Сериализация объектов в xml

Question

У меня есть методы, которые будут создавать объекты, а затем отправляются другому методу, который будет выполнять некоторые операции, и, наконец, он будет сериализован в XML-файл. проблема. xml имеет только последний объект, который был создан, ему не хватает первых 15 из них.

public void reserveRoom() 
    { 
     reserveRoom("2345", "20160905", 1, "00001", "KB"); 
     reserveRoom("2345", "20160905", 4, "00002", "KB"); 
     reserveRoom("2345", "20160905", 5, "00003", "KB"); 
     reserveRoom("2345", "20160907", 3, "00004", "KB"); 
     reserveRoom("2345", "20160909", 4, "00005", "KB"); 
     reserveRoom("2345", "20160906", 5, "00006", "KB"); // fails - room not available 
     reserveRoom("2345", "20160905", 1, "00007", "QB"); 
     reserveRoom("2345", "20160905", 1, "00008", "KB"); // fails- room not available 
     reserveRoom("2345", "20160905", 4, "00009", "QB"); 
     reserveRoom("2345", "20170905", 1, "00010", "KB"); // fails- room not available 
     reserveRoom("7890", "20160915", 5, "00011", "QB"); 
     reserveRoom("7890", "20160925", 10, "00012", "QB"); // fails- room not available 
     reserveRoom("7890", "20160907", 3, "00013", "QB"); 
     reserveRoom("7890", "20160909", 3, "00014", "KB"); 
     reserveRoom("7890", "20160905", 1, "00015", "AB"); // fails- unknown room type 
     reserveRoom("1234", "20160905", 1, "00016", "KB"); 
    } 

public void reserveRoom(string h, string d, int n, string c, string t) 
    { 
     ReservationType reservation = new ReservationType(); 

     reservation.hotelId = h; 
     reservation.StartDate = d; 
     reservation.numDays = n; 
     reservation.customerId = c; 
     reservation.roomType = t; 

     // Console.Write(reservation.customerId + ": " + Environment.NewLine); 

     bool result = ReserveRoom(reservation); 


    } 

public bool ReserveRoom(ReservationType reservation) 
    { 
     ReservationType RequestReservation = new ReservationType(); 
     RequestReservation = reservation; //List that contains the test cases. 
     string reservationid = "0001"; 
     int number = int.Parse(reservationid); 
     int count = 0; 

     List<String> DateList = CreateDateList(RequestReservation.StartDate, RequestReservation.numDays); 

      foreach (Inventory inventory in RoomInventory) 
      { 
       if (RequestReservation.hotelId == inventory.HotelId) 
        { 

        if (RequestReservation.roomType == inventory.RoomType) 
        { 
         for (int i = 0; i < DateList.Count; i++) 
         { 
          if (DateList[i] == inventory.Date && inventory.Quantity > 0) 
          { 
           count++; 
           inventory.Quantity--; 
           if (RequestReservation.numDays == count) 
           { 
            RequestReservation.reservationId = "000"+(number + 1).ToString(); 
            RequestReservation.result = ReservationType.ReservationResultType.Success; 
            count = 0; 


           foreach (Hotels h in LHotels) 
            { 
             foreach (Room rm in h.RoomList) 
             { 
              if (RequestReservation.hotelId == h.HotelId && RequestReservation.roomType == rm.RoomType) 
              { 

               RequestReservation.cost = (RequestReservation.numDays * rm.Rate); 
              }            
             } 
            } 

           } // 4th if , numDays == Count 

           else 
           {          
            RequestReservation.reservationId = null; 
            RequestReservation.result = ReservationType.ReservationResultType.RoomNotAvailable;          
           } 

          } //3rd if, DateList check 

         }//Forloop, Datelist 

        }// 2nd if, roomtype check 

       }//first if ,hotelId check 

      number++; 

     } // Inventory List Foreach loop 



       serial = new XmlSerializer(RequestReservation.GetType()); 
       sw = new StreamWriter(ReservationFilename); 
       serial.Serialize(sw, RequestReservation); 
       sw.Close(); 

     return true; 
    }// End of ReserveRoom Method 

Answer 1

Каждый раз, когда вы звоните sw = new StreamWriter(ReservationFilename); StreamWriter переписать все данные, которые были ранее записаны на файл. С объектом FileStream вы можете открыть файл в режиме Append.

Кроме того, не забудьте утилизировать свой StreamWriter после использования его, используя using блок:

serial = new XmlSerializer(RequestReservation.GetType()); 

using (FileStream fs = new FileStream(ReservationFilename, FileMode.Append, FileAccess.Write)) 
using (StreamWriter sw = new StreamWriter(fs)) 
{ 
    serial.Serialize(sw, RequestReservation); // Serialize at the end of the file 
} // Closing and disposing FileStream and StreamWriter