public delegate void ThingToTryDeletage();

public static void TryNTimes(ThingToTryDelegate, int N, int sleepTime)
{
   while(true)
   {
      try
      {
        ThingToTryDelegate();
      } catch {

            if( --N == 0) throw;
          else Thread.Sleep(time);          
      }
}

我将在接受的答案中添加以下代码

public static class Retry<TException> where TException : Exception //ability to pass the exception type
    {
        //same code as the accepted answer ....

        public static T Do<T>(Func<T> action, TimeSpan retryInterval, int retryCount = 3)
        {
            var exceptions = new List<Exception>();

            for (int retry = 0; retry < retryCount; retry++)
            {
                try
                {
                    return action();
                }
                catch (TException ex) //Usage of the exception type
                {
                    exceptions.Add(ex);
                    Thread.Sleep(retryInterval);
                }
            }

            throw new AggregateException(String.Format("Failed to excecute after {0} attempt(s)", retryCount), exceptions);
        }
    }

基本上，上面的代码使Retry类成为泛型，这样您就可以传递想要捕获的异常类型进行重试。

现在，以几乎相同的方式使用它，但指定异常类型

Retry<EndpointNotFoundException>.Do(() => SomeFunctionThatCanFail(), TimeSpan.FromSeconds(1));

我使用Polly实现了该模式的两个实现。其一是异步。

我的同步方法是基于Erik Bergstedt的回答

public static T Retry<T>(Func<T> action, TimeSpan retryWait, int retryCount = 0)
{
    PolicyResult<T> policyResult = Policy
        .Handle<ApiException>(ex => ex.ResponseCode == (int)HttpStatusCode.TooManyRequests)
        .WaitAndRetry(retryCount, retryAttempt => retryWait)
        .ExecuteAndCapture(action);

    if (policyResult.Outcome == OutcomeType.Failure)
    {
        throw policyResult.FinalException;
    }

    return policyResult.Result;
}

异步:

public static async Task<T> RetryAsync<T>(Func<Task<T>> action, TimeSpan retryWait, int retryCount = 0)
{
    PolicyResult<T> policyResult = await Policy
        .Handle<ApiException>(ex => ex.ResponseCode == (int)HttpStatusCode.TooManyRequests)
        .WaitAndRetryAsync(retryCount, retryAttempt => retryWait)
        .ExecuteAndCaptureAsync(action);

    if (policyResult.Outcome == OutcomeType.Failure)
    {
        throw policyResult.FinalException;
    }

    return policyResult.Result;
}

允许传入异常类型以及异常类型的lambda也很容易。

对于那些既想对任何异常进行重试，又想显式设置异常类型的人，可以使用以下方法:

public class RetryManager 
{
    public void Do(Action action, 
                    TimeSpan interval, 
                    int retries = 3)
    {
        Try<object, Exception>(() => {
            action();
            return null;
        }, interval, retries);
    }

    public T Do<T>(Func<T> action, 
                    TimeSpan interval, 
                    int retries = 3)
    {
        return Try<T, Exception>(
              action
            , interval
            , retries);
    }

    public T Do<E, T>(Func<T> action, 
                       TimeSpan interval, 
                       int retries = 3) where E : Exception
    {
        return Try<T, E>(
              action
            , interval
            , retries);
    }

    public void Do<E>(Action action, 
                       TimeSpan interval, 
                       int retries = 3) where E : Exception
    {
        Try<object, E>(() => {
            action();
            return null;
        }, interval, retries);
    }

    private T Try<T, E>(Func<T> action, 
                       TimeSpan interval, 
                       int retries = 3) where E : Exception
    {
        var exceptions = new List<E>();

        for (int retry = 0; retry < retries; retry++)
        {
            try
            {
                if (retry > 0)
                    Thread.Sleep(interval);
                return action();
            }
            catch (E ex)
            {
                exceptions.Add(ex);
            }
        }

        throw new AggregateException(exceptions);
    }
}

我是递归和扩展方法的粉丝，所以这里是我的观点:

public static void InvokeWithRetries(this Action @this, ushort numberOfRetries)
{
    try
    {
        @this();
    }
    catch
    {
        if (numberOfRetries == 0)
            throw;

        InvokeWithRetries(@this, --numberOfRetries);
    }
}

我需要传递一些参数给我的方法来重试，并有一个结果值;所以我需要一个表达。 我建立了这个类来做这个工作(它是受布什金的启发) 你可以这样使用它:

static void Main(string[] args)
{
    // one shot
    var res = Retry<string>.Do(() => retryThis("try"), 4, TimeSpan.FromSeconds(2), fix);

    // delayed execute
    var retry = new Retry<string>(() => retryThis("try"), 4, TimeSpan.FromSeconds(2), fix);
    var res2 = retry.Execute();
}

static void fix()
{
    Console.WriteLine("oh, no! Fix and retry!!!");
}

static string retryThis(string tryThis)
{
    Console.WriteLine("Let's try!!!");
    throw new Exception(tryThis);
}

public class Retry<TResult>
{
    Expression<Func<TResult>> _Method;
    int _NumRetries;
    TimeSpan _RetryTimeout;
    Action _OnFailureAction;

    public Retry(Expression<Func<TResult>> method, int numRetries, TimeSpan retryTimeout, Action onFailureAction)
    {
        _Method = method;
        _NumRetries = numRetries;
        _OnFailureAction = onFailureAction;
        _RetryTimeout = retryTimeout;
    }

    public TResult Execute()
    {
        TResult result = default(TResult);
        while (_NumRetries > 0)
        {
            try
            {
                result = _Method.Compile()();
                break;
            }
            catch
            {
                _OnFailureAction();
                _NumRetries--;
                if (_NumRetries <= 0) throw; // improved to avoid silent failure
                Thread.Sleep(_RetryTimeout);
            }
        }
        return result;
    }

    public static TResult Do(Expression<Func<TResult>> method, int numRetries, TimeSpan retryTimeout, Action onFailureAction)
    {
        var retry = new Retry<TResult>(method, numRetries, retryTimeout, onFailureAction);
        return retry.Execute();
    }
}

ps。 b布什金的解再重试一次=D