Javascript 原型(prototype)和访问类的问题

Question

Family = function(name) {
  this._Name = name;
}

Family.prototype = {
  getName: function() {
    return this._Name;
  },
  People: function(num) {
    this._Number = num;
  }
}

Family.People.prototype = {
  clearNumber: function() {
    this._Number = 0;
  }
}

People 是一个嵌套类。它的父类是 Family。

我得到的错误是 Family.People 未定义。有人可以更正上面的代码吗？

Answer 1

工作代码

// function doesn't need "new" operator
var Family = function(name) { this._Name = name; };

Family.prototype = {
    getName: function() { return this._Name; }, // missing comma
    People: function(num) {
        this._Number = num;
    }
};

// work with prototypes
Family.prototype.People.prototype = {
    clearNumber: function() { this._Number = 0; }
};

这会起作用。但是你必须知道，当你打电话时:

var f = new Family("Doe");

f.People 只是一个对象构造函数，而不是某个其他对象的实例。您还必须实例化它:

f.members = new f.People(3);

你的实例中有一个构造函数，这很令人困惑。

更好的方法

因此，如果您以这种方式编写原型(prototype)可能会更好:

var Family = function(name) {
    this._Name = name;
    this.getName = function() { return this._Name; };
};

Family.People = function(num) {
    this._Number = num;
    this.clearNumber = function() { this._Number = 0; };
};

这实际上是在一个类中创建一个类(而不是在实例中)。所以上面的行会这样称呼:

var f = new Family("Doe");
f.members = new Family.People(3);

f 实例的向下钻取如下所示:

f
  _Name
  getName()
  members
    _Number
    clearNumber()

私有(private)变量

var Family = function(name) {
    var _name = name;
    this.getName = function() { return _name; };
};

Family.People = function(num) {
    var _num = num;
    this.getNumber = function() { return _num; }
    this.clearNumber = function() { _num = 0; };
};

通过这种方式，我们将变量设为私有(private)并且只能在内部访问，因此无法在外部对其进行操作。您必须始终使用函数来操作它们。这使得它更加健壮，尤其是当存在与变量值相关的某些业务规则时。

var f = new Family("Doe");
f._name; // this is undefined because "_name" is private closure variable

向下钻取 f 实例现在看起来更像是一个类对象实例:

f
  getName()
  members
    getNumber()
    clearNumber()