1

在将对象序列化为 XML 时,我收到以下异常消息:

{"Type 'Alerter.EmailSender' with data contract name  'EmailSender:http://schemas.datacontract.org/2004/07/Alerter' is not expected. Consider using a DataContractResolver or add any types not known statically to the list of known types - for example, by using the KnownTypeAttribute attribute or by adding them to the list of known types passed to DataContractSerializer."}

这是我试图将其对象序列化为 XML 文件的类:

namespace Alerter
{
  [DataContract]
  public class EmailSender : IAction
  {
    private EmailSetting _emailSetting;
    private SmtpClient _smtpClient;
    [DataMember]
    public bool IncludeFullDetails
    {
        get;
        set;
    }
    [DataMember]
    public string[] Receivers
    {
        get;
        set;
    }

    public EmailSender()
    {
        _emailSetting = new EmailSetting();
        SetupClient();
    }

    private void SetupClient()
    {
        // Some Logic
    }


    public void Report(LogDictionary logDictionary)
    {
       // Some Logic
    }
   }
 }

这是我用于序列化的代码:

 using (FileStream writer = new FileStream(fileName, FileMode.Create))
        {
            DataContractSerializer ser =
                new DataContractSerializer(typeof(List<Rule>));
            ser.WriteObject(writer, list);
        }

我感谢您的帮助。

4

1 回答 1

2

确保您已使用以下命令将此类指定EmailSender为序列化程序的已知类型proper constructor

DataContractSerializer ser = new DataContractSerializer(
    typeof(List<Rule>), 
    new[] { typeof(EmailSender) }
);
ser.WriteObject(writer, list);

你需要这个的原因是因为可能在Rule类的对象图中你只使用了IAction所有成员的接口,而序列化器甚至不知道EmailSender实现的存在。

Rule您应该对对象图中静态未知的所有其他类型执行相同的操作。

于 2013-02-17T13:13:27.873 回答