如何用深度合并代替浅合并?

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

如果您想合并多个普通对象(不要修改输入对象)。基于对象。分配polyfill

如果你想合并有限的深度

2021-08-15 07:09:20

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*/}
}

2017-10-27 10:47:40

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。

2015-01-31 05:35:26

ES5的一个简单解决方案(覆盖现有值):

2018-05-08 07:25:28

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。

2019-11-23 14:07:25

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;

};

2019-12-08 19:28:55

如何用深度合并代替浅合并?

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