您如何获得方法的参数NAMES。这些示例说明如何获取参数的[[values,而不是NAMES。我想看到parma = 99,parmb =1。不只是99,1。 using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
using System.Threading.Tasks;
using System.Diagnostics;
using PostSharp.Aspects;
namespace GettingParmNames
{
public class Program
{
static void Main(string[] args)
{
Foo myfoo = new Foo();
int sum = myfoo.DoA(99, 1);
Console.WriteLine(sum.ToString());
Console.ReadKey();
}
}
public class Foo
{
[ExceptionAspect]
public int DoA(int parma, int parmb)
{
int retVal;
try
{
retVal = parma + parmb;
if (parma == 99)
{
throw new Exception("Fake Exception");
}
}
catch (Exception ex)
{
retVal = -1;
throw new Exception("There was a problem");
}
return retVal;
}
}
[Serializable]
public class ExceptionAspect : OnExceptionAspect
{
public override void OnException(MethodExecutionArgs args)
{
string parameterValues = "";
foreach (object arg in args.Arguments)
{
if (parameterValues.Length > 0)
{
parameterValues += ", ";
}
if (arg != null)
{
parameterValues += arg.ToString();
}
else
{
parameterValues += "null";
}
}
Console.WriteLine("Exception {0} in {1}.{2} invoked with arguments {3}", args.Exception.GetType().Name, args.Method.DeclaringType.FullName, args.Method.Name, parameterValues );
}
}
}
如何获得方法的参数名称。这些示例向您展示了如何获取参数的值,而不是名称。我想看到parma = 99,parmb =1。不只是99,1.使用...
OnException
在args.Method.GetParameters()
方法中访问方法参数的信息。但是通常出于性能原因,最好在编译时初始化数据-在CompileTimeInitialize方法覆盖中。