Node.js中如何合并两个复杂对象详解

站长资源 2025/12/14 佚名

2 1538 2

南强小屋 Design By 杰米

前言

相信大家都知道在通常情况下，在Node.js中我们可以通过underscore的extend或者lodash的merge来合并两个对象，但是对于像下面这种复杂的对象，要如何来应对呢？下面来一起学习学习吧。

Node.js合并两个复杂对象

例如我有以下两个object：

var obj1 = {
 "name" : "myname",
 "status" : 0,
 "profile": { "sex":"m", "isactive" : true},
 "strarr":["one", "three"],
 "objarray": [
 {
  "id": 1,
  "email": "a1@me.com",
  "isactive":true
 },
 {
  "id": 2,
  "email": "a2@me.com",
  "isactive":false
 }
 ]
};

var obj2 = {
 "name" : "myname",
 "status" : 1,
 "newfield": 1,
 "profile": { "isactive" : false, "city": "new York"},
 "strarr":["two"],
 "objarray": [
 {
  "id": 1,
  "isactive":false
 },
 {
  "id": 2,
  "email": "a2modified@me.com"
 },
 {
  "id": 3,
  "email": "a3new@me.com",
  "isactive" : true
 }
 ]
};

希望合并之后的结果输出成下面这样：

{ name: 'myname',
 status: 1,
 profile: { sex: 'm', isactive: false, city: 'new York' },
 strarr: [ 'one', 'three', 'two' ],
 objarray: 
 [ { id: 1, email: 'a1@me.com', isactive: false },
 { id: 2, email: 'a2modified@me.com', isactive: false },
 { id: 3, email: 'a3new@me.com', isactive: true } ],
newfield: 1 }

通过underscore或者lodash现有的方法我们无法实现上述结果，那只能自己写代码来实现了。

function mergeObjs(def, obj) {
 if (!obj) {
 return def;
 } else if (!def) {
 return obj;
 }

 for (var i in obj) {
 // if its an object
 if (obj[i] != null && obj[i].constructor == Object)
 {
  def[i] = mergeObjs(def[i], obj[i]);
 }
 // if its an array, simple values need to be joined. Object values need to be remerged.
 else if(obj[i] != null && (obj[i] instanceof Array) && obj[i].length > 0)
 {
  // test to see if the first element is an object or not so we know the type of array we're dealing with.
  if(obj[i][0].constructor == Object)
  {
  var newobjs = [];
  // create an index of all the existing object IDs for quick access. There is no way to know how many items will be in the arrays.
  var objids = {}
  for(var x= 0, l= def[i].length ; x < l; x++ )
  {
   objids[def[i][x].id] = x;
  }

  // now walk through the objects in the new array
  // if the ID exists, then merge the objects.
  // if the ID does not exist, push to the end of the def array
  for(var x= 0, l= obj[i].length; x < l; x++)
  {
   var newobj = obj[i][x];
   if(objids[newobj.id] !== undefined)
   {
   def[i][x] = mergeObjs(def[i][x],newobj);
   }
   else {
   newobjs.push(newobj);
   }
  }

  for(var x= 0, l = newobjs.length; x<l; x++) {
   def[i].push(newobjs[x]);
  }
  }
  else {
  for(var x=0; x < obj[i].length; x++)
  {
   var idxObj = obj[i][x];
   if(def[i].indexOf(idxObj) === -1) {
    def[i].push(idxObj);
   }
  }
  }
 }
 else
 {
  def[i] = obj[i];
 }
 }
 return def;}

将上述代码稍作改进，我们可以实现在合并过程中将Number类型的值自动相加。

function merge(def, obj) {
 if (!obj) {
  return def;
 }
 else if (!def) {
  return obj;
 }

 for (var i in obj) {
  // if its an object
  if (obj[i] != null && obj[i].constructor == Object)
  {
   def[i] = merge(def[i], obj[i]);
  }
  // if its an array, simple values need to be joined. Object values need to be re-merged.
  else if(obj[i] != null && (obj[i] instanceof Array) && obj[i].length > 0)
  {
   // test to see if the first element is an object or not so we know the type of array we're dealing with.
   if(obj[i][0].constructor == Object)
   {
    var newobjs = [];
    // create an index of all the existing object IDs for quick access. There is no way to know how many items will be in the arrays.
    var objids = {}
    for(var x= 0, l= def[i].length ; x < l; x++ )
    {
     objids[def[i][x].id] = x;
    }

