我在那天写这门课是为了做我认为的事情，类似于你想做的事情。您可以对希望序列化为 XML 的对象使用此类的方法。例如，给定一个员工...
使用实用程序；使用 System.Xml.Serialization；
[XmlRoot("Employee")] public class Employee { private String name = "Steve";
[XmlElement("Name")]
public string Name { get { return name; } set{ name = value; } }
public static void Main(String[] args)
{
Employee e = new Employee();
XmlObjectSerializer.Save("c:\steve.xml", e);
}
}
此代码应输出：
<Employee>
<Name>Steve</Name>
</Employee>
对象类型 (Employee) 必须是可序列化的。尝试 [Serializable(true)]。我在某个地方有此代码的更好版本，我在编写它时只是在学习。不管怎样，看看下面的代码。我在某个项目中使用它，所以它肯定有效。
using System;
using System.IO;
using System.Xml.Serialization;
namespace Utilities
{
/// <summary>
/// Opens and Saves objects to Xml
/// </summary>
/// <projectIndependent>True</projectIndependent>
public static class XmlObjectSerializer
{
/// <summary>
/// Serializes and saves data contained in obj to an XML file located at filePath <para></para>
/// </summary>
/// <param name="filePath">The file path to save to</param>
/// <param name="obj">The object to save</param>
/// <exception cref="System.IO.IOException">Thrown if an error occurs while saving the object. See inner exception for details</exception>
public static void Save(String filePath, Object obj)
{
// allows access to the file
StreamWriter oWriter = null;
try
{
// Open a stream to the file path
oWriter = new StreamWriter(filePath);
// Create a serializer for the object's type
XmlSerializer oSerializer = new XmlSerializer(obj.GetType());
// Serialize the object and write to the file
oSerializer.Serialize(oWriter.BaseStream, obj);
}
catch (Exception ex)
{
// throw any errors as IO exceptions
throw new IOException("An error occurred while saving the object", ex);
}
finally
{
// if a stream is open
if (oWriter != null)
{
// close it
oWriter.Close();
}
}
}
/// <summary>
/// Deserializes saved object data of type T in an XML file
/// located at filePath
/// </summary>
/// <typeparam name="T">Type of object to deserialize</typeparam>
/// <param name="filePath">The path to open the object from</param>
/// <returns>An object representing the file or the default value for type T</returns>
/// <exception cref="System.IO.IOException">Thrown if the file could not be opened. See inner exception for details</exception>
public static T Open<T>(String filePath)
{
// gets access to the file
StreamReader oReader = null;
// the deserialized data
Object data;
try
{
// Open a stream to the file
oReader = new StreamReader(filePath);
// Create a deserializer for the object's type
XmlSerializer oDeserializer = new XmlSerializer(typeof(T));
// Deserialize the data and store it
data = oDeserializer.Deserialize(oReader.BaseStream);
//
// Return the deserialized object
// don't cast it if it's null
// will be null if open failed
//
if (data != null)
{
return (T)data;
}
else
{
return default(T);
}
}
catch (Exception ex)
{
// throw error
throw new IOException("An error occurred while opening the file", ex);
}
finally
{
// Close the stream
oReader.Close();
}
}
}
}