骨幹如何正確構建json

Question

我想構建一個酒店和房間的應用程序。
每個酒店能有更多的房間，我檢索XML中的外部服務器這個數據，我分析它，現在我已經分成兩個數組：酒店和房間是這樣的：
hotel.json

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

房.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" 
    } 
] 

進入我的骨幹應用程序，我必須做一些控制器和一些解析來檢索其酒店的房間。
我想知道，如果是更好地爲骨幹，構建一個JSON這樣的：

[ 
     { 
     "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" 
     } 
    ] 

這是對骨幹網的效率和解析方面更好的模式？ 我想到了第一個案例，但構建應用程序後，我不知道。

Answer 1

我會建議保持數據結構平坦，因爲Backbone並不真正支持嵌套集合而無需額外的工作。使數據模型保持平坦還將使您更容易映射到REST端點（即「/ hotels/1/rooms」，「rooms/1」等）。

只是爲了演示的複雜性，這裏是怎麼一會要收集到的模型相關聯的例子：

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; 
    } 
}); 

配有一臺數據結構，你可以做這樣的事情：

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}); 
    } 
});