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
// 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,
},
});
});
});
我不喜欢现有的解决方案。所以，我开始写我自己的。
Object.prototype.merge = function(object) {
for (const key in object) {
if (object.hasOwnProperty(key)) {
if (typeof this[key] === "object" && typeof object[key] === "object") {
this[key].merge(object[key]);
continue;
}
this[key] = object[key];
}
}
return this;
}
我希望这能帮助那些努力理解正在发生的事情的人。我在这里看到了很多无意义的变量。
谢谢
有一些维护良好的库已经做到了这一点。npm注册表中的一个例子是merge-deep
我在加载缓存redux状态时遇到了这个问题。如果我只是加载缓存的状态，我会遇到错误的新应用程序版本与更新的状态结构。
前面已经提到过，lodash提供了merge函数，我使用了这个函数:
const currentInitialState = configureState().getState();
const mergedState = _.merge({}, currentInitialState, cachedState);
const store = configureState(mergedState);
许多答案使用数十行代码，或者需要向项目添加一个新库，但如果您使用递归，这只是4行代码。
函数合并(当前，更新){ for (Object.keys(updates)的key) { if (!current. hasownproperty (key) || typeof updates[key] !== 'object') current[key] = updates[key]; Else merge(current[key]， updates[key]); ｝ 返回当前; ｝ console.log(合并({答:{:1}},{:{b: 1}}));
数组处理:上面的版本用新值覆盖旧的数组值。如果你想保留旧的数组值并添加新的，只需在else语句上方添加一个else If (current[key] instanceof array && updates[key] instanceof array) current[key] = current[key].concat(updates[key])块，你就都设置好了。