    // now walk through the objects in the new array
    // if the ID exists, then merge the objects.
    // if the ID does not exist, push to the end of the def array
    for(var x= 0, l= obj[i].length; x < l; x++)
    {
     var newobj = obj[i][x];
     if(objids[newobj.id] !== undefined)
     {
      def[i][x] = merge(def[i][x],newobj);
     }
     else {
      newobjs.push(newobj);
     }
    }

    for(var x= 0, l = newobjs.length; x<l; x++) {
     def[i].push(newobjs[x]);
    }
   }
   else {
    for(var x=0; x < obj[i].length; x++)
    {
     var idxObj = obj[i][x];
     if(def[i].indexOf(idxObj) === -1) {
      def[i].push(idxObj);
     }
    }
   }
  }
  else
  {
   if (isNaN(obj[i]) || i.indexOf('_key') > -1){
    def[i] = obj[i];
   }
   else{
    def[i] += obj[i];
   }
  }
 }
 return def;
}

例如有以下两个对象：

var data1 = {
 "_id" : "577327c544bd90be508b46cc",
 "channelId_info" : [
 {
  "channelId_key" : "0",
  "secondLevel_group" : [
  {
   "secondLevel_key" : "568cc36c44bd90625a045c60",
   "sender_group" : [
   {
    "sender_key" : "577327c544bd90be508b46cd",
    "sender_sum" : 40.0
   }
   ],
   "senders_sum" : 40.0
  }
  ],
  "channelId_sum" : 40.0
 }
 ],
 "car_sum" : 40.0
};

var data2 = {
 "_id" : "577327c544bd90be508b46cc",
 "channelId_info" : [
 {
  "channelId_key" : "0",
  "secondLevel_group" : [
  {
   "secondLevel_key" : "568cc36c44bd90625a045c60",
   "sender_group" : [
   {
    "sender_key" : "577327c544bd90be508b46cd",
    "sender_sum" : 20.0
   },
   {
    "sender_key" : "5710bcc7e66620fd4bc0914f",
    "sender_sum" : 5.0
   }
   ],
   "senders_sum" : 25.0
  },
  {
   "secondLevel_key" : "55fbeb4744bd9090708b4567",
   "sender_group" : [
   {
    "sender_key" : "5670f993a2f5dbf12e73b763",
    "sender_sum" : 10.0
   }
   ],
   "senders_sum" : 10.0
  }
  ],
  "channelId_sum" : 35.0
 },
 {
  "channelId_key" : "1",
  "secondLevel_group" : [
  {
   "secondLevel_key" : "568cc36c44bd90625a045c60",
   "sender_group" : [
   {
    "sender_key" : "577327c544bd90be508b46cd",
    "sender_sum" : 20.0
   }
   ],
   "senders_sum" : 20.0
  }
  ],
  "channelId_sum" : 20.0
 }
 ],
 "car_sum" : 55.0
};

合并之后的结果如下：

{
 "_id": "577327c544bd90be508b46cc",
 "channelId_info": [
  {
   "channelId_key": "0",
   "secondLevel_group": [
    {
     "secondLevel_key": "568cc36c44bd90625a045c60",
     "sender_group": [
      {
       "sender_key": "577327c544bd90be508b46cd",
       "sender_sum": 60
      },
      {
       "sender_key": "5710bcc7e66620fd4bc0914f",
       "sender_sum": 5
      }
     ],
     "senders_sum": 65
    },
    {
     "secondLevel_key": "55fbeb4744bd9090708b4567",
     "sender_group": [
      {
       "sender_key": "5670f993a2f5dbf12e73b763",
       "sender_sum": 10
      }
     ],
     "senders_sum": 10
    }
   ],
   "channelId_sum": 75
  },
  {
   "channelId_key": "1",
   "secondLevel_group": [
    {
     "secondLevel_key": "568cc36c44bd90625a045c60",
     "sender_group": [
      {
       "sender_key": "577327c544bd90be508b46cd",
       "sender_sum": 20
      }
     ],
     "senders_sum": 20
    }
   ],
   "channelId_sum": 20
  }
 ],
 "car_sum": 95
}

总结

以上就是这篇文章的全部内容了，文中提到的上述代码在日常工作中很有用，值得大家收藏！希望本文的内容对大家的学习或者工作能带来一定的帮助。

nodejs,对象合并,node,合并对象,nodejs,遍历对象

标签：

nodejs,对象合并,node,合并对象,nodejs,遍历对象

南强小屋 Design By 杰米

广告合作：本站广告合作请联系QQ：858582 申请时备注：广告合作（否则不回）
免责声明：本站文章均来自网站采集或用户投稿，网站不提供任何软件下载或自行开发的软件！如有用户或公司发现本站内容信息存在侵权行为，请邮件告知！ 858582#qq.com

南强小屋 Design By 杰米

评论“Node.js中如何合并两个复杂对象详解”

暂无Node.js中如何合并两个复杂对象详解的评论...

Node.js中如何合并两个复杂对象详解

nodejs,对象合并,node,合并对象,nodejs,遍历对象

Javascript 实现全屏滚动实例代码

javascript 闭包详解及简单实例应用

评论“Node.js中如何合并两个复杂对象详解”

《魔兽世界》大逃杀！60人新游玩模式《强袭风暴》3月21日上线

友情链接