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
下面的函数对对象进行深度复制，它涵盖了复制原语、数组以及对象
function mergeDeep (target, source) {
if (typeof target == "object" && typeof source == "object") {
for (const key in source) {
if (source[key] === null && (target[key] === undefined || target[key] === null)) {
target[key] = null;
} else if (source[key] instanceof Array) {
if (!target[key]) target[key] = [];
//concatenate arrays
target[key] = target[key].concat(source[key]);
} else if (typeof source[key] == "object") {
if (!target[key]) target[key] = {};
this.mergeDeep(target[key], source[key]);
} else {
target[key] = source[key];
}
}
}
return target;
}
其他回答
我把这里所有的答案都看了一遍，然后拼凑出了一个我自己的答案。现有的大多数答案都不是我想要的方式。
这对于2021年来说是相当可怕的，所以任何改善的建议，我都洗耳恭听!
这是在Typescript中
type Props = Record<string, any>
export const deepMerge = (target: Props, ...sources: Props[]): Props => {
if (!sources.length) {
return target
}
Object.entries(sources.shift() ?? []).forEach(([key, value]) => {
if (!target[key]) {
Object.assign(target, { [key]: {} })
}
if (
value.constructor === Object ||
(value.constructor === Array && value.find(v => v.constructor === Object))
) {
deepMerge(target[key], value)
} else if (value.constructor === Array) {
Object.assign(target, {
[key]: value.find(v => v.constructor === Array)
? target[key].concat(value)
: [...new Set([...target[key], ...value])],
})
} else {
Object.assign(target, { [key]: value })
}
})
return target
}
平面数组使用[…]删除重复值。新的(…)]。
嵌套数组使用concat连接。
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>
// copies all properties from source object to dest object recursively
export function recursivelyMoveProperties(source, dest) {
for (const prop in source) {
if (!source.hasOwnProperty(prop)) {
continue;
}
if (source[prop] === null) {
// property is null
dest[prop] = source[prop];
continue;
}
if (typeof source[prop] === 'object') {
// if property is object let's dive into in
if (Array.isArray(source[prop])) {
dest[prop] = [];
} else {
if (!dest.hasOwnProperty(prop)
|| typeof dest[prop] !== 'object'
|| dest[prop] === null || Array.isArray(dest[prop])
|| !Object.keys(dest[prop]).length) {
dest[prop] = {};
}
}
recursivelyMoveProperties(source[prop], dest[prop]);
continue;
}
// property is simple type: string, number, e.t.c
dest[prop] = source[prop];
}
return dest;
}
单元测试:
describe('recursivelyMoveProperties', () => {
it('should copy properties correctly', () => {
const source: any = {
propS1: 'str1',
propS2: 'str2',
propN1: 1,
propN2: 2,
propA1: [1, 2, 3],
propA2: [],
propB1: true,
propB2: false,
propU1: null,
propU2: null,
propD1: undefined,
propD2: undefined,
propO1: {
subS1: 'sub11',
subS2: 'sub12',
subN1: 11,
subN2: 12,
subA1: [11, 12, 13],
subA2: [],
subB1: false,
subB2: true,
subU1: null,
subU2: null,
subD1: undefined,
subD2: undefined,
},
propO2: {
subS1: 'sub21',
subS2: 'sub22',
subN1: 21,
subN2: 22,
subA1: [21, 22, 23],
subA2: [],
subB1: false,
subB2: true,
subU1: null,
subU2: null,
subD1: undefined,
subD2: undefined,
},
};
let dest: any = {
propS2: 'str2',
propS3: 'str3',
propN2: -2,
propN3: 3,
propA2: [2, 2],
propA3: [3, 2, 1],
propB2: true,
propB3: false,
propU2: 'not null',
propU3: null,
propD2: 'defined',
propD3: undefined,
propO2: {
subS2: 'inv22',
subS3: 'sub23',
subN2: -22,
subN3: 23,
subA2: [5, 5, 5],
subA3: [31, 32, 33],
subB2: false,
subB3: true,
subU2: 'not null --- ',
subU3: null,
subD2: ' not undefined ----',
subD3: undefined,
},
propO3: {
subS1: 'sub31',
subS2: 'sub32',
subN1: 31,
subN2: 32,
subA1: [31, 32, 33],
subA2: [],
subB1: false,
subB2: true,
subU1: null,
subU2: null,
subD1: undefined,
subD2: undefined,
},
};
dest = recursivelyMoveProperties(source, dest);
expect(dest).toEqual({
propS1: 'str1',
propS2: 'str2',
propS3: 'str3',
propN1: 1,
propN2: 2,
propN3: 3,
propA1: [1, 2, 3],
propA2: [],
propA3: [3, 2, 1],
propB1: true,
propB2: false,
propB3: false,
propU1: null,
propU2: null,
propU3: null,
propD1: undefined,
propD2: undefined,
propD3: undefined,
propO1: {
subS1: 'sub11',
subS2: 'sub12',
subN1: 11,
subN2: 12,
subA1: [11, 12, 13],
subA2: [],
subB1: false,
subB2: true,
subU1: null,
subU2: null,
subD1: undefined,
subD2: undefined,
},
propO2: {
subS1: 'sub21',
subS2: 'sub22',
subS3: 'sub23',
subN1: 21,
subN2: 22,
subN3: 23,
subA1: [21, 22, 23],
subA2: [],
subA3: [31, 32, 33],
subB1: false,
subB2: true,
subB3: true,
subU1: null,
subU2: null,
subU3: null,
subD1: undefined,
subD2: undefined,
subD3: undefined,
},
propO3: {
subS1: 'sub31',
subS2: 'sub32',
subN1: 31,
subN2: 32,
subA1: [31, 32, 33],
subA2: [],
subB1: false,
subB2: true,
subU1: null,
subU2: null,
subD1: undefined,
subD2: undefined,
},
});
});
});
(本机解决方案)如果你知道你想要深度合并的属性，那么
const x = { a: { a: 1 } }
const y = { a: { b: 1 } }
Object.assign(y.a, x.a);
Object.assign(x, y);
// output: a: {b: 1, a: 1}
有时候你并不需要深度合并，即使你这样认为。例如，如果您有一个带有嵌套对象的默认配置，并且您希望用自己的配置对其进行深入扩展，您可以为此创建一个类。概念很简单:
function AjaxConfig(config) {
// Default values + config
Object.assign(this, {
method: 'POST',
contentType: 'text/plain'
}, config);
// Default values in nested objects
this.headers = Object.assign({}, this.headers, {
'X-Requested-With': 'custom'
});
}
// Define your config
var config = {
url: 'https://google.com',
headers: {
'x-client-data': 'CI22yQEI'
}
};
// Extend the default values with your own
var fullMergedConfig = new AjaxConfig(config);
// View in DevTools
console.log(fullMergedConfig);
您可以将其转换为函数(而不是构造函数)。
