Question

这是我的代码，它加载一个dll文件并在xml中写入有关该文件的信息：

MyOptions ass = MyOptionsList[MyOptionsList.Count - 1];
AppDomain.CurrentDomain.AssemblyResolve += CurrentDomain_AssemblyResolve;
StringWriter sw = new StringWriter();
XmlSerializer serializer = new XmlSerializer(ass.GetType(),
    new Type[] { typeof(MyOptions) });
serializer.Serialize(sw, ass);
//more code writing in an xml file
AppDomain.CurrentDomain.AssemblyResolve -= CurrentDomain_AssemblyResolve;

Assembly CurrentDomain_AssemblyResolve(object sender, ResolveEventArgs args)
{
    return Assembly.LoadFrom(correctPath + ".dll");
}

问题发生在第serializer.Serialize(sw, ass);行。内部异常表示我无法从一个dll转换为另一个dll。两个dll文件都具有相同的名称。一个在correctPath，第二个在另一条路径上。如何解决这个问题？

MyOptions是一个类库项目，我正在加载的dll正在扩展。

Answer 1

我只是尝试了下面没有问题，我认为你的问题必须在其他地方。

public class MyOptions
{
  public int n;
}

static Assembly CurrentDomain_AssemblyResolve(object sender, ResolveEventArgs args)
{
  return Assembly.LoadFrom(@"C:\windwos\test.dll");
}

static void Main(string[] args)
{
  MyOptions ass = new MyOptions();
  AppDomain.CurrentDomain.AssemblyResolve += CurrentDomain_AssemblyResolve;
  StringWriter sw = new StringWriter();
  XmlSerializer serializer = new XmlSerializer(ass.GetType(), new Type[] { typeof(MyOptions) });
  serializer.Serialize(sw, ass);
  //more code writing in an xml file
  AppDomain.CurrentDomain.AssemblyResolve -= CurrentDomain_AssemblyResolve;
}

请注意，不会调用事件代码。

装配无法施放

1 个答案: