NetDataContractSerializer生成无效的XML

时间:2013-02-20 13:56:48

标签: c#-4.0 serialization xml-serialization netdatacontractserializer

我的NetDataContractSerializer似乎很混乱:XML的结尾出现了两次:

<?xml version="1.0" encoding="utf-8"?>
<Project xmlns:i="http://www.w3.org/2001/XMLSchema-instance" z:Id="1" 

[...]
        <d2p1:anyType i:nil="true" />
      </d2p1:_items>
      <d2p1:_size>2</d2p1:_size>
      <d2p1:_version>2</d2p1:_version>
    </d2p1:items>
  </ProjectParts>
  <ProjectPath z:Id="31">D:\t10\</ProjectPath>
</Project>ze>
              <d2p1:_version>3</d2p1:_version>
            </d2p1:items>
            <d2p1:_monitor xmlns:d7p1="http://schemas.datacontract.org/2004/07/System.Collections.ObjectModel" z:Id="33">
              <d7p1:_busyCount>0</d7p1:_busyCount>
            </d2p1:_monitor>
          </Elements>
          <Project z:Ref="1" i:nil="true" xmlns="http://schemas.datacontract.org/2004/07/Modules.WorkspaceManager.Types" />
        </d2p1:anyType>
        <d2p1:anyType i:nil="true" />
        <d2p1:anyType i:nil="true" />
      </d2p1:_items>
      <d2p1:_size>2</d2p1:_size>
      <d2p1:_version>2</d2p1:_version>
    </d2p1:items>
  </ProjectParts>
  <ProjectPath z:Id="34">D:\t10\</ProjectPath>
</Project>

正如你所看到的,有一些严重的口吃。它偶尔发生,我无法重现错误。有任何想法吗?可能是因为在写入文件时在VS中打开了文件吗?

我像这样序列化我的对象:

private void SerializeToFile(object objectToSerialize)
    {
        Stream stream = null;

        try
        {
            stream = File.Open(_fileName, FileMode.OpenOrCreate, FileAccess.Write);
            using (var writer = XmlWriter.Create(stream, new XmlWriterSettings { Indent = true }))
            {
                NetDataContractSerializer serializer = new NetDataContractSerializer();
                serializer.WriteObject(writer, objectToSerialize);
            }
        }

        finally
        {
            if (stream != null) stream.Close();
        }
    }

序列化的类看起来像这样:

[DataContract(IsReference = true)]
public class Project : IProject
{
    [DataMember] public string ProjectPath { get; set; }
    [DataMember] public string ProjectName { get; set; }
    [DataMember] public Collection<IProjectPart> ProjectParts { get; set; }

    public T GetPart<T>() where T : IProjectPart
    {
        return ProjectParts.OfType<T>().First();
    }

    public void RegisterPart<T>(T part) where T : IProjectPart
    {
        if (ProjectParts.Any(p => p.GetType().IsInstanceOfType(part))) throw new InvalidOperationException("Part already registered.");
        ProjectParts.Add(part);
        part.Project = this;
    }

    public void Load()
    {
        foreach (var projectPart in ProjectParts)
        {
            projectPart.Load();
        }
    }

    public void Unload()
    {
        foreach (var projectPart in ProjectParts)
        {
            projectPart.Unload();
        }
    }

    public void Save()
    {
        foreach (var projectPart in ProjectParts)
        {
            projectPart.Save();
        }
    }

    public Project()
    {
        ProjectParts = new Collection<IProjectPart>();
    }
}

谢谢!

1 个答案:

答案 0 :(得分:1)

问题很简单 - 当您反复对序列进行序列化时,可以使用不同大小的IProjectPart集合进行序列化。 File.Open方法不会清除先前内容中的文件,因此请执行以下步骤:

i)使用两个IProjectPart实例序列化对象 - 假设它需要10行xml文件

ii)使用集合中的一个IProjectPart实例再次序列化对象 - 这次需要8行xml文件

iii)第9行和第10行将填充旧的xml数据,因为它们在序列化尝试之间未被清​​除 - 因此存在一些重复的垃圾状xml数据。

亲自尝试一下,您将看到这些多个标签的确切生成方式。

注意:8行和10行是我实现的近似值

注意2:我建议对序列化方法中的流使用using语句(对于所有IDisposable对象):

private void SerializeToFile(object objectToSerialize)
{
    using(var stream = File.Open(_fileName, FileMode.OpenOrCreate, FileAccess.Write))
    {
        using (var writer = XmlWriter.Create(stream, new XmlWriterSettings { Indent = true }))
        {
            NetDataContractSerializer serializer = new NetDataContractSerializer();
            serializer.WriteObject(writer, objectToSerialize);
        }
    }
}