C#静态类型不能用作参数

Question

public static void SendEmail(String from, String To, String Subject, String HTML, String AttachmentPath = null, String AttachmentName = null, MediaTypeNames AttachmentType = null)
{
    ....

    // Add an attachment if required
    if (AttachmentPath != null)
    {
        var ct = new ContentType(MediaTypeNames.Text.Plain);
        using (var a = new Attachment(AttachmentPath, ct)
                    {
                        Name = AttachmentName,
                        NameEncoding = Encoding.UTF8,
                        TransferEncoding = TransferEncoding.Base64
                    })
        {
            mailMessage.Attachments.Add(a);
        }
    }

    ....
}

正如您所看到的那样MediaTypeNames AttachmentType抛出错误:

'System.Net.Mime.MediaTypeNames': static types cannot be used as parameters

处理这个问题的最佳方法是什么？

Answer 1

您不能将静态类型作为参数传递给方法,因为它必须实例化,并且您无法创建static类的实例.

Answer 2

不建议这样做,但您可以模拟使用静态类作为参数.像这样创建一个Instance类:

public class Instance
{

    public Type StaticObject { get; private set; }

    public Instance(Type staticType)
    {
        StaticObject = staticType;
    }

    public object Call(string name, params object[] parameters)
    {
        MethodInfo method = StaticObject.GetMethod(name);
        return method.Invoke(StaticObject, parameters);
    }

    public object Call(string name)
    {
        return Call(name, null);
    }

}

然后你的函数将使用静态类:

    private static void YourFunction(Instance instance)
    {
        instance.Call("TheNameOfMethodToCall", null);
    }

例如.电话:

• 第一个参数是要调用的静态类的方法的名称
• 第二个参数是要传递给方法的参数列表.

并使用这样的:

    static void Main(string[] args)
    {

        YourFunction(new Instance(typeof(YourStaticClass)));

        Console.ReadKey();

    }