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
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 }));
其他回答
我知道这是一个老问题，但在ES2015/ES6中我能想到的最简单的解决方案实际上很简单，使用Object.assign()，
希望这能有所帮助:
/**
* Simple object check.
* @param item
* @returns {boolean}
*/
export function isObject(item) {
return (item && typeof item === 'object' && !Array.isArray(item));
}
/**
* Deep merge two objects.
* @param target
* @param ...sources
*/
export function mergeDeep(target, ...sources) {
if (!sources.length) return target;
const source = sources.shift();
if (isObject(target) && isObject(source)) {
for (const key in source) {
if (isObject(source[key])) {
if (!target[key]) Object.assign(target, { [key]: {} });
mergeDeep(target[key], source[key]);
} else {
Object.assign(target, { [key]: source[key] });
}
}
}
return mergeDeep(target, ...sources);
}
使用示例:
mergeDeep(this, { a: { b: { c: 123 } } });
// or
const merged = mergeDeep({a: 1}, { b : { c: { d: { e: 12345}}}});
console.dir(merged); // { a: 1, b: { c: { d: [Object] } } }
你将在下面的答案中找到一个不可更改的版本。
注意，这将导致循环引用上的无限递归。这里有一些关于如何检测循环引用的很好的答案，如果你认为你会面临这个问题。
这是我刚刚写的另一个支持数组的程序。它把它们连接起来。
function isObject(obj) {
return obj !== null && typeof obj === 'object';
}
function isPlainObject(obj) {
return isObject(obj) && (
obj.constructor === Object // obj = {}
|| obj.constructor === undefined // obj = Object.create(null)
);
}
function mergeDeep(target, ...sources) {
if (!sources.length) return target;
const source = sources.shift();
if(Array.isArray(target)) {
if(Array.isArray(source)) {
target.push(...source);
} else {
target.push(source);
}
} else if(isPlainObject(target)) {
if(isPlainObject(source)) {
for(let key of Object.keys(source)) {
if(!target[key]) {
target[key] = source[key];
} else {
mergeDeep(target[key], source[key]);
}
}
} else {
throw new Error(`Cannot merge object with non-object`);
}
} else {
target = source;
}
return mergeDeep(target, ...sources);
};
// 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,
},
});
});
});
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 }));
当涉及到宿主对象或比值包更复杂的任何类型的对象时，这个问题就不那么简单了
do you invoke a getter to obtain a value or do you copy over the property descriptor? what if the merge target has a setter (either own property or in its prototype chain)? Do you consider the value as already-present or call the setter to update the current value? do you invoke own-property functions or copy them over? What if they're bound functions or arrow functions depending on something in their scope chain at the time they were defined? what if it's something like a DOM node? You certainly don't want to treat it as simple object and just deep-merge all its properties over into how to deal with "simple" structures like arrays or maps or sets? Consider them already-present or merge them too? how to deal with non-enumerable own properties? what about new subtrees? Simply assign by reference or deep clone? how to deal with frozen/sealed/non-extensible objects?
另一件需要记住的事情是:包含循环的对象图。这通常不难处理——简单地保留一组已经访问过的源对象——但经常被遗忘。
您可能应该编写一个深度合并函数，它只期望原始值和简单对象(结构化克隆算法最多可以处理的那些类型)作为合并源。如果遇到它不能处理的东西，或者只是通过引用而不是深度合并进行赋值，则抛出。
换句话说，没有一种适合所有人的算法，您要么必须使用自己的算法，要么寻找恰好涵盖您的用例的库方法。
