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

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

其他回答

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*/}
}

我的用例是将默认值合并到配置中。如果我的组件接受一个具有深度嵌套结构的配置对象,并且我的组件定义了默认配置,那么我希望在配置中为未提供的所有配置选项设置默认值。

使用示例:

export default MyComponent = ({config}) => {
  const mergedConfig = mergeDefaults(config, {header:{margins:{left:10, top: 10}}});
  // Component code here
}

这允许我传递一个空配置或空配置,或一个部分配置,并让所有未配置的值回落到它们的默认值。

我的mergeDefaults实现如下所示:

export default function mergeDefaults(config, defaults) {
  if (config === null || config === undefined) return defaults;
  for (var attrname in defaults) {
    if (defaults[attrname].constructor === Object) config[attrname] = mergeDefaults(config[attrname], defaults[attrname]);
    else if (config[attrname] === undefined) config[attrname] = defaults[attrname];
  }
  return config;
}


这些是单元测试

import '@testing-library/jest-dom/extend-expect';
import mergeDefaults from './mergeDefaults';

describe('mergeDefaults', () => {
  it('should create configuration', () => {
    const config = mergeDefaults(null, { a: 10, b: { c: 'default1', d: 'default2' } });
    expect(config.a).toStrictEqual(10);
    expect(config.b.c).toStrictEqual('default1');
    expect(config.b.d).toStrictEqual('default2');
  });
  it('should fill configuration', () => {
    const config = mergeDefaults({}, { a: 10, b: { c: 'default1', d: 'default2' } });
    expect(config.a).toStrictEqual(10);
    expect(config.b.c).toStrictEqual('default1');
    expect(config.b.d).toStrictEqual('default2');
  });
  it('should not overwrite configuration', () => {
    const config = mergeDefaults({ a: 12, b: { c: 'config1', d: 'config2' } }, { a: 10, b: { c: 'default1', d: 'default2' } });
    expect(config.a).toStrictEqual(12);
    expect(config.b.c).toStrictEqual('config1');
    expect(config.b.d).toStrictEqual('config2');
  });
  it('should merge configuration', () => {
    const config = mergeDefaults({ a: 12, b: { d: 'config2' } }, { a: 10, b: { c: 'default1', d: 'default2' }, e: 15 });
    expect(config.a).toStrictEqual(12);
    expect(config.b.c).toStrictEqual('default1');
    expect(config.b.d).toStrictEqual('config2');
    expect(config.e).toStrictEqual(15);
  });
});

有一个lodash包专门处理对象的深度克隆。这样做的好处是不需要包含整个lodash库。

它叫lodash.clonedeep

在nodejs中,这种用法是这样的

var cloneDeep = require('lodash.clonedeep');
 
const newObject = cloneDeep(oldObject);

在ReactJS中,用法是

import cloneDeep from 'lodash/cloneDeep';

const newObject = cloneDeep(oldObject);

查看这里的文档。如果您对它的工作原理感兴趣,请查看这里的源文件

我使用lodash:

import _ = require('lodash');
value = _.merge(value1, value2);

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。