修改

Question

假设我有来自服务器的JSON响应，其结构如下

   var data={
    "Data1": {
      "height": 39, 
      "weight": 62, 
      "shape": {
        "length": 19, 
        "width": 72
      }, 
      "color": "#00ff00", 
      "radius": 9.5, 
      "color_srv": "#ffff00"
    }, 
    "Data2": {
      "height": 0, 
      "weight": 40, 
      "shape": {
        "length": 19, 
        "width": 72
      }, 
      "color": "#000000", 
      "radius": 2.5, 
      "color_srv": "#ff0000"
    }
  }

我希望这个数据字典在一个字典中与某些数据分成两部分，同时保持结构。对于例如

  var data_height = {
    "Data1":{
      "height": 39,  
      "shape": {
        "length": 19, 
        "width": 72
      }, 
      "color": "#00ff00", 
      "radius": 9.5, 
    },
    "Data2":{
      "height": 0,  
      "shape": {
        "length": 19, 
        "width": 72
      }, 
      "color": "#000000", 
      "radius": 2.5, 
    }
  }
  var data_weight = {
    "Data1":{
      "weight": 39,  
      "shape": {
        "length": 19, 
        "width": 72
      }, 
      "color_srv": "#00ff00", 
      "radius": 9.5, 
    },
    "Data2":{
      "weight": 0,  
      "shape": {
        "length": 19, 
        "width": 72
      }, 
      "color_srv": "#000000", 
      "radius": 2.5, 
    }
  }

上面两个字典有不同的用途，所以在得到统一结果之后我怎么想把后端的单个数据分成两个不同的字典。

修改

这是我尝试过的，但它会抛出错误

解决方案1：

    var serve={},live={};
      for(d in data){
        pname = d.split(':')[0];
        serve['pname'].radius= data[d].radius;
        serve['pname'].center= data[d].center;
        serve['pname'].color= data[d].color_srv;
        live['pname'].radius= data[d].radius;
        live['pname'].center= data[d].center;
        live['pname'].color= data[d].color;
        serve['pname'].numbers= data[d].serving;
        live['pname'].numbers= data[d].living;
        serve['pname'].place= pname;
        live['pname'].place= pname;
      }

EDIT2

解决方案2：

  var serve={},live={};
    for(d in data){
      pname = d.split(':')[0];
      serve['radius']= data[d].radius;
      serve['center']= data[d].center;
      serve['color']= data[d].color_srv;
      live['radius']= data[d].radius;
      live['center']= data[d].center;
      live['color']= data[d].color;
      serve['numbers']= data[d].serving;
      live['numbers']= data[d].living;
      serve['place']= pname;
      live['plcae']= pname;
    }

上述两种解决方案似乎都不起作用。

Answer 1

正如Nina所说，只需克隆对象并删除每个对象不需要的属性。在这里，我使用了reduce一个带有data_height和data_height属性的初始对象。

var clone = function (obj) { return JSON.parse(JSON.stringify(obj)); }

var output = Object.keys(data).reduce(function (p, c) {
  var obj = data[c];
  p.data_height[c] = clone(obj);
  delete p.data_height[c].weight;
  delete p.data_height[c].color_srv;
  p.data_weight[c] = clone(obj);
  delete p.data_weight[c].height;
  delete p.data_weight[c].color;
  return p;
}, { data_height: {}, data_weight: {} });

输出

{
  "data_height": {
    "Data1": {
      "height": 39,
      "shape": {
        "length": 19,
        "width": 72
      },
      "color": "#00ff00",
      "radius": 9.5
    },
    "Data2": {
      "height": 0,
      "shape": {
        "length": 19,
        "width": 72
      },
      "color": "#000000",
      "radius": 2.5
    }
  },
  "data_weight": {
    "Data1": {
      "weight": 62,
      "shape": {
        "length": 19,
        "width": 72
      },
      "radius": 9.5,
      "color_srv": "#ffff00"
    },
    "Data2": {
      "weight": 40,
      "shape": {
        "length": 19,
        "width": 72
      },
      "radius": 2.5,
      "color_srv": "#ff0000"
    }
  }
}

DEMO

从javascript中的单个JSON响应创建多个字典？

修改

EDIT2

1 个答案: