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?
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
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。