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

var x = {'e': 2, 'd': 8, 'b': 5};

const y = {};
for(let key in x) {
    y[key] = x[key];
}
console.log(y); // =>>> {e: 2, d: 8, b: 5}

const z = {};
Object.keys(x).forEach(key => {
    z[key] = x[key];
});
console.log(z); // =>>> {e: 2, d: 8, b: 5}

const w = {};
for(let i = 0; i < Object.keys(x).length; i++) {
    w[Object.keys(x)[i]] = x[Object.keys(x)[i]];
}
console.log(w); // =>>> {e: 2, d: 8, b: 5}

const v = {};
for(let key of Object.keys(x)) {
    v[key] = x[key];
}
console.log(v); // =>>> {e: 2, d: 8, b: 5}

x['q'] = 100;   // Altering x will not affect the other objects

console.log(x); // =>>> {e: 2, d: 8, b: 5, q: 100}
console.log(y); // =>>> {e: 2, d: 8, b: 5}
console.log(z); // =>>> {e: 2, d: 8, b: 5}
console.log(w); // =>>> {e: 2, d: 8, b: 5}
console.log(v); // =>>> {e: 2, d: 8, b: 5}

其他回答

在JavaScript中复制对象的方法

使用排列(…)语法使用Object.assign()方法使用JSON.stringify()和JSON.parse()方法

const person = {
    firstName: 'John',
    lastName: 'Doe'
};

// using spread ...
let p1 = {
    ...person
};

// using  Object.assign() method
let p2 = Object.assign({}, person);

// using JSON
let p3 = JSON.parse(JSON.stringify(person));

Jan Turo的上述答案非常接近,由于兼容性问题,可能是在浏览器中使用的最佳选择,但这可能会导致一些奇怪的枚举问题。例如,执行:

for ( var i in someArray ) { ... }

在遍历数组元素后,将clone()方法赋给i。下面是一个避免枚举并适用于node.js的改编:

Object.defineProperty( Object.prototype, "clone", {
    value: function() {
        if ( this.cloneNode )
        {
            return this.cloneNode( true );
        }

        var copy = this instanceof Array ? [] : {};
        for( var attr in this )
        {
            if ( typeof this[ attr ] == "function" || this[ attr ] == null || !this[ attr ].clone )
            {
                copy[ attr ] = this[ attr ];
            }
            else if ( this[ attr ] == this )
            {
                copy[ attr ] = copy;
            }
            else
            {
                copy[ attr ] = this[ attr ].clone();
            }
        }
        return copy;
    }
});

Object.defineProperty( Date.prototype, "clone", {
    value: function() {
        var copy = new Date();
        copy.setTime( this.getTime() );
        return copy;
    }
});

Object.defineProperty( Number.prototype, "clone", { value: function() { return this; } } );
Object.defineProperty( Boolean.prototype, "clone", { value: function() { return this; } } );
Object.defineProperty( String.prototype, "clone", { value: function() { return this; } } );

这避免了使clone()方法可枚举,因为defineProperty()默认为false。

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。

function clone(obj) {
    if(obj == null || typeof(obj) != 'object')
        return obj;    
    var temp = new obj.constructor(); 
    for(var key in obj)
        temp[key] = clone(obj[key]);    
    return temp;
}