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)

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}), {})
        }),
        {}
    );

(本机解决方案)如果你知道你想要深度合并的属性，那么

const x = { a: { a: 1 } }
const y = { a: { b: 1 } }
Object.assign(y.a, x.a);
Object.assign(x, y);
// output: a: {b: 1, a: 1}

我试着写一个对象。基于Object的pollyfill的assignDeep。在mdn上赋值。

(ES5)

Object.assignDeep = function (target, varArgs) { // .length of function is 2 'use strict'; if (target == null) { // TypeError if undefined or null throw new TypeError('Cannot convert undefined or null to object'); } var to = Object(target); for (var index = 1; index < arguments.length; index++) { var nextSource = arguments[index]; if (nextSource != null) { // Skip over if undefined or null for (var nextKey in nextSource) { // Avoid bugs when hasOwnProperty is shadowed if (Object.prototype.hasOwnProperty.call(nextSource, nextKey)) { if (typeof to[nextKey] === 'object' && to[nextKey] && typeof nextSource[nextKey] === 'object' && nextSource[nextKey]) { Object.assignDeep(to[nextKey], nextSource[nextKey]); } else { to[nextKey] = nextSource[nextKey]; } } } } } return to; }; console.log(Object.assignDeep({},{a:{b:{c:1,d:1}}},{a:{b:{c:2,e:2}}}))

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*/}
}

如果你正在使用ImmutableJS，你可以使用mergeDeep:

fromJS(options).mergeDeep(options2).toJS();