覆盖javascript原型时调用父函数[重复]答案

【问题标题】：Call parent function when overriding javascript prototype [duplicate]覆盖javascript原型时调用父函数[重复]
【发布时间】：2014-11-20 08:22:59
【问题描述】：

我有以下元代码

var Parent = function(){}
Parent.prototype.doSomething = function(){
  console.log("As a parent I did like a parent");
}

var Child = function(){}
Child.prototype = new Parent();
Child.prototype.doSomething = function(){
  console.log("As a child I did like a child");
  //This is where I am not sure what to put
}

我想要的 2 行

As a child I did like a child
As a parent I did like a parent

当然第一个很简单，但我不确定如何/是否可以在父函数被覆盖后调用它。

【问题讨论】：

这里有一个例子：jsfiddle.net/zu097ep4

标签： javascript

【解决方案1】：

您可以通过执行以下操作来保存基本方法：

var Parent = function () {}
Parent.prototype.doSomething = function () {
    alert("As a parent I did like a parent");
}

var Child = function () {}
Child.prototype = new Parent();
Child.prototype.doSomething = (function () {
    // since this is an IIFE - Child.prototype.doSomething will refer to the base
    // implementation. We haven't assigned this new one yet!
    var parent_doSomething = Child.prototype.doSomething;   
    return function () {
        alert("As a child I did like a child");
        parent_doSomething();
    }
})();

var c = new Child();
c.doSomething();

这样做的好处是不必担心父母是谁。尽管您可能应该检查父级是否具有doSomething 方法。

【讨论】：

【解决方案2】：

一种方法是调用Parent.prototype.method.call(this, arg1, arg2, ...)。您可以在HERE 中阅读有关超级调用的更多信息。

var Child = function(){}
Child.prototype = new Parent();

Child.prototype.doSomething = function(){
  console.log("As a child I did like a child");
  Parent.prototype.doSomething.call(this); //with this line
}

【讨论】：