两个对象。assign和Object spread只做浅合并。

这个问题的一个例子:

// No object nesting
const x = { a: 1 }
const y = { b: 1 }
const z = { ...x, ...y } // { a: 1, b: 1 }

输出是您所期望的。然而,如果我尝试这样做:

// Object nesting
const x = { a: { a: 1 } }
const y = { a: { b: 1 } }
const z = { ...x, ...y } // { a: { b: 1 } }

而不是

{ a: { a: 1, b: 1 } }

你得到

{ a: { b: 1 } }

X被完全覆盖,因为扩展语法只覆盖了一层。这与Object.assign()相同。

有办法做到这一点吗?


当前回答

我用es6做这个方法进行深度赋值。

function isObject(item) {
  return (item && typeof item === 'object' && !Array.isArray(item) && item !== null)
}

function deepAssign(...objs) {
    if (objs.length < 2) {
        throw new Error('Need two or more objects to merge')
    }

    const target = objs[0]
    for (let i = 1; i < objs.length; i++) {
        const source = objs[i]
        Object.keys(source).forEach(prop => {
            const value = source[prop]
            if (isObject(value)) {
                if (target.hasOwnProperty(prop) && isObject(target[prop])) {
                    target[prop] = deepAssign(target[prop], value)
                } else {
                    target[prop] = value
                }
            } else if (Array.isArray(value)) {
                if (target.hasOwnProperty(prop) && Array.isArray(target[prop])) {
                    const targetArray = target[prop]
                    value.forEach((sourceItem, itemIndex) => {
                        if (itemIndex < targetArray.length) {
                            const targetItem = targetArray[itemIndex]

                            if (Object.is(targetItem, sourceItem)) {
                                return
                            }

                            if (isObject(targetItem) && isObject(sourceItem)) {
                                targetArray[itemIndex] = deepAssign(targetItem, sourceItem)
                            } else if (Array.isArray(targetItem) && Array.isArray(sourceItem)) {
                                targetArray[itemIndex] = deepAssign(targetItem, sourceItem)
                            } else {
                                targetArray[itemIndex] = sourceItem
                            }
                        } else {
                            targetArray.push(sourceItem)
                        }
                    })
                } else {
                    target[prop] = value
                }
            } else {
                target[prop] = value
            }
        })
    }

    return target
}

其他回答

有一个lodash包专门处理对象的深度克隆。这样做的好处是不需要包含整个lodash库。

它叫lodash.clonedeep

在nodejs中,这种用法是这样的

var cloneDeep = require('lodash.clonedeep');
 
const newObject = cloneDeep(oldObject);

在ReactJS中,用法是

import cloneDeep from 'lodash/cloneDeep';

const newObject = cloneDeep(oldObject);

查看这里的文档。如果您对它的工作原理感兴趣,请查看这里的源文件

2022年更新:

我创建mergician是为了满足评论中讨论的各种合并/克隆需求。它基于与我最初的答案相同的概念(如下),但提供了可配置的选项:

Unlike native methods and other merge/clone utilities, Mergician provides advanced options for customizing the merge/clone process. These options make it easy to inspect, filter, and modify keys and properties; merge or skip unique, common, and universal keys (i.e., intersections, unions, and differences); and merge, sort, and remove duplicates from arrays. Property accessors and descriptors are also handled properly, ensuring that getter/setter functions are retained and descriptor values are defined on new merged/cloned objects.

值得注意的是,mergician比lodash等类似工具要小得多(1.5k min+gzip)。合并(5.1k min+gzip)。

GitHub: https://github.com/jhildenbiddle/mergician NPM: https://www.npmjs.com/package/mergician 文档:https://jhildenbiddle.github.io/mergician/


最初的回答:

由于这个问题仍然存在,这里有另一种方法:

ES6/2015 不可变(不修改原始对象) 处理数组(连接它们)

/** * Performs a deep merge of objects and returns new object. Does not modify * objects (immutable) and merges arrays via concatenation. * * @param {...object} objects - Objects to merge * @returns {object} New object with merged key/values */ function mergeDeep(...objects) { const isObject = obj => obj && typeof obj === 'object'; return objects.reduce((prev, obj) => { Object.keys(obj).forEach(key => { const pVal = prev[key]; const oVal = obj[key]; if (Array.isArray(pVal) && Array.isArray(oVal)) { prev[key] = pVal.concat(...oVal); } else if (isObject(pVal) && isObject(oVal)) { prev[key] = mergeDeep(pVal, oVal); } else { prev[key] = oVal; } }); return prev; }, {}); } // Test objects const obj1 = { a: 1, b: 1, c: { x: 1, y: 1 }, d: [ 1, 1 ] } const obj2 = { b: 2, c: { y: 2, z: 2 }, d: [ 2, 2 ], e: 2 } const obj3 = mergeDeep(obj1, obj2); // Out console.log(obj3);

你可以使用Lodash合并:

Var对象= { 'a': [{'b': 2}, {'d': 4}] }; Var other = { 'a': [{'c': 3}, {'e': 5}] }; console.log(_。合并(对象,其他)); / / = > {a: [{b: 2,“c”:3},{' d ': 4,“e”:5}]} < script src = " https://cdnjs.cloudflare.com/ajax/libs/lodash.js/4.17.21/lodash.min.js " > < /脚本>

这里,直走;

一个简单的解决方案,工作像Object。仅赋值deep,适用于数组,无需任何修改。

function deepAssign(target, ...sources) { for (source of sources) { for (let k in source) { let vs = source[k], vt = target[k] if (Object(vs) == vs && Object(vt) === vt) { target[k] = deepAssign(vt, vs) continue } target[k] = source[k] } } return target } x = { a: { a: 1 }, b: [1,2] } y = { a: { b: 1 }, b: [3] } z = { c: 3, b: [,,,4] } x = deepAssign(x, y, z) console.log(JSON.stringify(x) === JSON.stringify({ "a": { "a": 1, "b": 1 }, "b": [ 1, 2, null, 4 ], "c": 3 }))

编辑: 我在别的地方回答过一种深度比较两个对象的新方法。 该方法也可以用于深度合并。如果你想要植入,请留言 https://stackoverflow.com/a/71177790/1919821

当涉及到宿主对象或比值包更复杂的任何类型的对象时,这个问题就不那么简单了

do you invoke a getter to obtain a value or do you copy over the property descriptor? what if the merge target has a setter (either own property or in its prototype chain)? Do you consider the value as already-present or call the setter to update the current value? do you invoke own-property functions or copy them over? What if they're bound functions or arrow functions depending on something in their scope chain at the time they were defined? what if it's something like a DOM node? You certainly don't want to treat it as simple object and just deep-merge all its properties over into how to deal with "simple" structures like arrays or maps or sets? Consider them already-present or merge them too? how to deal with non-enumerable own properties? what about new subtrees? Simply assign by reference or deep clone? how to deal with frozen/sealed/non-extensible objects?

另一件需要记住的事情是:包含循环的对象图。这通常不难处理——简单地保留一组已经访问过的源对象——但经常被遗忘。

您可能应该编写一个深度合并函数,它只期望原始值和简单对象(结构化克隆算法最多可以处理的那些类型)作为合并源。如果遇到它不能处理的东西,或者只是通过引用而不是深度合并进行赋值,则抛出。

换句话说,没有一种适合所有人的算法,您要么必须使用自己的算法,要么寻找恰好涵盖您的用例的库方法。