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
我们可以使用$.extend(true,object1,object2)进行深度合并。值true表示递归合并两个对象，修改第一个对象。
美元扩展(真的,目标,对象)
其他回答
这里是@Salakar的答案的一个不可变(不修改输入)版本。如果你在做函数式编程，这很有用。
export function isObject(item) {
return (item && typeof item === 'object' && !Array.isArray(item));
}
export default function mergeDeep(target, source) {
let output = Object.assign({}, target);
if (isObject(target) && isObject(source)) {
Object.keys(source).forEach(key => {
if (isObject(source[key])) {
if (!(key in target))
Object.assign(output, { [key]: source[key] });
else
output[key] = mergeDeep(target[key], source[key]);
} else {
Object.assign(output, { [key]: source[key] });
}
});
}
return output;
}
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
}
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;
}
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}}
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*/}
}
