2013-02-23 55 views
0

我一直试图小时做到这一点使用json.js但只是太多的东西,似乎很简单。我有这样的例子数据:联接和Javascript中数组聚集

var hotels = [ 
     { id: 101, Name: "Hotel 101", WebFacilities: [8, 9, 10] }, 
     { id: 102, Name: "Hotel 101", WebFacilities: [8] }, 
     { id: 103, Name: "Hotel 101", WebFacilities: [8, 10] } 
    ]; 

    var facilities = [ 
     { id: 8, Name: "Facility 8" }, 
     { id: 9, Name: "Facility 9" }, 
     { id: 10, Name: "Facility 10" } 
    ]; 

我想要得到这样的:

var selectedFacilities = [ 
     { id: 8, Name: "Facility 8", Count: 3 }, 
     { id: 9, Name: "Facility 9", Count: 1 }, 
     { id: 10, Name: "Facility 10", Count: 2 } 
    ]; 

我该怎么办呢?

回答

2

所以看起来你想指望有多少各设施的。

下面是使用C#编写查询的一种方法:

var hotelFacilities = 
    from hotel in hotels 
    from id in hotel.WebFacilities 
    group id by id; 

var query = 
    from facility in facilities 
    join g in hotelFacilities on facility.id equals g.Key 
    select new 
    { 
     id = facility.id, 
     Name = facility.Name, 
     Count = g.Count(), 
    }; 

现在,如果你能想象这种使用方法的语法,这几乎是1:1转换到linq.js版本。

注意编译器翻译上面的方式通常会包括前面SelectMany()调用中的GroupBy()调用。然而这样写会让编写linq.js等效的查询更容易,更尴尬。

var hotelFacilities = hotels 
    .SelectMany(hotel => hotel.WebFacilities) 
    .GroupBy(id => id); 

var query = facilities 
    .Join(
     hotelFacilities, 
     facility => facility.id, 
     g => g.Key, 
     (facility, g) => new 
     { 
      id = facility.id, 
      Name = facility.Name, 
      Count = g.Count(), 
     } 
    ); 

和等价的linq.js查询。

var hotelFacilities = Enumerable.From(hotels) 
    .SelectMany("hotel => hotel.WebFacilities") 
    .GroupBy("id => id") 
    .ToArray(); 

var query = Enumerable.From(facilities) 
    .Join(
     hotelFacilities, 
     "facility => facility.id", 
     "g => g.Key()", 
     "(facility, g) => { id: facility.id, Name: facility.Name, Count: g.Count() }" 
    ).ToArray(); 
+0

优秀的答案我已经更新的ID。谢谢! – 2013-03-10 14:52:30

0

使用此:

var selectedFacilities = facilities; 

for(var i = 0; i < facilities.length; i++) { 
    for(var j = 0; j < hotels.length; j++) { 
     if(hotels[j]["id"] == facilities[i]["id"]) { 
      // Add data 
      selectedFacilities[i]["Count"] = hotels[i]["WebFacilities"].length; 
     } else { 
      selectedFacilities[i]["Count"] = 0; 
     } 
    } 

} 
+0

这看起来不正确。我认为这是我的错误,因为没有说清楚。这个想法是获得每个设施的酒店数量。所以它更清晰 – 2013-02-23 18:31:34