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

这里的大多数示例似乎太复杂了,我使用的是我创建的TypeScript中的一个,我认为它应该涵盖大多数情况(我将数组作为常规数据处理,只是替换它们)。

const isObject = (item: any) => typeof item === 'object' && !Array.isArray(item);

export const merge = <A = Object, B = Object>(target: A, source: B): A & B => {
  const isDeep = (prop: string) =>
    isObject(source[prop]) && target.hasOwnProperty(prop) && isObject(target[prop]);
  const replaced = Object.getOwnPropertyNames(source)
    .map(prop => ({ [prop]: isDeep(prop) ? merge(target[prop], source[prop]) : source[prop] }))
    .reduce((a, b) => ({ ...a, ...b }), {});

  return {
    ...(target as Object),
    ...(replaced as Object)
  } as A & B;
};

在纯JS中也是如此,以防万一:

const isObject = item => typeof item === 'object' && !Array.isArray(item);

const merge = (target, source) => {
  const isDeep = prop => 
    isObject(source[prop]) && target.hasOwnProperty(prop) && isObject(target[prop]);
  const replaced = Object.getOwnPropertyNames(source)
    .map(prop => ({ [prop]: isDeep(prop) ? merge(target[prop], source[prop]) : source[prop] }))
    .reduce((a, b) => ({ ...a, ...b }), {});

  return {
    ...target,
    ...replaced
  };
};

下面是我的测试用例,向您展示如何使用它

describe('merge', () => {
  context('shallow merges', () => {
    it('merges objects', () => {
      const a = { a: 'discard' };
      const b = { a: 'test' };
      expect(merge(a, b)).to.deep.equal({ a: 'test' });
    });
    it('extends objects', () => {
      const a = { a: 'test' };
      const b = { b: 'test' };
      expect(merge(a, b)).to.deep.equal({ a: 'test', b: 'test' });
    });
    it('extends a property with an object', () => {
      const a = { a: 'test' };
      const b = { b: { c: 'test' } };
      expect(merge(a, b)).to.deep.equal({ a: 'test', b: { c: 'test' } });
    });
    it('replaces a property with an object', () => {
      const a = { b: 'whatever', a: 'test' };
      const b = { b: { c: 'test' } };
      expect(merge(a, b)).to.deep.equal({ a: 'test', b: { c: 'test' } });
    });
  });

  context('deep merges', () => {
    it('merges objects', () => {
      const a = { test: { a: 'discard', b: 'test' }  };
      const b = { test: { a: 'test' } } ;
      expect(merge(a, b)).to.deep.equal({ test: { a: 'test', b: 'test' } });
    });
    it('extends objects', () => {
      const a = { test: { a: 'test' } };
      const b = { test: { b: 'test' } };
      expect(merge(a, b)).to.deep.equal({ test: { a: 'test', b: 'test' } });
    });
    it('extends a property with an object', () => {
      const a = { test: { a: 'test' } };
      const b = { test: { b: { c: 'test' } } };
      expect(merge(a, b)).to.deep.equal({ test: { a: 'test', b: { c: 'test' } } });
    });
    it('replaces a property with an object', () => {
      const a = { test: { b: 'whatever', a: 'test' } };
      const b = { test: { b: { c: 'test' } } };
      expect(merge(a, b)).to.deep.equal({ test: { a: 'test', b: { c: 'test' } } });
    });
  });
});

如果你认为我缺少一些功能,请告诉我。

其他回答

许多答案使用数十行代码,或者需要向项目添加一个新库,但如果您使用递归,这只是4行代码。

函数合并(当前,更新){ for (Object.keys(updates)的key) { if (!current. hasownproperty (key) || typeof updates[key] !== 'object') current[key] = updates[key]; Else merge(current[key], updates[key]); } 返回当前; } console.log(合并({答:{:1}},{:{b: 1}}));

数组处理:上面的版本用新值覆盖旧的数组值。如果你想保留旧的数组值并添加新的,只需在else语句上方添加一个else If (current[key] instanceof array && updates[key] instanceof array) current[key] = current[key].concat(updates[key])块,你就都设置好了。

2022年更新:

我创建mergician是为了满足评论中讨论的各种合并/克隆需求。它基于与我最初的答案相同的概念(如下),但提供了可配置的选项:

Unlike native methods and other merge/clone utilities, Mergician provides advanced options for customizing the merge/clone process. These options make it easy to inspect, filter, and modify keys and properties; merge or skip unique, common, and universal keys (i.e., intersections, unions, and differences); and merge, sort, and remove duplicates from arrays. Property accessors and descriptors are also handled properly, ensuring that getter/setter functions are retained and descriptor values are defined on new merged/cloned objects.

值得注意的是,mergician比lodash等类似工具要小得多(1.5k min+gzip)。合并(5.1k min+gzip)。

GitHub: https://github.com/jhildenbiddle/mergician NPM: https://www.npmjs.com/package/mergician 文档:https://jhildenbiddle.github.io/mergician/


最初的回答:

由于这个问题仍然存在,这里有另一种方法:

ES6/2015 不可变(不修改原始对象) 处理数组(连接它们)

/** * Performs a deep merge of objects and returns new object. Does not modify * objects (immutable) and merges arrays via concatenation. * * @param {...object} objects - Objects to merge * @returns {object} New object with merged key/values */ function mergeDeep(...objects) { const isObject = obj => obj && typeof obj === 'object'; return objects.reduce((prev, obj) => { Object.keys(obj).forEach(key => { const pVal = prev[key]; const oVal = obj[key]; if (Array.isArray(pVal) && Array.isArray(oVal)) { prev[key] = pVal.concat(...oVal); } else if (isObject(pVal) && isObject(oVal)) { prev[key] = mergeDeep(pVal, oVal); } else { prev[key] = oVal; } }); return prev; }, {}); } // Test objects const obj1 = { a: 1, b: 1, c: { x: 1, y: 1 }, d: [ 1, 1 ] } const obj2 = { b: 2, c: { y: 2, z: 2 }, d: [ 2, 2 ], e: 2 } const obj3 = mergeDeep(obj1, obj2); // Out console.log(obj3);

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

我发现只有2行解决方案得到深度合并在javascript。一定要告诉我你的结果。

const obj1 = { a: { b: "c", x: "y" } }
const obj2 = { a: { b: "d", e: "f" } }
temp = Object.assign({}, obj1, obj2)
Object.keys(temp).forEach(key => {
    temp[key] = (typeof temp[key] === 'object') ? Object.assign(temp[key], obj1[key], obj2[key]) : temp[key])
}
console.log(temp)

临时对象将打印{a: {b: 'd', e: 'f', x: 'y'}}

https://lodash.com/docs/4.17.15#defaultsDeep

注意:此方法会使源发生突变。

_.defaultsDeep({ 'a': { 'b': 2 } }, { 'a': { 'b': 1, 'c': 3 } });
// => { 'a': { 'b': 2, 'c': 3 } }