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?
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。