

  var j = 'jsonp([1,2,3]);'

  console.log(unwrap(j)); // Return undefined

More info: I'm running this in a node.js scraper, using request library.

Here's a jsfiddle https://jsfiddle.net/bortao/3nc967wd/

to answer why it returns undefined ... because 'jsonp([1,2,3]);' should be 'return jsonp([1,2,3]);' - you need to return values from functions if you want functions to return values — Jaromanda X
– Jaromanda X, Commented Jul 27, 2018 at 4:47
Further to what @JaromandaX said: var f = new Function("jsonp", "return " + jsonp); — nnnnnn
– nnnnnn, Commented Jul 27, 2018 at 4:50
either or @nnnnnn same result :p ahh, yes, but I know the difference - good pickup — Jaromanda X
– Jaromanda X, Commented Jul 27, 2018 at 4:51
Sure @JaromandaX, but I figured the j variable was standing in for the scraper utility so it would make sense to have the return part within the unwrap() and separate to that string. — nnnnnn
– nnnnnn, Commented Jul 27, 2018 at 4:53
yep, as I said, I see now how that makes more sense even though the end result is identical :p — Jaromanda X
– Jaromanda X, Commented Jul 27, 2018 at 4:53

CertainPerformance · Accepted Answer · 2018-07-27 04:50:29Z

4

Just slice the string to remove the jsonp( and );, and then you can JSON.parse it:

function unwrap(jsonp) {
  return JSON.parse(jsonp.slice(6, jsonp.length - 2));
}

var j = 'jsonp([1,2,3]);'

console.log(unwrap(j)); // returns the whole array
console.log(unwrap(j)[2]); // returns the third item in the array

Note that new Function is just as bad as eval.

edited Jul 27, 2018 at 4:50

answered Jul 27, 2018 at 4:45

CertainPerformance

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1 Comment

Jaromanda X Over a year ago

but op was expecting 3 as the output :p

Hamed Motallebi · Accepted Answer · 2018-07-27 06:50:45Z

1

Just a little changes and it'll work fine:

function unwrap(jsonp) {
    var f = new Function("jsonp", `return ${jsonp}`);
    console.log(f.toString())
    return f(unwrapper);
}

function unwrapper(param) {
    console.log(param[2]); // This works!
    return param[2];
}

var j = 'jsonp([1,2,3]);'
console.log(unwrap(j)); // Return undefined

without return your anonymous function is like this :

function anonymous(jsonp) {
    jsonp([1,2,3]);
}

because this function doesn't return so the output will be undefined.

answered Jul 27, 2018 at 6:50

Hamed Motallebi

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