我有一个对象x。我想将其复制为对象y,这样对y的更改不会修改x。我意识到复制从内置JavaScript对象派生的对象会导致额外的、不需要的财产。这不是问题,因为我正在复制我自己的一个文字构造对象。

如何正确克隆JavaScript对象?


当前回答

我最喜欢的优雅JS对象克隆解决方案是

function CloneObject() {}
function cloneObject(o) {
   CloneObject.prototype = o;
   return new CloneObject();
}

使用cloneObject(object)获取JS对象的克隆。

与许多复制解决方案不同,此克隆在克隆对象中保持原型关系。

其他回答

使用lodash_.cloneDeep()。

浅拷贝:lodash_.clone()

只需复制参考即可进行浅层复制。

let obj1 = {
    a: 0,
    b: {
        c: 0,
        e: {
            f: 0
        }
    }
};
let obj3 = _.clone(obj1);
obj1.a = 4;
obj1.b.c = 4;
obj1.b.e.f = 100;

console.log(JSON.stringify(obj1));
//{"a":4,"b":{"c":4,"e":{"f":100}}}

console.log(JSON.stringify(obj3));
//{"a":0,"b":{"c":4,"e":{"f":100}}}

深度复制:lodash_.cloneDeep()

取消引用字段:而不是复制对象的引用

let obj1 = {
    a: 0,
    b: {
        c: 0,
        e: {
            f: 0
        }
    }
};
let obj3 = _.cloneDeep(obj1);
obj1.a = 100;
obj1.b.c = 100;
obj1.b.e.f = 100;

console.log(JSON.stringify(obj1));
{"a":100,"b":{"c":100,"e":{"f":100}}}

console.log(JSON.stringify(obj3));
{"a":0,"b":{"c":0,"e":{"f":0}}}

根据Apple JavaScript编码指南:

// Create an inner object with a variable x whose default
// value is 3.
function innerObj()
{
        this.x = 3;
}
innerObj.prototype.clone = function() {
    var temp = new innerObj();
    for (myvar in this) {
        // this object does not contain any objects, so
        // use the lightweight copy code.
        temp[myvar] = this[myvar];
    }
    return temp;
}

// Create an outer object with a variable y whose default
// value is 77.
function outerObj()
{
        // The outer object contains an inner object.  Allocate it here.
        this.inner = new innerObj();
        this.y = 77;
}
outerObj.prototype.clone = function() {
    var temp = new outerObj();
    for (myvar in this) {
        if (this[myvar].clone) {
            // This variable contains an object with a
            // clone operator.  Call it to create a copy.
            temp[myvar] = this[myvar].clone();
        } else {
            // This variable contains a scalar value,
            // a string value, or an object with no
            // clone function.  Assign it directly.
            temp[myvar] = this[myvar];
        }
    }
    return temp;
}

// Allocate an outer object and assign non-default values to variables in
// both the outer and inner objects.
outer = new outerObj;
outer.inner.x = 4;
outer.y = 16;

// Clone the outer object (which, in turn, clones the inner object).
newouter = outer.clone();

// Verify that both values were copied.
alert('inner x is '+newouter.inner.x); // prints 4
alert('y is '+newouter.y); // prints 16

史蒂夫

如果对象中没有循环依赖关系,我建议使用其他答案之一或jQuery的复制方法,因为它们看起来都很有效。

如果存在循环依赖关系(即,两个子对象彼此链接),那么你就有点糟糕了,因为(从理论角度)没有办法优雅地解决这个问题。

let clone = Object.assign( Object.create( Object.getPrototypeOf(obj)), obj)

ES6解决方案,如果您想要(浅层)克隆类实例而不仅仅是属性对象。

正如此链接所示,使用以下代码:

let clone = Object.create(Object.getPrototypeOf(obj),
 Object.getOwnPropertyDescriptors(obj));