如何连接两个对象

我有这个：

      var parent= parents.json
      var child=  childrens.json
      var obj=parent["children"];
      var obj2=child["children"];
      for(i in obj){
      for(c in obj2){
      if(obj[i].name == obj2[c].parent){
     //here is the push of the child associated to the parent,  but how and
     //also how create the result.json I need to create other object to put all together.
      } 
      }
      }

尝试此代码

var obj1 = { "name": "ParentsLevel2", "Level": 1, "children": [ { "name": "analytics", "level": 2, }, { "name": "animate", "level": 2 } ]
};
var obj2 = { "name": "childrensLevel3", "Level": 1, "children": [ { "name": "analytics1", "level": 3, "parent": "analytics" }, { "name": "analytics2", "level": 3, "parent": "analytics" }, { "name": "animate1", "level": 3, "parent": "animate" }, { "name": "animate2", "level": 3, "parent": "animate" } ]};
var obj = {};
obj [0] = obj 1;
obj [1] = obj 2;
console.log(obj);

解决方案：

 var parentJson={
 "name": "ParentsLevel2",
 "Level": 1,
 "children": [
    {
        "name": "analytics",
        "level": 2,
         "children":[]
    },
    {
        "name": "animate",
        "level": 2,
        "children":[]
    }
]
};
var sonJson={
"name": "ChildrenLevel3",
"Level": 1,
"children": [
    {
        "name": "analytics1",
        "level": 3,
        "parent": "analytics"
    },
     {
        "name": "analytics2",
        "level": 3,
        "parent": "analytics"
    }
]
  };

  //push a new object to the parentJson
 //parentJson["children"].push("object to push");
   //push a new object to the sonJson
  //sonJson["children"].push("object to push");

//join 
  var objSon=sonJson["children"];
  var oP=parentJson["children"];
  for(p in oP){
  for(s in objSon){
  if(oP[p].name==objSon[s].parent){
     alert("match");
   oP[p].children.push(objSon[s]);
  }
 }
}

//create the result
var result=[];
var obj=oP;
for(i in obj){
var temp=[];
temp.push("FirstName");
temp.push(obj[i].name);
temp.push("Level");
temp.push(obj[i].level);
temp.push("Hijos")
temp.push(obj[i].children);
result.push(temp);
 $("#D3").html(JSON.stringify(result));

  $("#D3").prepend("result=")

jsfiddle解决方案