如果沒有任何內建一個,你可以嘗試:

function clone(obj) {
    if (obj === null || typeof (obj) !== 'object' || 'isActiveClone' in obj)
        return obj;

    if (obj instanceof Date)
        var temp = new obj.constructor(); //or new Date(obj);
    else
        var temp = obj.constructor();

    for (var key in obj) {
        if (Object.prototype.hasOwnProperty.call(obj, key)) {
            obj['isActiveClone'] = null;
            temp[key] = clone(obj[key]);
            delete obj['isActiveClone'];
        }
    }
    return temp;
}

这里是一个全面的克隆()方法,可以克隆任何JavaScript对象,它处理几乎所有情况:

function clone(src, deep) {

    var toString = Object.prototype.toString;
    if (!src && typeof src != "object") {
        // Any non-object (Boolean, String, Number), null, undefined, NaN
        return src;
    }

    // Honor native/custom clone methods
    if (src.clone && toString.call(src.clone) == "[object Function]") {
        return src.clone(deep);
    }

    // DOM elements
    if (src.nodeType && toString.call(src.cloneNode) == "[object Function]") {
        return src.cloneNode(deep);
    }

    // Date
    if (toString.call(src) == "[object Date]") {
        return new Date(src.getTime());
    }

    // RegExp
    if (toString.call(src) == "[object RegExp]") {
        return new RegExp(src);
    }

    // Function
    if (toString.call(src) == "[object Function]") {

        //Wrap in another method to make sure == is not true;
        //Note: Huge performance issue due to closures, comment this :)
        return (function(){
            src.apply(this, arguments);
        });
    }

    var ret, index;
    //Array
    if (toString.call(src) == "[object Array]") {
        //[].slice(0) would soft clone
        ret = src.slice();
        if (deep) {
            index = ret.length;
            while (index--) {
                ret[index] = clone(ret[index], true);
            }
        }
    }
    //Object
    else {
        ret = src.constructor ? new src.constructor() : {};
        for (var prop in src) {
            ret[prop] = deep
                ? clone(src[prop], true)
                : src[prop];
        }
    }
    return ret;
};

因为回归只是太昂贵的JavaScript,我发现的大多数答案是使用回归,而JSON方法将错过非JSON转换部分(功能等)。所以我做了一些研究,并发现这个拖拉机技术避免它。

/*
 * Trampoline to avoid recursion in JavaScript, see:
 *     https://www.integralist.co.uk/posts/functional-recursive-javascript-programming/
 */
function trampoline() {
    var func = arguments[0];
    var args = [];
    for (var i = 1; i < arguments.length; i++) {
        args[i - 1] = arguments[i];
    }

    var currentBatch = func.apply(this, args);
    var nextBatch = [];

    while (currentBatch && currentBatch.length > 0) {
        currentBatch.forEach(function(eachFunc) {
            var ret = eachFunc();
            if (ret && ret.length > 0) {
                nextBatch = nextBatch.concat(ret);
            }
        });

        currentBatch = nextBatch;
        nextBatch = [];
    }
};

/*
 *  Deep clone an object using the trampoline technique.
 *
 *  @param target {Object} Object to clone
 *  @return {Object} Cloned object.
 */
function clone(target) {
    if (typeof target !== 'object') {
        return target;
    }
    if (target == null || Object.keys(target).length == 0) {
        return target;
    }

    function _clone(b, a) {
        var nextBatch = [];
        for (var key in b) {
            if (typeof b[key] === 'object' && b[key] !== null) {
                if (b[key] instanceof Array) {
                    a[key] = [];
                }
                else {
                    a[key] = {};
                }
                nextBatch.push(_clone.bind(null, b[key], a[key]));
            }
            else {
                a[key] = b[key];
            }
        }
        return nextBatch;
    };

    var ret = target instanceof Array ? [] : {};
    (trampoline.bind(null, _clone))(target, ret);
    return ret;
};

扩展操作器... (原始序列 - 仅) 序列(0) (原始序列 - 仅) 序列() (原始序列 - 仅) concat() (原始序列 - 仅) 定制功能,如下所示(每个序列) jQuery 的 $.extend() (每个序列) JSON.parse(JSON.stringify()) (原始和字面序列 - 仅) Underscore 的 _.clone() (原始和字面序列 - 仅) Lodash 的 _.cloneDeep() (每个序列)

let arr1a = [1, 'a', true];

let arr1b = [...arr1a];

而且在哪裡 slice() 比 concat( 有更好的性能: https://jsbench.me/x5ktn7o94d/

let arr1c = arr1a.splice(0);
let arr1d = arr1a.slice();
let arr1e = arr1a.concat();

let arr2a = [1, 'a', true, {}, []];
let arr2b = JSON.parse(JSON.stringify(arr2a));

let arr3a = [1, 'a', true, {}, [], new Object()];

function copy(aObject) {
  // Prevent undefined objects
  // if (!aObject) return aObject;

  let bObject = Array.isArray(aObject) ? [] : {};

  let value;
  for (const key in aObject) {

    // Prevent self-references to parent object
    // if (Object.is(aObject[key], aObject)) continue;
    
    value = aObject[key];

    bObject[key] = (typeof value === "object") ? copy(value) : value;
  }

  return bObject;
}

let arr3b = copy(arr3a);

或使用第三方实用功能:

let arr3c = $.extend(true, [], arr3a); // jQuery Extend
let arr3d = _.cloneDeep(arr3a); // Lodash

注意: jQuery 的 $.extend 也比 JSON.parse(JSON.stringify() 表现更好):

对于未来的参考,可以使用此代码

第6章:

_clone: function(obj){
    let newObj = {};
    for(let i in obj){
        if(typeof(obj[i]) === 'object' && Object.keys(obj[i]).length){
            newObj[i] = clone(obj[i]);
        } else{
            newObj[i] = obj[i];
        }
    }
    return Object.assign({},newObj);
}

第5章:

function clone(obj){
let newObj = {};
for(let i in obj){
    if(typeof(obj[i]) === 'object' && Object.keys(obj[i]).length){
        newObj[i] = clone(obj[i]);
    } else{
        newObj[i] = obj[i];
    }
}
return Object.assign({},newObj);

) )

E.G

var obj ={a:{b:1,c:3},d:4,e:{f:6}}
var xc = clone(obj);
console.log(obj); //{a:{b:1,c:3},d:4,e:{f:6}}
console.log(xc); //{a:{b:1,c:3},d:4,e:{f:6}}

xc.a.b = 90;
console.log(obj); //{a:{b:1,c:3},d:4,e:{f:6}}
console.log(xc); //{a:{b:90,c:3},d:4,e:{f:6}}

没有触摸原型遗产,你可以如下深孤独的物体和岩石;

函数 objectClone(o){ var ot = Array.isArray(o); return o!== null && typeof o === “object”? Object.keys(o).reduce((r,k) => o[k]!== null && typeof o[k] === “object”? (r[k] = objectClone(o[k]),r) : (r[k] = o[k],r), ot?