https://stackoverflow.com/questions/34517538/setting-an-es6-class-getter-to-enumerablees6
1. 我要 getter 沒有要暴露this
class A { get age() { return 11; } } console.log(Object.keys(new A())); // []
翻譯 es5es5
var A = /** @class */ (function () { function A() { } Object.defineProperty(A.prototype, "age", { get: function () { return 11; }, enumerable: true, configurable: true }); return A; }()); console.log(Object.keys(new A())); // []
因爲定義到了 prototype 上,因此 object keys 就找不到了。spa
2. 我要 getter 要暴露prototype
class A { constructor() { Object.defineProperty(this, 'age', { get: function () { return 11; }, enumerable: true }) } readonly age: number; // 爲了智能提示 } console.log(Object.keys(new A())); // ['age']
直接定義到了對象上邊而不是 prototype 翻譯
3. 我要 getter setter 不要暴露所有code
by default 是會出現 private _age 的對象
class A { private _age: number; get age() { return this._age; } set age(value: number) { this._age = value; } } const a = new A(); a.age = 11 console.log(Object.keys(a)); // ['_age']
添加 define blog
class A { constructor() { Object.defineProperty(this, '_age', { enumerable: false, writable: true }); } private _age: number; get age() { return this._age; } set age(value: number) { this._age = value; } } const a = new A(); a.age = 15; console.log(Object.keys(a)); console.log(a.age);
4. 我要 getter setter 暴露但不要暴露 private get
class A { constructor() { Object.defineProperty(this, '_age', { enumerable: false, writable: true }); Object.defineProperty(this, 'age', { enumerable: true, get: function () { return this._age; }, set: function (value: number) { this._age = value; } }); } private _age: number; age: number; } const a = new A(); a.age = 15; console.log(Object.keys(a)); console.log(a.age);