Get name of instance of object in JavaScript

Question

I have a code like the one bellow and want to detect the instance name of my exScript.

In this case it would be exScript123.

eecore = {
    something: 1,
    // ...
    someelse: function() { /* whatever */ };
};

var exScript = (function (undefined) {
    function exScript(inputOptions) {
        this.version = "0.0";
    }
    exScript.prototype.init = function () {
        // some code here
    };
    return exScript;
})();

eecore.exScript123 = new exScript();
eecore.exScript123.init();

I have been experimenting for the last hour with arguments.calle.name and this.parent.name but they do not seem to work in my case. I keep getting undefined.

The same object reference can be shared by several variables. In that case, if what you want to do was possible, which variable name should be returned? — Frédéric Hamidi
– Frédéric Hamidi, Commented Aug 9, 2013 at 10:26
As I mentioned in the question. In my case it should return exScript123. I need to get exScript123 inside of the init prototype. In order to set a cookie that holds some data uniquely for this instance so it does not conflict with others. — transilvlad
– transilvlad, Commented Aug 9, 2013 at 10:27
But what if you had eecore.exScript123 = eecore.foo = eecore.bar = new exScript()? Should exScript123, foo or bar be returned (as they all point to the same object)? (That's actually a rhetorical question, what you want to achieve is not possible in the first place.) — Frédéric Hamidi
– Frédéric Hamidi, Commented Aug 9, 2013 at 10:29
The instance doesn't really have any inherent way of knowing what variable(s) it's being referenced by. Why not accept a name argument on the constructor so the instance has its own record of it? — Jonathan Lonowski
– Jonathan Lonowski, Commented Aug 9, 2013 at 10:30

Community · Accepted Answer · 2017-05-23 12:28:25Z

3

Slightly modified version of this code:

function objectName(x, context, path) {

    function search(x, context, path) {
        if(x === context)
            return path;
        if(typeof context != "object" || seen.indexOf(context) >= 0)
            return;
        seen.push(context);
        for(var p in context) {
            var q = search(x, context[p], (path ? path + "." : "") + p);
            if(q)
                return q;
        }
    }

    var seen = [];
    return search(x, context || window, path || "");
}

In your init function

    exScript.prototype.init = function () {
        console.log(objectName(this, eecore))
    };

correctly prints exScript123.

As pointed out in the comments, this is unreliable and a strange idea in general. You might want to clarify why you need that - surely there are better ways.

edited May 23, 2017 at 12:28

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answered Aug 9, 2013 at 10:35

georg

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2 Comments

transilvlad Over a year ago

Not exactly a simple solution but thanks for answering. I have found a better solution for my issue.

transilvlad Over a year ago

I simply set a variable exScriptInstance in window that I increment each time a new instance is instantiated and store the incremented value inside a variable of the instance.

HMR · Accepted Answer · 2013-08-09 10:41:23Z

Not sure why you would have a need to know the/a varialbe name that points to the instance of your object. As stated before you can have several variables with different names point to the same instance.

As mentioned before; if you need a unique id or name then create that as a property of the exScript instance when you create it.

If you want to make sure that init only executes when the this constext is an instance of exScript you can do the following:

var exScript = (function (undefined) {
    function exScript(inputOptions) {
        this.version = "0.0";
    }
    exScript.prototype.init = function () {
      // you can use (this instanceof exScript)
      // that will go up the prototype chain and see
      // if the current object is an instance of or inherits from
      // exScript
      // simple check: see if this is a direct instance of exScript
      if(this.constructor!==exScript){
        console.log("no good");
        return;
      }
      console.log("ok init");
      // some code here
    };
    return exScript;
})();

var e = new exScript();
e.init();//= ok init
setTimeout(e.init,100);//= no good

shmuli · Accepted Answer · 2015-04-06 04:32:30Z

0

Simple code example:

 function Parent(){
        // custom properties
    }

    Parent.prototype.getInstanceName = function(){
        for (var instance in window){
            if (window[instance] === this){
                return instance;
            }
        }
    };

    var child = new Parent();

    console.log(child.getInstanceName()); // outputs: "child"

answered Apr 6, 2015 at 4:32

shmuli

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1 Comment

shmuli Over a year ago

@tntu Note: This implementation works if the instance was created in the root.

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