javascript从多个对象继承

发布于 2024-10-12 23:39:05 字数 419 浏览 10 评论 0原文

我不太熟悉 javascript 继承，我试图让一个对象从另一个对象继承，并定义它自己的方法：

function Foo() {}
Foo.prototype = {
    getColor: function () {return this.color;},
};
function FooB() {}
FooB.prototype = new Foo();
FooB.prototype = {
    /* other methods here */
};

var x = new FooB().getColor();

但是，第二个对象覆盖第一个对象(FooB.prototype = new Foo( ) 被取消)。有什么办法可以解决这个问题，还是我走错了方向？

预先感谢，对于任何不好的术语表示抱歉。

原文

I'm not very well aquainted with javascript inheritance, and I'm trying to make one object inherit from another, and define its own methods:

function Foo() {}
Foo.prototype = {
    getColor: function () {return this.color;},
};
function FooB() {}
FooB.prototype = new Foo();
FooB.prototype = {
    /* other methods here */
};

var x = new FooB().getColor();

However, the second one overwrites the first one(FooB.prototype = new Foo() is cancelled out). Is there any way to fix this problem, or am I going in the wrong direction?

Thanks in advance, sorry for any bad terminology.

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时常饿 2024-10-19 23:39:06

每个对象只能有一个原型，所以如果你想在继承（复制）它之后添加原型，你必须扩展它而不是分配一个新的原型。例子：

function Foo() {}

Foo.prototype = {
    x: function(){ alert('x'); },
    y: function(){ alert('y'); }
};

function Foo2() {}

Foo2.prototype = new Foo();
Foo2.prototype.z = function() { alert('z'); };

var a = new Foo();
a.x();
a.y();
var b = new Foo2();
b.x();
b.y();
b.z();

Each object can only have one prototype, so if you want to add to the prototype after inheriting (copying) it, you have to expand it instead of assigning a new prototype. Example:

function Foo() {}

Foo.prototype = {
    x: function(){ alert('x'); },
    y: function(){ alert('y'); }
};

function Foo2() {}

Foo2.prototype = new Foo();
Foo2.prototype.z = function() { alert('z'); };

var a = new Foo();
a.x();
a.y();
var b = new Foo2();
b.x();
b.y();
b.z();

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昨迟人 2024-10-19 23:39:06

一种解决方案是：

function FooB() {}
var p = new Foo();
p.methodA = function(){...}
p.methodB = function(){...}
p.methodC = function(){...}
...

FooB.prototype = p;

更新：关于使用现有对象进行扩展。您始终可以将一个对象的现有属性复制到另一个对象：

FooB.prototype = new Foo();
var proto = {
     /*...*/
};

for(var prop in proto) {
    FooB.prototype[prop] = proto[prop];
}

只要 proto 是一个“普通”对象（即不从另一个对象继承）就可以了。否则，您可能需要添加 if(proto.hasOwnProperty(prop)) 来仅添加非继承属性。

One solution would be:

function FooB() {}
var p = new Foo();
p.methodA = function(){...}
p.methodB = function(){...}
p.methodC = function(){...}
...

FooB.prototype = p;

Update: Regarding expanding with an existing object. You can always copy the existing properties of one object to another one:

FooB.prototype = new Foo();
var proto = {
     /*...*/
};

for(var prop in proto) {
    FooB.prototype[prop] = proto[prop];
}

As long as proto is a "plain" object (i.e. that does not inherit from another object) it is fine. Otherwise you might want to add if(proto.hasOwnProperty(prop)) to only add non-inherited properties.

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剪不断理还乱 2024-10-19 23:39:06

您可以使用 extend 函数将新成员复制到原型对象。

function FooB() {}
FooB.prototype = new FooA();

extend(FooB.prototype, {
    /* other methods here */
});

延长

/**
 * Copies members from an object to another object.
 * @param {Object} target the object to be copied onto
 * @param {Object} source the object to copy from
 * @param {Boolean} deep  whether the copy is deep or shallow
 */
function extend(target, source, deep) {
    for (var i in source) {
        if (deep || Object.hasOwnProperty.call(source, i)) {
            target[i] = source[i];
        }
    }
    return target;
}

You can use an extend function which copies the new members to the prototype object.

function FooB() {}
FooB.prototype = new FooA();

extend(FooB.prototype, {
    /* other methods here */
});

extend

/**
 * Copies members from an object to another object.
 * @param {Object} target the object to be copied onto
 * @param {Object} source the object to copy from
 * @param {Boolean} deep  whether the copy is deep or shallow
 */
function extend(target, source, deep) {
    for (var i in source) {
        if (deep || Object.hasOwnProperty.call(source, i)) {
            target[i] = source[i];
        }
    }
    return target;
}

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