JavaScript - 获取作为特定类的实例的所有对象

Question

I am not sure it is possible but I will try my luck.

Is it possible in JavaScript to find all objects that are instances of a specific class?

For example:

var obj1 = new MyClass();
var obj2 = new MyClass();
var obj3 = new MyClass();

I want to specify the class name "MyClass" and to get "obj1,obj2,obj3" in response .

How can I do that?

Answer 1

You can do it like this:

 function MyClass(){}; MyClass.prototype.test = function(){ console.log('TEST'); }; function MyClassFactory(){ this.instances = []; }; MyClassFactory.prototype.create = function(){ let tmp = new MyClass(); this.instances.push(tmp); return tmp; }; MyClassFactory.prototype.get = function(i){ return this.instances[i]; }; MyClassFactory.prototype.getAll = function(){ return this.instances; }; let factory = new MyClassFactory(); let obj1 = factory.create(); let obj2 = factory.create(); let obj3 = factory.create(); let test1 = factory.get(0); let test2 = factory.getAll(); for(let t of test2){ t.test(); } test1.test();

Answer 2

As far as I know, this doesn't work. You could save your objects in an array and iterate over it to retrieve all objects of class MyClass .

var objects = [new MyClass(), new MyClass(), new MyClass()];
function getObjectsOfClass(objects, clazz) {
   var objArr = [];
   for (var i = 0; i < objects.length; i++) {
       if (objects[i] instanceof clazz) {
           objArr.push(objects[i];
       }
   }
   return objArr;
}

Answer 3

Imagine this situation.

function M(){}
var m = new M();

Then you van do something like

for(let f in window){ if(window[f] instanceof M){console.log("founded"+f);}}

And you will find your object.

Hope that helps

Answer 4

You can only if you made some modification to the constructor

example:

var MyClassInstances = [];
var MyClass = function() { //constructor
   MyClassInstances.push(this); //push the new instance
}

thats it!. But be aware that it might become memory leak. so, you have to remove the reference from MyClassInstances when the objects are not needed.