如何遍历对象原型链中的所有属性?

问题描述

我有一个es6-class实例，我需要获取其所有属性(以及继承的属性).有没有一种方法可以遍历原型链?

I have an es6-class instance and I need to get all its properties (and inherited properties too). Is there a way to do this without traversing prototype chain?

class A {
  get a() {
    return 123;
  }
}

class B extends A {
  get b() {
    return 456;
  }
}

const b = new B();

for (let prop in b) {
  console.log(prop); //nothing
}

console.log(Object.keys(b)); //empty array
console.log(Object.getOwnPropertyNames(b)); //empty array
console.log(Reflect.ownKeys(b)); //empty array
console.log(Object.keys(Object.getPrototypeOf(b))); //empty array
console.log(Object.getOwnPropertyNames(Object.getPrototypeOf(b))); //["contructor", "b"] -- without "a"
console.log(Reflect.ownKeys(Object.getPrototypeOf(b))); //["contructor", "b"] -- without "a"

推荐答案

如果它们是不可枚举的，则不是，就像您的b属性一样.要枚举不可枚举的属性(！)，必须使用getOwnPropertyNames(和getOwnPropertySymbols)，要包含继承的属性，必须遍历原型链.

Not if they're non-enumerable, as your b property is. To enumerate non-enumerable properties (!), you have to use getOwnPropertyNames (and getOwnPropertySymbols), and to include inherited ones, you have to loop through the prototype chain.

这不是问题:

class A {
  get a() {
    return 123;
  }
}

class B extends A {
  get b() {
    return 456;
  }
}

const b = new B();

let allNames = new Set();
for (let o = b; o != Object.prototype; o = Object.getPrototypeOf(o)) {
  for (let name of Object.getOwnPropertyNames(o)) {
    allNames.add(name);
  }
}
console.log(Array.from(allNames));

请注意，我假设您想跳过Object.prototype上的那些内容，例如toString，hasOwnProperty等.如果要包含这些内容，请将循环条件更改为o != null(如果您喜欢这种情况，只需将o更改为).

Note that I assumed you wanted to skip the ones on Object.prototype like toString, hasOwnProperty, and such. If you want to include those, change the loop condition to o != null (or just o if you like that sort of thing).

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