将 XML 反序列化为类

我有正在反序列化的 XML，这是我的 XML

<?xml version=\"1.0\" encoding=\"utf-16\"?>
<UserInfo xmlns:xsi='http://www.w3.org/2001/XMLSchema-instance' xmlns:xsd='http://www.w3.org/2001/XMLSchema'>  
<UserName>First_User</UserName>  
<Age>25</Age>  
</UserInfo>

我有这门课

namespace MyProject{
    [System.SerializableAttribute()]
    [System.Xml.Serialization.XmlRootAttribute(Namespace = "http:/MyProject.ServiceContracts/2006/11", IsNullable = false)]

    [DataContract]
    public class UserInfo
    {
       private string username;

        private string age;

        [DataMember]
        public string UserName
        {
            get
            {
                return this.username;
            }
            set
            {
                this.username = value;
            }
        }

           [DataMember]
        public string Age
        {
            get
            {
                return this.age;
            }
            set
            {
                this.age = value;
            }
        }

    }
    }

我正在这样做

     XmlSerializer xmlSerSale = new XmlSerializer(typeof(UserInfo));

                        StringReader stringReader = new StringReader(myXML);

                        info = (UserInfo)xmlSerSale.Deserialize(stringReader);

                        stringReader.Close();

它给了我例外，

{"<UserInformation xmlns=''> was not expected."}

我该如何解决这个问题？解决这个问题的替代方法？ 我必须在 Web 服务中使用它

您在 xml 元描述符中声明名称空间：

[XmlRoot(Namespace = "http:/MyProject.ServiceContracts/2006/11", IsNullable = false)]
public class UserInfo
{
  [XmlElement]
  public string UserName { get; set; }
  [XmlElement]
  public string Age { get; set; }
}

当您执行此操作时，您还必须在 xml 中包含此名称空间：

var foo = @"<?xml version=""1.0"" encoding=""utf-16""?>
<UserInfo xmlns='http:/MyProject.ServiceContracts/2006/11' 
xmlns:xsi='http://www.w3.org/2001/XMLSchema-instance' 
xmlns:xsd='http://www.w3.org/2001/XMLSchema'>  
<UserName>First_User</UserName>  
<Age>25</Age>  
</UserInfo>";

var xmlSerSale = new XmlSerializer(typeof(UserInfo));
using (var stringReader = new StringReader(foo))
{
  var info = (UserInfo)xmlSerSale.Deserialize(stringReader);
}    

另请注意[DataContract] [DataMember]属性被忽略XmlSerializer.

更新：如果您无法更改 xml，则必须从中删除名称空间描述符XmlRoot属性。 IE。：

[XmlRoot(IsNullable = false)]
public class UserInfo
{
  [XmlElement]
  public string UserName { get; set; }
  [XmlElement]
  public string Age { get; set; }
}