Xml Десериализация к классу ошибке

Question

Я пытаюсь десериализациями этого XML-строка ответа на класс:

<?xml version="1.0" encoding="UTF-8"?> 
<result command="searchhotels" tID="1463758732000001" ip="70.112.14.31" date="2016-05-20 15:39:00" version="2.0" elapsedTime="8.5034000873566"> 
    <currencyShort>USD</currencyShort> 
    <hotels count="6"> 
     <hotel runno="0" preferred="no" cityname="DUBAI" order="3" hotelid="856915"> 
      <from>5549.2769 
       <formatted>5,549.28</formatted> 
      </from> 
      <fromMinimumSelling>6975 
       <formatted>6,975.00</formatted> 
      </fromMinimumSelling> 
      <availability>available</availability> 
     </hotel> 
     <hotel runno="1" preferred="no" cityname="DUBAI" order="3" hotelid="30924"> 
      <from>5819.9734 
       <formatted>5,819.97</formatted> 
      </from> 
      <fromMinimumSelling>7316 
       <formatted>7,316.00</formatted> 
      </fromMinimumSelling> 
      <availability>available</availability> 
     </hotel> 
     <hotel runno="2" preferred="no" cityname="DUBAI" order="3" hotelid="994075"> 
      <from>5459.0448 
       <formatted>5,459.04</formatted> 
      </from> 
      <fromMinimumSelling>6851 
       <formatted>6,851.00</formatted> 
      </fromMinimumSelling> 
      <availability>available</availability> 
     </hotel> 
     <hotel runno="3" preferred="no" cityname="DUBAI" order="3" hotelid="911115"> 
      <from>6418 
       <formatted>6,418.00</formatted> 
      </from> 
      <fromMinimumSelling>7976 
       <formatted>7,976.00</formatted> 
      </fromMinimumSelling> 
      <availability>available</availability> 
     </hotel> 
     <hotel runno="4" preferred="no" cityname="DUBAI" order="3" hotelid="994005"> 
      <from>4821.246 
       <formatted>4,821.25</formatted> 
      </from> 
      <fromMinimumSelling>6107 
       <formatted>6,107.00</formatted> 
      </fromMinimumSelling> 
      <availability>available</availability> 
     </hotel> 
     <hotel runno="5" preferred="no" cityname="DUBAI" order="3" hotelid="446025"> 
      <from>11173.763 
       <formatted>11,173.76</formatted> 
      </from> 
      <fromMinimumSelling>13864 
       <formatted>13,864.00</formatted> 
      </fromMinimumSelling> 
      <availability>available</availability> 
     </hotel> 
    </hotels> 
    <successful>TRUE</successful> 
</result> 

Вот класс им пытается десериализации строки:

public class DOTWSearchResponse 
    { 
    [XmlRoot(ElementName = "from")] 
    public class From 
    { 
     [XmlElement(ElementName = "formatted")] 
     public string Formatted { get; set; } 
    } 

    [XmlRoot(ElementName = "fromMinimumSelling")] 
    public class FromMinimumSelling 
    { 
     [XmlElement(ElementName = "formatted")] 
     public string Formatted { get; set; } 
    } 

    [XmlRoot(ElementName = "hotel")] 
    public class Hotel 
    { 
     [XmlElement(ElementName = "from")] 
     public From From { get; set; } 
     [XmlElement(ElementName = "fromMinimumSelling")] 
     public FromMinimumSelling FromMinimumSelling { get; set; } 
     [XmlElement(ElementName = "availability")] 
     public string Availability { get; set; } 
     [XmlAttribute(AttributeName = "runno")] 
     public string Runno { get; set; } 
     [XmlAttribute(AttributeName = "preferred")] 
     public string Preferred { get; set; } 
     [XmlAttribute(AttributeName = "cityname")] 
     public string Cityname { get; set; } 
     [XmlAttribute(AttributeName = "order")] 
     public string Order { get; set; } 
     [XmlAttribute(AttributeName = "hotelid")] 
     public string Hotelid { get; set; } 
    } 

    [XmlRoot(ElementName = "hotels")] 
    public class Hotels 
    { 
     [XmlElement(ElementName = "hotel")] 
     public List<Hotel> Hotel { get; set; } 
     [XmlAttribute(AttributeName = "count")] 
     public string Count { get; set; } 
    } 

    [XmlRoot(ElementName = "result")] 
    public class Result 
    { 
     [XmlElement(ElementName = "currencyShort")] 
     public string CurrencyShort { get; set; } 
     [XmlElement(ElementName = "hotels")] 
     public Hotels Hotels { get; set; } 
     [XmlElement(ElementName = "successful")] 
     public string Successful { get; set; } 
     [XmlAttribute(AttributeName = "command")] 
     public string Command { get; set; } 
     [XmlAttribute(AttributeName = "tID")] 
     public string TID { get; set; } 
     [XmlAttribute(AttributeName = "ip")] 
     public string Ip { get; set; } 
     [XmlAttribute(AttributeName = "date")] 
     public string Date { get; set; } 
     [XmlAttribute(AttributeName = "version")] 
     public string Version { get; set; } 
     [XmlAttribute(AttributeName = "elapsedTime")] 
     public string ElapsedTime { get; set; } 
    } 
} 

Вот код, который бросает ошибку:

 try 
     { 
      XmlSerializer serializer = new XmlSerializer(typeof(TResponse)); 
      StringReader reader = new StringReader(response); 

      TResponse deserialized = (TResponse)serializer.Deserialize(reader); 
      return deserialized; 

     } 
     catch (Exception exception) 
     { 
      throw exception; 
     } 

и ошибка исключения брошенную является: {"<result xmlns=''> was not expected."} Я всегда alot проблемы десериализации, когда дело доходит до xml ... Я знаю, что там много, я здесь отсутствует. Помогите?

Answer 1

Привет вы должны десериализации ввести результат вместо TResponse что-то вроде ниже

 TextReader sr = new StringReader(response); 
     var deserializer = new XmlSerializer(typeof(Result)); 
     Object storage = (Result)deserializer.Deserialize(sr); 

переменной Storage теперь будут иметь объекты заселяется

Answer 2

Попробуйте это

XmlSerializer serializer = new XmlSerializer(typeof(DOTWSearchResponse.Result)); 
       StringReader reader = new StringReader(response); 

       DOTWSearchResponse.Result deserialized = (DOTWSearchResponse.Result)serializer.Deserialize(reader);