Parse JSON string to JavaScript object with function name

Question

I have a JSON string stored in a data attribute.

{
    "active": true,
    "icons": {
        "activeHeader": "ui-icon-alert"
    },
    "animate": {
        "duration": 1000,
        "always": dMethod
    }
}

And I have a function which named dMethod:

function dMethod() {
  alert("DONE");
}

When I try to parse the string via JSON.parse I get an error said invalid character. I check and the dMethod is defined when the parse method was running and if I removed the ,"always":dMethod part then the parser worked correctly. I can't use quotation marks around the dMethod because then the type will be string type instead of object function.
Any help would be appreciated.
Thanks,
Péter
EDIT:
Thanks you for all the answers. I make some clarification so maybe better you understand the problem. I make a really simple js library to make jqueryui unobstructive:

var juiObjects = ["accordion", "autocomplete", "button", "datepicker", "dialog", "menu", "progressbar", "slider", "spinner", "tabs", "tooltip"];

$(document).ready(function() {
  for (var i = 0; i < juiObjects.length; i++) {
    var attributeName = "data-" + juiObjects[i];
    $("["+ attributeName + "]").each(function () {
      var optionsValue = $(this).attr(attributeName);
      var options = JSON.parse(optionsValue);
      $(this)[juiObjects[i]](options);
    });
  }
});

I had to choice between JSON.parse and eval. But I think eval wouldn't be so good choice. And try to keep the "library" as simple as possible. But it looks like I have sparete the code along the widgets.

mkorman · Accepted Answer · 2013-08-12 10:31:10Z

1

Functions are not valid data types in JSON (see http://en.wikipedia.org/wiki/JSON#Data_types.2C_syntax_and_example).

I think you have to deserialize it as a string, then post-process your object and set "always" to your method.

answered Aug 12, 2013 at 10:31

mkorman

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1 Comment

Péter Over a year ago

Thanks. I made som progress but stuck on some javascript magic. I made another question (try not spam this one) If you intrested: stackoverflow.com/questions/18189367/…

MrCode · Accepted Answer · 2013-08-12 10:28:15Z

1

It can be done by quoting the dMethod, by executing the function on the window object using the [] syntax:

function dMethod() {
  alert("DONE");
}

var json = '{"active":true,"icons":{"activeHeader":"ui-icon-alert"},"animate":{"duration":1000,"always":"dMethod"}}'; // quoted

var obj = JSON.parse(json);
window[obj.animate.always]();

answered Aug 12, 2013 at 10:28

MrCode

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Comments

Krasimir · Accepted Answer · 2013-08-12 10:30:52Z

1

JSON.parse expects a valid JSON string. So, if you want to use it you should quote the dMethod function. Isn't it possible to replace the string "dMethod" with the real function after the parsing.

answered Aug 12, 2013 at 10:30

Krasimir

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Guffa · Accepted Answer · 2013-08-12 10:37:28Z

1

You can't parse that string as JSON, because it's not valid JSON.

You can turn the string into an object by executing it using the eval function, but of course the usual warnings about executing anything dynamically applies. If you don't have full control over what's in the string, it might be possible to use for cross site scripting.

var obj = eval(json);
obj.always();

answered Aug 12, 2013 at 10:37

Guffa

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Putra Ardiansyah · Accepted Answer · 2016-07-27 22:46:47Z

This is how to serialize object with its functions:

   JSON.stringify(YOUR_OBJECT, function (key, value) {
        if (typeof value === 'function') {
            return value.toString();
        }
        return value;
    });

and this is how to deserialize it back:

JSON.parse(YOUR_JSON_STRING, function (key, value) {
     if (value 
         && typeof value === "string" 
         && value.substr(0,8) == "function") {
         var startBody = value.indexOf('{') + 1;
         var endBody = value.lastIndexOf('}');
         var startArgs = value.indexOf('(') + 1;
         var endArgs = value.indexOf(')');

         return new Function(value.substring(startArgs, endArgs)
                           , value.substring(startBody, endBody));
     }
     return value;
 });

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