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

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*/}
}

其他回答

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}}

ES5的一个简单解决方案(覆盖现有值):

function merge(current, update) { Object.keys(update).forEach(function(key) { // if update[key] exist, and it's not a string or array, // we go in one level deeper if (current.hasOwnProperty(key) && typeof current[key] === 'object' && !(current[key] instanceof Array)) { merge(current[key], update[key]); // if update[key] doesn't exist in current, or it's a string // or array, then assign/overwrite current[key] to update[key] } else { current[key] = update[key]; } }); return current; } var x = { a: { a: 1 } } var y = { a: { b: 1 } } console.log(merge(x, y));

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);
};

我不喜欢现有的解决方案。所以,我开始写我自己的。

Object.prototype.merge = function(object) {
    for (const key in object) {
        if (object.hasOwnProperty(key)) {
            if (typeof this[key] === "object" && typeof object[key] === "object") {
                this[key].merge(object[key]);

                continue;
            }

            this[key] = object[key];
        }
    }

    return this;
}

我希望这能帮助那些努力理解正在发生的事情的人。我在这里看到了很多无意义的变量。

谢谢

我使用下面的短函数进行深度合并对象。 这对我来说很有效。 作者在这里完全解释了它是如何工作的。

/*!
 * Merge two or more objects together.
 * (c) 2017 Chris Ferdinandi, MIT License, https://gomakethings.com
 * @param   {Boolean}  deep     If true, do a deep (or recursive) merge [optional]
 * @param   {Object}   objects  The objects to merge together
 * @returns {Object}            Merged values of defaults and options
 * 
 * Use the function as follows:
 * let shallowMerge = extend(obj1, obj2);
 * let deepMerge = extend(true, obj1, obj2)
 */

var extend = function () {

    // Variables
    var extended = {};
    var deep = false;
    var i = 0;

    // Check if a deep merge
    if ( Object.prototype.toString.call( arguments[0] ) === '[object Boolean]' ) {
        deep = arguments[0];
        i++;
    }

    // Merge the object into the extended object
    var merge = function (obj) {
        for (var prop in obj) {
            if (obj.hasOwnProperty(prop)) {
                // If property is an object, merge properties
                if (deep && Object.prototype.toString.call(obj[prop]) === '[object Object]') {
                    extended[prop] = extend(extended[prop], obj[prop]);
                } else {
                    extended[prop] = obj[prop];
                }
            }
        }
    };

    // Loop through each object and conduct a merge
    for (; i < arguments.length; i++) {
        merge(arguments[i]);
    }

    return extended;

};