有时,我需要在放弃之前将一个操作重试几次。我的代码是:

int retries = 3;
while(true) {
  try {
    DoSomething();
    break; // success!
  } catch {
    if(--retries == 0) throw;
    else Thread.Sleep(1000);
  }
}

我想在一个通用的重试函数中重写这个:

TryThreeTimes(DoSomething);

这在c#中可行吗?TryThreeTimes()方法的代码是什么?


当前回答

允许函数和重试消息

public static T RetryMethod<T>(Func<T> method, int numRetries, int retryTimeout, Action onFailureAction)
{
 Guard.IsNotNull(method, "method");            
 T retval = default(T);
 do
 {
   try
   {
     retval = method();
     return retval;
   }
   catch
   {
     onFailureAction();
      if (numRetries <= 0) throw; // improved to avoid silent failure
      Thread.Sleep(retryTimeout);
   }
} while (numRetries-- > 0);
  return retval;
}

其他回答

用c#、Java或其他语言简单地完成:

  internal class ShouldRetryHandler {
    private static int RETRIES_MAX_NUMBER = 3;
    private static int numberTryes;

    public static bool shouldRetry() {
        var statusRetry = false;

        if (numberTryes< RETRIES_MAX_NUMBER) {
            numberTryes++;
            statusRetry = true;
            //log msg -> 'retry number' + numberTryes

        }

        else {
            statusRetry = false;
            //log msg -> 'reached retry number limit' 
        }

        return statusRetry;
    }
}

并在你的代码中简单地使用它:

 void simpleMethod(){
    //some code

    if(ShouldRetryHandler.shouldRetry()){
    //do some repetitive work
     }

    //some code    
    }

或者你可以在递归方法中使用它:

void recursiveMethod(){
    //some code

    if(ShouldRetryHandler.shouldRetry()){
    recursiveMethod();
     }

    //some code    
    }

您还可以考虑添加要重试的异常类型。例如,这是您想要重试的超时异常吗?数据库异常?

RetryForExcpetionType(DoSomething, typeof(TimeoutException), 5, 1000);

public static void RetryForExcpetionType(Action action, Type retryOnExceptionType, int numRetries, int retryTimeout)
{
    if (action == null)
        throw new ArgumentNullException("action");
    if (retryOnExceptionType == null)
        throw new ArgumentNullException("retryOnExceptionType");
    while (true)
    {
        try
        {
            action();
            return;
        }
        catch(Exception e)
        {
            if (--numRetries <= 0 || !retryOnExceptionType.IsAssignableFrom(e.GetType()))
                throw;

            if (retryTimeout > 0)
                System.Threading.Thread.Sleep(retryTimeout);
        }
    }
}

您可能还注意到,所有其他示例在测试retries == 0时都存在类似的问题,要么重试无穷大,要么在给定负值时无法引发异常。Sleep(-1000)在上面的catch块中也会失败。这取决于你期望人们有多“愚蠢”,但防御性编程永远不会伤害到你。

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。

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);          
      }
}