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
6年后更新:现在我认为下面的方法非常糟糕。为了创建重试逻辑，我们应该考虑使用Polly这样的库。
重试方法的异步实现:
public static async Task<T> DoAsync<T>(Func<dynamic> action, TimeSpan retryInterval, int retryCount = 3)
{
var exceptions = new List<Exception>();
for (int retry = 0; retry < retryCount; retry++)
{
try
{
return await action().ConfigureAwait(false);
}
catch (Exception ex)
{
exceptions.Add(ex);
}
await Task.Delay(retryInterval).ConfigureAwait(false);
}
throw new AggregateException(exceptions);
}
重点:我使用。configureawait (false);和Func<dynamic>代替Func<T>
其他回答
使用c# 6.0保持简单
public async Task<T> Retry<T>(Func<T> action, TimeSpan retryInterval, int retryCount)
{
try
{
return action();
}
catch when (retryCount != 0)
{
await Task.Delay(retryInterval);
return await Retry(action, retryInterval, --retryCount);
}
}
指数回退是比简单地尝试x次更好的重试策略。您可以使用Polly这样的库来实现它。
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);
}
}
以最新的方式实现了LBushkin的答案:
public static async Task Do(Func<Task> task, TimeSpan retryInterval, int maxAttemptCount = 3)
{
var exceptions = new List<Exception>();
for (int attempted = 0; attempted < maxAttemptCount; attempted++)
{
try
{
if (attempted > 0)
{
await Task.Delay(retryInterval);
}
await task();
return;
}
catch (Exception ex)
{
exceptions.Add(ex);
}
}
throw new AggregateException(exceptions);
}
public static async Task<T> Do<T>(Func<Task<T>> task, TimeSpan retryInterval, int maxAttemptCount = 3)
{
var exceptions = new List<Exception>();
for (int attempted = 0; attempted < maxAttemptCount; attempted++)
{
try
{
if (attempted > 0)
{
await Task.Delay(retryInterval);
}
return await task();
}
catch (Exception ex)
{
exceptions.Add(ex);
}
}
throw new AggregateException(exceptions);
}
要使用它:
await Retry.Do([TaskFunction], retryInterval, retryAttempts);
而函数[TaskFunction]可以是Task<T>，也可以只是Task。
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。
