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

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));

其他回答

我已经实现了一个异步版本的接受的答案,就像这样-它似乎工作得很好-有评论吗?


        public static async Task DoAsync(
            Action action,
            TimeSpan retryInterval,
            int maxAttemptCount = 3)
        {
            DoAsync<object>(() =>
            {
                action();
                return null;
            }, retryInterval, maxAttemptCount);
        }

        public static async Task<T> DoAsync<T>(
            Func<Task<T>> action,
            TimeSpan retryInterval,
            int maxAttemptCount = 3)
        {
            var exceptions = new List<Exception>();

            for (int attempted = 0; attempted < maxAttemptCount; attempted++)
            {
                try
                {
                    if (attempted > 0)
                    {
                        Thread.Sleep(retryInterval);
                    }
                    return await action();
                }
                catch (Exception ex)
                {
                    exceptions.Add(ex);
                }
            }
            throw new AggregateException(exceptions);
        }

并且,简单地像这样称呼它:

var result = await Retry.DoAsync(() => MyAsyncMethod(), TimeSpan.FromSeconds(5), 4);

重试助手:一个通用的java实现,它包含可返回类型和无效类型的重试。

import java.util.function.Supplier;
import org.slf4j.Logger;
import org.slf4j.LoggerFactory;

public class RetryHelper {
  private static final Logger log = LoggerFactory.getLogger(RetryHelper.class);
  private int retryWaitInMS;
  private int maxRetries;

  public RetryHelper() {
    this.retryWaitInMS = 300;
    this.maxRetries = 3;
  }

  public RetryHelper(int maxRetry) {
    this.maxRetries = maxRetry;
    this.retryWaitInMS = 300;
  }

  public RetryHelper(int retryWaitInSeconds, int maxRetry) {
    this.retryWaitInMS = retryWaitInSeconds;
    this.maxRetries = maxRetry;
  }

  public <T> T retryAndReturn(Supplier<T> supplier) {
    try {
      return supplier.get();
    } catch (Exception var3) {
      return this.retrySupplier(supplier);
    }
  }

  public void retry(Runnable runnable) {
    try {
      runnable.run();
    } catch (Exception var3) {
      this.retrySupplier(() -> {
        runnable.run();
        return null;
      });
    }

  }

  private <T> T retrySupplier(Supplier<T> supplier) {
    log.error("Failed <TASK>, will be retried " + this.maxRetries + " times.");
    int retryCounter = 0;

    while(retryCounter < this.maxRetries) {
      try {
        return supplier.get();
      } catch (Exception var6) {
        ++retryCounter;
        log.error("<TASK> failed on retry: " + retryCounter + " of " + this.maxRetries + " with error: " + var6.getMessage());
        if (retryCounter >= this.maxRetries) {
          log.error("Max retries exceeded.");
          throw var6;
        }

        try {
          Thread.sleep((long)this.retryWaitInMS);
        } catch (InterruptedException var5) {
          var5.printStackTrace();
        }
      }
    }

    return supplier.get();
  }

  public int getRetryWaitInMS() {
    return this.retryWaitInMS;
  }

  public int getMaxRetries() {
    return this.maxRetries;
  }
}

用法:

    try {
      returnValue = new RetryHelper().retryAndReturn(() -> performSomeTask(args));
      //or no return type:
      new RetryHelper().retry(() -> mytask(args));
    } catch(Exception ex){
      log.error(e.getMessage());
      throw new CustomException();
    }

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

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);          
      }
}

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。