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
下面是TypeScript的实现:
export const mergeObjects = <T extends object = object>(target: T, ...sources: T[]): T => {
if (!sources.length) {
return target;
}
const source = sources.shift();
if (source === undefined) {
return target;
}
if (isMergebleObject(target) && isMergebleObject(source)) {
Object.keys(source).forEach(function(key: string) {
if (isMergebleObject(source[key])) {
if (!target[key]) {
target[key] = {};
}
mergeObjects(target[key], source[key]);
} else {
target[key] = source[key];
}
});
}
return mergeObjects(target, ...sources);
};
const isObject = (item: any): boolean => {
return item !== null && typeof item === 'object';
};
const isMergebleObject = (item): boolean => {
return isObject(item) && !Array.isArray(item);
};
和单元测试:
describe('merge', () => {
it('should merge Objects and all nested Ones', () => {
const obj1 = { a: { a1: 'A1'}, c: 'C', d: {} };
const obj2 = { a: { a2: 'A2'}, b: { b1: 'B1'}, d: null };
const obj3 = { a: { a1: 'A1', a2: 'A2'}, b: { b1: 'B1'}, c: 'C', d: null};
expect(mergeObjects({}, obj1, obj2)).toEqual(obj3);
});
it('should behave like Object.assign on the top level', () => {
const obj1 = { a: { a1: 'A1'}, c: 'C'};
const obj2 = { a: undefined, b: { b1: 'B1'}};
expect(mergeObjects({}, obj1, obj2)).toEqual(Object.assign({}, obj1, obj2));
});
it('should not merge array values, just override', () => {
const obj1 = {a: ['A', 'B']};
const obj2 = {a: ['C'], b: ['D']};
expect(mergeObjects({}, obj1, obj2)).toEqual({a: ['C'], b: ['D']});
});
it('typed merge', () => {
expect(mergeObjects<TestPosition>(new TestPosition(0, 0), new TestPosition(1, 1)))
.toEqual(new TestPosition(1, 1));
});
});
class TestPosition {
constructor(public x: number = 0, public y: number = 0) {/*empty*/}
}
其他回答
简单递归解
使用对象。条目，遍历其中一个对象。如果条目不存在，则添加该条目;如果条目是对象，则递归。
常量x = {a: {a: 1}} Const y = {a: {b: 1}} JSON.parse(JSON.stringify(y)) const mergeIntoZ = (firstObj, secondObj) => { Object.entries (firstObj) .forEach(([key, value]) => { if (secondObj[key] === undefined) { secondObj[key] = value } else if (typeof value === 'object') { mergeIntoZ (firstObj(关键),secondObj[主要]) ｝ }) ｝ mergeIntoZ (x, z) console.log (z)
有人知道深度合并在ES6/ES7规范中存在吗?
对象。赋值文档建议它不做深度克隆。
如果你正在使用ImmutableJS，你可以使用mergeDeep:
fromJS(options).mergeDeep(options2).toJS();
Ramda是一个很好的javascript函数库，它有mergeDeepLeft和mergeDeepRight。这些方法都能解决这个问题。请在这里查看文档:https://ramdajs.com/docs/#mergeDeepLeft
对于问题中的具体例子，我们可以使用:
import { mergeDeepLeft } from 'ramda'
const x = { a: { a: 1 } }
const y = { a: { b: 1 } }
const z = mergeDeepLeft(x, y)) // {"a":{"a":1,"b":1}}
// 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,
},
});
});
});
