To get derived class name from Base class method

Question

I'm trying to get the name of derived type from base class method.I have classes as below:

export class Base{   
    public Add(value: any) {

    }
}

class Derived extends Base{
    public Add(value: any) {
        Base.prototype.Add.call(this, value);
    } 
}

I tried using $.type(this) inside Add() of Base class which gives type as object.But while debugging code type of this is shown as "Object (Derived)".

Is there any way to get this??

Answer 1

You need to use the name property of constructor, like so:

class Base {   
    public Add(value: any) {
        alert(this.constructor["name"]); // alerts 'Derived' in this example
    }
}

class Derived extends Base {
    public Add(value: any) {
        super.Add(value); // use super to call methods on the base class
    }
}

(new Derived).Add('someValue');

Answer 2

First, you've not defined inheritance in your snippet. Use extends to create class inheritance in Typescript.

class Derived extends Base {
..

Second - to access base class you can use super keyword:

public Add(value: any) {
   super.Add.call(this, value);
} 

Answer 3

Each Object has a constructor function and you can look at its name property.

class Alpha { }
var a = new Alpha();
alert((<any>a).constructor.name);  // Alpha

link

You'll note I had to cast a to any. The constructor property isn't defined in lib.d.ts and I assume that's because its use is discouraged. Minification will also change the result. I would only use this for debugging purposes.