这里，直走;

一个简单的解决方案，工作像Object。仅赋值deep，适用于数组，无需任何修改。

function deepAssign(target, ...sources) { for (source of sources) { for (let k in source) { let vs = source[k], vt = target[k] if (Object(vs) == vs && Object(vt) === vt) { target[k] = deepAssign(vt, vs) continue } target[k] = source[k] } } return target } x = { a: { a: 1 }, b: [1,2] } y = { a: { b: 1 }, b: [3] } z = { c: 3, b: [,,,4] } x = deepAssign(x, y, z) console.log(JSON.stringify(x) === JSON.stringify({ "a": { "a": 1, "b": 1 }, "b": [ 1, 2, null, 4 ], "c": 3 }))

编辑: 我在别的地方回答过一种深度比较两个对象的新方法。 该方法也可以用于深度合并。如果你想要植入，请留言 https://stackoverflow.com/a/71177790/1919821

与减少

export const merge = (objFrom, objTo) => Object.keys(objFrom)
    .reduce(
        (merged, key) => {
            merged[key] = objFrom[key] instanceof Object && !Array.isArray(objFrom[key])
                ? merge(objFrom[key], merged[key] ?? {})
                : objFrom[key]
            return merged
        }, { ...objTo }
    )

test('merge', async () => {
    const obj1 = { par1: -1, par2: { par2_1: -21, par2_5: -25 }, arr: [0,1,2] }
    const obj2 = { par1: 1, par2: { par2_1: 21 }, par3: 3, arr: [3,4,5] }
    const obj3 = merge3(obj1, obj2)
    expect(obj3).toEqual(
        { par1: -1, par2: { par2_1: -21, par2_5: -25 }, par3: 3, arr: [0,1,2] }
    )
})

// 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,
        },
      });
    });
  });

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。

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}}

用例:合并默认配置

如果我们以以下形式定义配置:

const defaultConf = {
    prop1: 'config1',
    prop2: 'config2'
}

我们可以这样定义更具体的配置:

const moreSpecificConf = {
    ...defaultConf,
    prop3: 'config3'
}

但是如果这些配置包含嵌套结构，这种方法就不再适用了。

因此，我写了一个函数，它只合并{key: value，…}并替换其余的。

const isObject = (val) => val === Object(val);

const merge = (...objects) =>
    objects.reduce(
        (obj1, obj2) => ({
            ...obj1,
            ...obj2,
            ...Object.keys(obj2)
                .filter((key) => key in obj1 && isObject(obj1[key]) && isObject(obj2[key]))
                .map((key) => ({[key]: merge(obj1[key], obj2[key])}))
                .reduce((n1, n2) => ({...n1, ...n2}), {})
        }),
        {}
    );