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。

这通常不是最有效的解决方案,但它是我需要的。

function clone(obj, clones) {
    // Makes a deep copy of 'obj'. Handles cyclic structures by
    // tracking cloned obj's in the 'clones' parameter. Functions 
    // are included, but not cloned. Functions members are cloned.
    var new_obj,
        already_cloned,
        t = typeof obj,
        i = 0,
        l,
        pair; 

    clones = clones || [];

    if (obj === null) {
        return obj;
    }

    if (t === "object" || t === "function") {

        // check to see if we've already cloned obj
        for (i = 0, l = clones.length; i < l; i++) {
            pair = clones[i];
            if (pair[0] === obj) {
                already_cloned = pair[1];
                break;
            }
        }

        if (already_cloned) {
            return already_cloned; 
        } else {
            if (t === "object") { // create new object
                new_obj = new obj.constructor();
            } else { // Just use functions as is
                new_obj = obj;
            }

            clones.push([obj, new_obj]); // keep track of objects we've cloned

            for (key in obj) { // clone object members
                if (obj.hasOwnProperty(key)) {
                    new_obj[key] = clone(obj[key], clones);
                }
            }
        }
    }
    return new_obj || obj;
}

自行车测试...

a = []
a.push("b", "c", a)
aa = clone(a)
aa === a //=> false
aa[2] === a //=> false
aa[2] === a[2] //=> false
aa[2] === aa //=> true

功能测试...

f = new Function
f.a = a
ff = clone(f)
ff === f //=> true
ff.a === a //=> false

我知道这是一个古老的帖子,但我以为这可能有助于下一个人,谁在跳动。

只要你不分配一个对象到任何东西,它保留没有参考在记忆中,所以要创建一个对象,你想在其他对象中分享,你将不得不创建一个工厂如下:

var a = function(){
    return {
        father:'zacharias'
    };
},
b = a(),
c = a();
c.father = 'johndoe';
alert(b.father);

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;
};

使用 Object.create() 获取原型和支持的例子,并使用 for() 圈获取可列的密钥:

function cloneObject(source) {
    var key,value;
    var clone = Object.create(source);

    for (key in source) {
        if (source.hasOwnProperty(key) === true) {
            value = source[key];

            if (value!==null && typeof value==="object") {
                clone[key] = cloneObject(value);
            } else {
                clone[key] = value;
            }
        }
    }

    return clone;
}

我的场景略有不同,我有一個物體,擁有<unk>的物體以及功能,因此,Object.assign() 和 JSON.stringify() 並沒有解決我的問題。

因此,我决定做一个简单的功能,使用内置的方法来复制一个对象,其字面特性,其粘着的对象和功能。

let deepCopy = (target, source) => {
    Object.assign(target, source);
    // check if there's any nested objects
    Object.keys(source).forEach((prop) => {
        /**
          * assign function copies functions and
          * literals (int, strings, etc...)
          * except for objects and arrays, so:
          */
        if (typeof(source[prop]) === 'object') {
            // check if the item is, in fact, an array
            if (Array.isArray(source[prop])) {
                // clear the copied referenece of nested array
                target[prop] = Array();
                // iterate array's item and copy over
                source[prop].forEach((item, index) => {
                    // array's items could be objects too!
                    if (typeof(item) === 'object') {
                        // clear the copied referenece of nested objects
                        target[prop][index] = Object();
                        // and re do the process for nested objects
                        deepCopy(target[prop][index], item);
                    } else {
                        target[prop].push(item);
                    }
                });
            // otherwise, treat it as an object
            } else {
                // clear the copied referenece of nested objects
                target[prop] = Object();
                // and re do the process for nested objects
                deepCopy(target[prop], source[prop]);
            }
        }
    });
};

下面是测试代码:

let a = {
    name: 'Human', 
    func: () => {
        console.log('Hi!');
    }, 
    prop: {
        age: 21, 
        info: {
            hasShirt: true, 
            hasHat: false
        }
    },
    mark: [89, 92, { exam: [1, 2, 3] }]
};

let b = Object();

deepCopy(b, a);

a.name = 'Alien';
a.func = () => { console.log('Wassup!'); };
a.prop.age = 1024;
a.prop.info.hasShirt = false;
a.mark[0] = 87;
a.mark[1] = 91;
a.mark[2].exam = [4, 5, 6];

console.log(a); // updated props
console.log(b);

对于效率相关的担忧,我认为这是我遇到的问题的最简单和最有效的解决方案,我会欣赏任何关于这个算法的评论,这可能会使它更有效。