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
这是一个廉价的深度合并，使用尽可能少的代码我能想到。当前一个属性存在时，每个源都会覆盖它。
const { keys } = Object;
const isObject = a => typeof a === "object" && !Array.isArray(a);
const merge = (a, b) =>
isObject(a) && isObject(b)
? deepMerge(a, b)
: isObject(a) && !isObject(b)
? a
: b;
const coalesceByKey = source => (acc, key) =>
(acc[key] && source[key]
? (acc[key] = merge(acc[key], source[key]))
: (acc[key] = source[key])) && acc;
/**
* Merge all sources into the target
* overwriting primitive values in the the accumulated target as we go (if they already exist)
* @param {*} target
* @param {...any} sources
*/
const deepMerge = (target, ...sources) =>
sources.reduce(
(acc, source) => keys(source).reduce(coalesceByKey(source), acc),
target
);
console.log(deepMerge({ a: 1 }, { a: 2 }));
console.log(deepMerge({ a: 1 }, { a: { b: 2 } }));
console.log(deepMerge({ a: { b: 2 } }, { a: 1 }));
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
}
我使用下面的短函数进行深度合并对象。 这对我来说很有效。 作者在这里完全解释了它是如何工作的。
/*!
* Merge two or more objects together.
* (c) 2017 Chris Ferdinandi, MIT License, https://gomakethings.com
* @param {Boolean} deep If true, do a deep (or recursive) merge [optional]
* @param {Object} objects The objects to merge together
* @returns {Object} Merged values of defaults and options
*
* Use the function as follows:
* let shallowMerge = extend(obj1, obj2);
* let deepMerge = extend(true, obj1, obj2)
*/
var extend = function () {
// Variables
var extended = {};
var deep = false;
var i = 0;
// Check if a deep merge
if ( Object.prototype.toString.call( arguments[0] ) === '[object Boolean]' ) {
deep = arguments[0];
i++;
}
// Merge the object into the extended object
var merge = function (obj) {
for (var prop in obj) {
if (obj.hasOwnProperty(prop)) {
// If property is an object, merge properties
if (deep && Object.prototype.toString.call(obj[prop]) === '[object Object]') {
extended[prop] = extend(extended[prop], obj[prop]);
} else {
extended[prop] = obj[prop];
}
}
}
};
// Loop through each object and conduct a merge
for (; i < arguments.length; i++) {
merge(arguments[i]);
}
return extended;
};
另一个使用递归的变体，希望你觉得有用。
const merge = (obj1, obj2) => {
const recursiveMerge = (obj, entries) => {
for (const [key, value] of entries) {
if (typeof value === "object") {
obj[key] = obj[key] ? {...obj[key]} : {};
recursiveMerge(obj[key], Object.entries(value))
else {
obj[key] = value;
}
}
return obj;
}
return recursiveMerge(obj1, Object.entries(obj2))
}
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>
