How to get the class prototype on the base of its name?

Question

This is my attempt:

// Environment detection (Browser or Node.js)
const isBrowser = new Function('try { return this === window } catch(e){ return false}')
// or
const isNode = new Function('try { return this === global }catch(e){ return false }')

// Get the window or global object
const getRoot = function(){
  if (isBrowser()){
    return (new Function('return window'))()
  }
  else{
    return (new Function('return global'))()
  }
}

// This is my class
class A {
  constructor(name,age){
    this.name = name
    this.age = age
    this.typeName = this.constructor.name
  }

  print(){
    return `${this.name}, ${this.age}`
  }
}

// Some JSON object (for example it is an DTO-object
// that I got from server or client). Each object has
// typeName property.
const obj = {name: 'Bob', age: 30, typeName: 'A'}

// I need to cast JSON-object to necessary type.
// I expected to get the class prototype on the base of its name

const root = getRoot() // window or global

// get class A
const _class = root[obj.typeName] // Oops... It is undefined

Object.setPrototypeOf(obj,_class.prototype)

console.log(obj.print())

UPD

I solved the problem:

const _proto = eval(`${obj.typeName}.prototype`);
Object.setPrototypeOf(obj,_proto)

You don't want to get the "constructor", you want to get the `A` object, which essentially means you want to use variable variables. See the duplicate. The sanest way is to use a mapping: `let map = {A: A}; new map[obj.typeName]();` — deceze, Dec 21 '17 at 09:35
Sorry. But then another approach: window.A = class A { constructor(name,age){ this.name = name this.age = age this.typeName = this.constructor.name } print(){ return `${this.name}, ${this.age}` } }; var s = new A('rest', '1'); console.log(window[s.typeName]); — Oleksandr Poshtaruk, Dec 21 '17 at 09:41

How to get the class prototype on the base of its name?

0 Answers0