什么是最有效的方式来克隆一个JavaScript对象?我已经看到obj = eval(uneval(o));被使用,但它是非标准的,仅支持Firefox.我做了事情,如obj = JSON.parse(JSON.stringify(o));但质疑效率。


当前回答

以下是 ConroyP 上面的答案的版本,即使制造商需要参数:

//If Object.create isn't already defined, we just do the simple shim,
//without the second argument, since that's all we need here
var object_create = Object.create;
if (typeof object_create !== 'function') {
    object_create = function(o) {
        function F() {}
        F.prototype = o;
        return new F();
    };
}

function deepCopy(obj) {
    if(obj == null || typeof(obj) !== 'object'){
        return obj;
    }
    //make sure the returned object has the same prototype as the original
    var ret = object_create(obj.constructor.prototype);
    for(var key in obj){
        ret[key] = deepCopy(obj[key]);
    }
    return ret;
}

此功能也在我的 simpleoo 图书馆中可用。

编辑:

下面是一个更坚实的版本(感谢Justin McCandless,现在支持自行车参考):

/**
 * Deep copy an object (make copies of all its object properties, sub-properties, etc.)
 * An improved version of http://keithdevens.com/weblog/archive/2007/Jun/07/javascript.clone
 * that doesn't break if the constructor has required parameters
 * 
 * It also borrows some code from http://stackoverflow.com/a/11621004/560114
 */ 
function deepCopy(src, /* INTERNAL */ _visited, _copiesVisited) {
    if(src === null || typeof(src) !== 'object'){
        return src;
    }

    //Honor native/custom clone methods
    if(typeof src.clone == 'function'){
        return src.clone(true);
    }

    //Special cases:
    //Date
    if(src instanceof Date){
        return new Date(src.getTime());
    }
    //RegExp
    if(src instanceof RegExp){
        return new RegExp(src);
    }
    //DOM Element
    if(src.nodeType && typeof src.cloneNode == 'function'){
        return src.cloneNode(true);
    }

    // Initialize the visited objects arrays if needed.
    // This is used to detect cyclic references.
    if (_visited === undefined){
        _visited = [];
        _copiesVisited = [];
    }

    // Check if this object has already been visited
    var i, len = _visited.length;
    for (i = 0; i < len; i++) {
        // If so, get the copy we already made
        if (src === _visited[i]) {
            return _copiesVisited[i];
        }
    }

    //Array
    if (Object.prototype.toString.call(src) == '[object Array]') {
        //[].slice() by itself would soft clone
        var ret = src.slice();

        //add it to the visited array
        _visited.push(src);
        _copiesVisited.push(ret);

        var i = ret.length;
        while (i--) {
            ret[i] = deepCopy(ret[i], _visited, _copiesVisited);
        }
        return ret;
    }

    //If we've reached here, we have a regular object

    //make sure the returned object has the same prototype as the original
    var proto = (Object.getPrototypeOf ? Object.getPrototypeOf(src): src.__proto__);
    if (!proto) {
        proto = src.constructor.prototype; //this line would probably only be reached by very old browsers 
    }
    var dest = object_create(proto);

    //add this object to the visited array
    _visited.push(src);
    _copiesVisited.push(dest);

    for (var key in src) {
        //Note: this does NOT preserve ES5 property attributes like 'writable', 'enumerable', etc.
        //For an example of how this could be modified to do so, see the singleMixin() function
        dest[key] = deepCopy(src[key], _visited, _copiesVisited);
    }
    return dest;
}

//If Object.create isn't already defined, we just do the simple shim,
//without the second argument, since that's all we need here
var object_create = Object.create;
if (typeof object_create !== 'function') {
    object_create = function(o) {
        function F() {}
        F.prototype = o;
        return new F();
    };
}

其他回答

A Recursive Deep Clone 比 JSON.parse(JSON.stringify(obj)) 提到的更快。

Jsperf 在这里排名第一: https://jsperf.com/deep-copy-vs-json-stringify-json-parse/5 Jsben 从上面的答案更新显示,一个重复的深度克隆打击所有其他提到的: http://jsben.ch/13YKQ

下面是快速参考的功能:

function cloneDeep (o) {
  let newO
  let i

  if (typeof o !== 'object') return o

  if (!o) return o

  if (Object.prototype.toString.apply(o) === '[object Array]') {
    newO = []
    for (i = 0; i < o.length; i += 1) {
      newO[i] = cloneDeep(o[i])
    }
    return newO
  }

  newO = {}
  for (i in o) {
    if (o.hasOwnProperty(i)) {
      newO[i] = cloneDeep(o[i])
    }
  }
  return newO
}

对象的克隆一直是JS的担忧,但在ES6之前,我列出了在下面的JavaScript中复制对象的不同方式,想象你有下面的对象,我希望有一个深刻的副本:

var obj = {a:1, b:2, c:3, d:4};

有几种方法可以复制这个对象,而不会改变其起源:

ES5+,使用一个简单的函数来为您进行复制:函数 deepCopyObj(obj) {如果(null == obj <unk> <unk> <unk> <unk> <unk> <unk> <unk> <unk> <unk> <unk> <unk> <unk> <unk> <unk> <unk> <unk> <unk> <unk> <unk> <unk> <unk> <unk> <unk> <unk> <unk>

希望这些帮助......

如何将对象的关键与其价值相结合?

function deepClone(o) {
    var keys = Object.keys(o);
    var values = Object.values(o);

    var clone = {};

    keys.forEach(function(key, i) {
        clone[key] = typeof values[i] == 'object' ? Object.create(values[i]) : values[i];
    });

    return clone;
}

注意: 这种方法不一定会做更深的复制,但它只会用一个内部对象的深度复制,这意味着当你给出像 {a: {b: {c: null}}} 这样的东西时,它只会克隆直接在它们内部的对象,所以 deepClone(a.b.c)技术上是对 a.b.c 的参考,而 deepClone(a.b)则是克隆,而不是参考。

代码:

// extends 'from' object with members from 'to'. If 'to' is null, a deep clone of 'from' is returned
function extend(from, to)
{
    if (from == null || typeof from != "object") return from;
    if (from.constructor != Object && from.constructor != Array) return from;
    if (from.constructor == Date || from.constructor == RegExp || from.constructor == Function ||
        from.constructor == String || from.constructor == Number || from.constructor == Boolean)
        return new from.constructor(from);

    to = to || new from.constructor();

    for (var name in from)
    {
        to[name] = typeof to[name] == "undefined" ? extend(from[name], null) : to[name];
    }

    return to;
}

测试:

var obj =
{
    date: new Date(),
    func: function(q) { return 1 + q; },
    num: 123,
    text: "asdasd",
    array: [1, "asd"],
    regex: new RegExp(/aaa/i),
    subobj:
    {
        num: 234,
        text: "asdsaD"
    }
}

var clone = extend(obj);

当您的对象被粘贴并包含数据对象、其他结构化对象或某种属性对象等时,使用 JSON.parse(JSON.stringify(object)) 或 Object.assign({}, obj) 或 $.extend(true, {}, obj) 不会工作。

var obj = {a: 25, b: {a: 1, b: 2}, c: new Date(), d: anotherNestedObject };
var A = _.cloneDeep(obj);

现在 A 将是您的新的 Obj 克隆,没有任何参考。