Question

我想构建一个酒店和房间的应用程序每个酒店都可以有更多的房间，我用XML从外部服务器检索这些数据，我解析它，现在我分成了两个阵列：酒店和这样的房间：
hotel.json

[
  {
    "id": "1", 
    "name": "Hotel1"
  }, 
  {
    "id": "2", 
    "name": "Hotel2"
  }, 
  {
    "id": "3", 
    "name": "Hotel3"
  }
]

rooms.json

[
  {
    "id" : "r1",
    "hotel_id" : "1",
    "name" : "Singola",
    "level" : "1"
  },
  {
    "id" : "r1_1",
    "hotel_id" : "1",
    "name" : "Doppia",
    "level" : "2"
  },
  {
    "id" : "r1_3",
    "hotel_id" : "1",
    "name" : "Doppia Uso singol",
    "level" : "1"
  },
  {
    "id" : "r2",
    "hotel_id" : "2",
    "name" : "Singola",
    "level" : "1"
  },
  {
    "id" : "r2_1",
    "hotel_id" : "2",
    "name" : "Tripla",
    "level" : "1"
  }
]

在我的骨干应用程序中，我必须制作一些控制器和一些解析来检索其酒店的房间我想知道骨干是否更适合构建像这样的Json：

[
      {
        "id": "1", 
        "name": "Hotel1",
        "rooms": [
                 {
                   "id" : "r1",
                   "hotel_id" : "1",
                   "name" : "Singola",
                   "level" : "1"
                 },
                 {
                   "id" : "r1_1",
                   "hotel_id" : "1",
                   "name" : "Doppia",
                   "level" : "2"
                 }
                 ]

      }, 
      {
        "id": "2", 
        "name": "Hotel2",
        "rooms": [
                 {
                   "id" : "r2",
                   "hotel_id" : "2",
                   "name" : "Singola",
                   "level" : "1"
                 },
                 {
                   "id" : "r2_1",
                   "hotel_id" : "1",
                   "name" : "Doppia",
                   "level" : "2"
                 }
                 ]
      }, 
      {
        "id": "3", 
        "name": "Hotel3"
      }
    ]

在效率和解析方面，哪种骨干模式更好？我认为第一种情况，但在构建应用程序后，我不确定。

Answer 1

我建议保持数据结构不变，因为Backbone并不真正支持嵌套集合而不需要额外的努力。保持数据模型平坦也将使您更容易映射到REST端点（即'/ hotels / 1 / rooms'，'rooms / 1'等）。

为了证明这些复杂性，下面是一个如何将集合与模型相关联的示例：

HotelModel = Backbone.Model.extend({
    initialize: function() {
        // because initialize is called after parse
        _.defaults(this, {
            rooms: new RoomCollection
        });
    },
    parse: function(response) {
        if (_.has(response, "rooms")) {
            this.rooms = new RoomCollection(response.rooms, {
                parse: true
            });
            delete response.rooms;
        }
        return response;
    },
    toJSON: function() {
        var json = _.clone(this.attributes);
        json.rooms = this.rooms.toJSON();
        return json;
    }
});

使用平面数据结构，您可以执行以下操作：

HotelModel = Backbone.Model.extend({
    idAttribute:'hotel_id',
    urlRoot:'/hotels'
});
RoomModel = Backbone.Model.extend({
    idAttribute:'room_id',
    urlRoot:'/rooms'
});

HotelCollection = Backbone.Collection.extend({
    url: '/hotels',
    model:HotelModel
});
RoomCollection = Backbone.Collection.extend({
    url: '/rooms',
    model:RoomModel,
    getByHotelId: function(hotelId){
        return this.findWhere({hotel_id:hotelId});
    }
});

Backbone如何正确构造json

1 个答案: