这会将obj合并为“默认”def。obj优先于两者中存在的任何对象，因为obj被复制到def中。还要注意，这是递归的。

function mergeObjs(def, obj) {
    if (typeof obj == 'undefined') {
        return def;
    } else if (typeof def == 'undefined') {
        return obj;
    }
    for (var i in obj) {
        if (obj[i] != null && obj[i].constructor == Object) {
            def[i] = mergeObjs(def[i], obj[i]);
        } else {
            def[i] = obj[i];
        }
    }
    return def;
}

a = {x : {y : [123]}}
b = {x : {z : 123}}
console.log(mergeObjs(a, b));
// {x: {y : [123], z : 123}}

我使用纯JavaScript中的以下内容。它从最右边的参数开始，并将它们组合到第一个参数。没有返回值，只修改了第一个参数，最左边的参数（第一个除外）在财产上具有最高的权重。

var merge = function() {
  var il = arguments.length;

  for (var i = il - 1; i > 0; --i) {
    for (var key in arguments[i]) {
      if (arguments[i].hasOwnProperty(key)) {
        arguments[0][key] = arguments[i][key];
      }
    }
  }
};

在一行代码中合并N个对象的财产

Object.assign方法是ECMAScript 2015（ES6）标准的一部分，它完全符合您的需要。（不支持IE）

var clone = Object.assign({}, obj);

assign（）方法用于将所有可枚举自身财产的值从一个或多个源对象复制到目标对象。

阅读更多。。。

支持旧浏览器的polyfill：

if (!Object.assign) {
  Object.defineProperty(Object, 'assign', {
    enumerable: false,
    configurable: true,
    writable: true,
    value: function(target) {
      'use strict';
      if (target === undefined || target === null) {
        throw new TypeError('Cannot convert first argument to object');
      }

      var to = Object(target);
      for (var i = 1; i < arguments.length; i++) {
        var nextSource = arguments[i];
        if (nextSource === undefined || nextSource === null) {
          continue;
        }
        nextSource = Object(nextSource);

        var keysArray = Object.keys(nextSource);
        for (var nextIndex = 0, len = keysArray.length; nextIndex < len; nextIndex++) {
          var nextKey = keysArray[nextIndex];
          var desc = Object.getOwnPropertyDescriptor(nextSource, nextKey);
          if (desc !== undefined && desc.enumerable) {
            to[nextKey] = nextSource[nextKey];
          }
        }
      }
      return to;
    }
  });
}

Object.assign(TargetObject, Obj1, Obj2, ...);

var obj1 = { food: 'pizza', car: 'ford' }
var obj2 = { animal: 'dog' }

// result
result: {food: "pizza", car: "ford", animal: "dog"}

使用jQuery.exde（）-Link

// Merge obj1 & obj2 to result
var result1 = $.extend( {}, obj1, obj2 );

使用_.merge（）-链接

// Merge obj1 & obj2 to result
var result2 = _.merge( {}, obj1, obj2 );

使用_.exde（）-链接

// Merge obj1 & obj2 to result
var result3 = _.extend( {}, obj1, obj2 );

使用Object.assign（）ECMAScript 2015（ES6）-Link

// Merge obj1 & obj2 to result
var result4 = Object.assign( {}, obj1, obj2 );

全部输出

obj1: { animal: 'dog' }
obj2: { food: 'pizza', car: 'ford' }
result1: {food: "pizza", car: "ford", animal: "dog"}
result2: {food: "pizza", car: "ford", animal: "dog"}
result3: {food: "pizza", car: "ford", animal: "dog"}
result4: {food: "pizza", car: "ford", animal: "dog"}

ECMAScript 2018标准方法

您可以使用对象扩散：

let merged = {...obj1, ...obj2};

merged现在是obj1和obj2的并集。obj2中的财产将覆盖obj1中的属性。

/** There's no limit to the number of objects you can merge.
 *  Later properties overwrite earlier properties with the same name. */
const allRules = {...obj1, ...obj2, ...obj3};

这里还有此语法的MDN文档。如果您正在使用babel，则需要@babel/plugin提议对象rest spread插件才能工作（该插件包含在ES2018中的@babel/preset-env中）。

ECMAScript 2015（ES6）标准方法

/* For the case in question, you would do: */
Object.assign(obj1, obj2);

/** There's no limit to the number of objects you can merge.
 *  All objects get merged into the first object. 
 *  Only the object in the first argument is mutated and returned.
 *  Later properties overwrite earlier properties with the same name. */
const allRules = Object.assign({}, obj1, obj2, obj3, etc);

（参见MDN JavaScript参考）

ES5及更早版本的方法

for (var attrname in obj2) { obj1[attrname] = obj2[attrname]; }

请注意，这将简单地将obj2的所有属性添加到obj1中，如果您仍然希望使用未修改的obj1，那么这可能不是您想要的。

如果你使用的是一个在你的原型上到处都是垃圾的框架，那么你必须通过hasOwnProperty这样的检查来获得更高的效率，但这段代码在99%的情况下都是有效的。

示例函数：

/**
 * Overwrites obj1's values with obj2's and adds obj2's if non existent in obj1
 * @param obj1
 * @param obj2
 * @returns obj3 a new object based on obj1 and obj2
 */
function merge_options(obj1,obj2){
    var obj3 = {};
    for (var attrname in obj1) { obj3[attrname] = obj1[attrname]; }
    for (var attrname in obj2) { obj3[attrname] = obj2[attrname]; }
    return obj3;
}