如何在读取xml文件时处理文件结尾

Question

所以我正在读取长度未知的xml文件,并将每个元素读入列表结构.现在,一旦我到达文件的末尾,我继续阅读,这会导致异常.现在我只是抓住这个例外并继续我的生活,但是有更清洁的方法吗？

try
{
    while(!textReader.EOF)
    {
        // Used to store info from each command as they are read from the xml file
        ATAPassThroughCommands command = new ATAPassThroughCommands ();                      
        // the following is just commands being read and their contents being saved
        XmlNodeType node = textReader.NodeType;                                             

        textReader.ReadStartElement( "Command" );
        node = textReader.NodeType;
        name = textReader.ReadElementString( "Name" );
        node = textReader.NodeType;
        CommandListContext.Add(name);
        command.m_Name = name;
        command.m_CMD = Convert .ToByte(textReader.ReadElementString("CMD" ),16);
        command.m_Feature = Convert .ToByte(textReader.ReadElementString("Feature" ),16);

        textReader.ReadEndElement(); //</command>
        m_ATACommands.Add(command);
    }
}
catch ( Exception ex)
{
    //</ATAPassThrough>   TODO: this is an ugly fix come up with something better later
    textReader.ReadEndElement(); 
    //cUtils.DisplayError(ex.Message);
}

xml文件:

<ATAPassThrough>
  <Command>
    <Name>Smart</Name>
    <CMD>B0</CMD>
    <Feature>D0</Feature>
  </Command>
  <Command>
    <Name>Identify</Name>
    <CMD>B1</CMD>
    <Feature>D0</Feature>
  </Command>
    .
    .
    .
    .
</ATAPassThrough>

Answer 1

我建议使用XDocument来读取XML数据...例如在您的情况下,因为您已经有一个XML的TextReader,您可以将它传递给XDocument.Load方法...您的上述整个函数看起来像这样.

var doc = XDocument.Load(textReader);
foreach (var commandXml in doc.Descendants("Command"))
{
    var command = new ATAPassThroughCommands();
    var name = commandXml.Descendants("Name").Single().Value;
    // I'm not sure what this does but it looks important...
    CommandListContext.Add(name); 
    command.m_Name = name;
    command.m_CMD = 
         Convert.ToByte(commandXml.Descendants("CMD").Single().Value, 16);
    command.m_Feature = 
         Convert.ToByte(commandXml.Descendants("Feature").Single().Value, 16);
    m_ATACommands.Add(command);
}

显着容易.让框架为您做繁重的工作.

Answer 2

如果你有正常和一致的XML,可能最简单的方法就是使用XML Serializer.

首先创建与您的XML匹配的对象

[Serializable()]
public class Command
{
  [System.Xml.Serialization.XmlElement("Name")]
  public string Name { get; set; }

  [System.Xml.Serialization.XmlElement("CMD")]
  public string Cmd { get; set; }

  [System.Xml.Serialization.XmlElement("Feature")]
  public string Feature { get; set; }
}

[Serializable()]
[System.Xml.Serialization.XmlRoot("ATAPassthrough")]
public class CommandCollection
{
  [XmlArrayItem("Command", typeof(Command))]
  public Command[] Command { get; set; }
}

一个返回的方法 CommandCollection

public class CommandSerializer
{
  public commands Deserialize(string path)
  {
    CommandCollection commands = null;

    XmlSerializer serializer = new XmlSerializer(typeof(CommandCollection ));

    StreamReader reader = new StreamReader(path);
    reader.ReadToEnd();
    commands = (CommandCollection)serializer.Deserialize(reader);
    reader.Close();

    return commands ;
  }
}

不确定这是否完全正确,我没有办法测试它,但应该非常接近.