对于不太复杂的对象，可以使用JSON：

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

objMerge = JSON.stringify(obj1) + JSON.stringify(obj2);

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

objMerge = objMerge.replace(/\}\{/, ","); //  \_ replace with comma for valid JSON

objMerge = JSON.parse(objMerge); // { food: 'pizza', animal: 'dog', car: 'chevy'}
// Of same keys in both objects, the last object's value is retained_/

请注意，在此示例中，“｝｛”不能出现在字符串中！

对于不太复杂的对象，可以使用JSON：

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

objMerge = JSON.stringify(obj1) + JSON.stringify(obj2);

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

objMerge = objMerge.replace(/\}\{/, ","); //  \_ replace with comma for valid JSON

objMerge = JSON.parse(objMerge); // { food: 'pizza', animal: 'dog', car: 'chevy'}
// Of same keys in both objects, the last object's value is retained_/

请注意，在此示例中，“｝｛”不能出现在字符串中！

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"}

Use:

//Takes any number of objects and returns one merged object
var objectMerge = function(){
    var out = {};
    if(!arguments.length)
        return out;
    for(var i=0; i<arguments.length; i++) {
        for(var key in arguments[i]){
            out[key] = arguments[i][key];
        }
    }
    return out;
}

测试方法：

console.log(objectMerge({a:1, b:2}, {a:2, c:4}));

结果是：

{ a: 2, b: 2, c: 4 }

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;
    }
  });
}

顺便说一句，你们正在做的是覆盖财产，而不是合并。。。

这就是JavaScript对象区域真正合并的方式：只有to对象中不是对象本身的键才会被from覆盖。其他一切都将被真正合并。当然，您可以将此行为更改为不覆盖任何存在的内容，例如仅当to[n]未定义时，等等…：

var realMerge = function (to, from) {

    for (n in from) {

        if (typeof to[n] != 'object') {
            to[n] = from[n];
        } else if (typeof from[n] == 'object') {
            to[n] = realMerge(to[n], from[n]);
        }
    }
    return to;
};

用法：

var merged = realMerge(obj1, obj2);