我需要重试某个方法，直到它返回一个非空的Guid。
有一个很棒的answer根据是否有异常重试;但是，我想概括这个类能够处理任何指定的条件。
当前用法将执行特定次数的操作，直到没有例外：
Retry.Do(() => SomeFunctionThatCanFail(), TimeSpan.FromSeconds(1));
或：
Retry.Do(SomeFunctionThatCanFail, TimeSpan.FromSeconds(1));
或：
int result = Retry.Do(SomeFunctionWhichReturnsInt, TimeSpan.FromSeconds(1), 4);
如何修改此类，使其根据我传入的函数的返回值重试？
例如，如果我想重试，直到我的函数返回3：
Retry.Do(() => SomeFunctionThatCanFail(), TimeSpan.FromSeconds(1)).Until(3);
这意味着每1秒执行一次SomeFunctionThatCanFail（），直到SomeFunctionThatCanFail（）= 3？
在满足条件之前，我如何概括Retry.Do的用法？
public static class Retry
{
public static void Do(
Action action,
TimeSpan retryInterval,
int retryCount = 3)
{
Do<object>(() =>
{
action();
return null;
}, retryInterval, retryCount);
}
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++) //I would like to change this logic so that it will retry not based on whether there is an exception but based on the return value of Action
{
try
{
if (retry > 0)
Thread.Sleep(retryInterval);
return action();
}
catch (Exception ex)
{
exceptions.Add(ex);
}
}
throw new AggregateException(exceptions);
}
}
答案 0 :(得分：5)
如何创建以下界面：
public interface IRetryCondition<TResult>
{
TResult Until(Func<TResult, bool> condition);
}
public class RetryCondition<TResult> : IRetryCondition<TResult>
{
private TResult _value;
private Func<IRetryCondition<TResult>> _retry;
public RetryCondition(TResult value, Func<IRetryCondition<TResult>> retry)
{
_value = value;
_retry = retry;
}
public TResult Until(Func<TResult, bool> condition)
{
return condition(_value) ? _value : _retry().Until(condition);
}
}
然后，您将更新Retry静态类：
public static class Retry
{
// This method stays the same
// Returning an IRetryCondition does not make sense in a "void" action
public static void Do(
Action action,
TimeSpan retryInterval,
int retryCount = 3)
{
Do<object>(() =>
{
action();
return null;
}, retryInterval, retryCount);
}
// Return an IRetryCondition<T> instance
public static IRetryCondition<T> Do<T>(
Func<T> action,
TimeSpan retryInterval,
int retryCount = 3)
{
var exceptions = new List<Exception>();
for (int retry = 0; retry < retryCount; retry++)
{
try
{
if (retry > 0)
Thread.Sleep(retryInterval);
// We return a retry condition loaded with the return value of action() and telling it to execute this same method again if condition is not met.
return new RetryCondition<T>(action(), () => Do(action, retryInterval, retryCount));
}
catch (Exception ex)
{
exceptions.Add(ex);
}
}
throw new AggregateException(exceptions);
}
}
您将能够实现以下目标：
int result = Retry.Do(() => SomeFunctionThatCanFail(), TimeSpan.FromSeconds(1)).Until(r => r == 3);
我试图想出一个更“面向功能”的解决方案（有点类似于LINQ）：
首先，我们将有两个用于执行操作的接口：
public interface IRetryResult
{
void Execute();
}
public interface IRetryResult<out TResult>
{
TResult Execute();
}
然后，我们需要两个接口来配置重试操作：
public interface IRetryConfiguration : IRetryResult
{
IRetryConfiguration Times(int times);
IRetryConfiguration Interval(TimeSpan interval);
}
public interface IRetryConfiguration<out TResult> : IRetryResult<TResult>
{
IRetryConfiguration<TResult> Times(int times);
IRetryConfiguration<TResult> Interval(TimeSpan interval);
IRetryConfiguration<TResult> Until(Function<TResult, bool> condition);
}
最后，我们需要两个接口的两个实现：
public class ActionRetryConfiguration : IRetryConfiguration
{
private readonly Action _action;
private readonly int? _times;
private readonly TimeSpan? _interval;
public ActionRetryConfiguration(Action action, int? times, TimeSpan? interval)
{
_action = action;
_times = times;
_interval = interval;
}
public void Execute()
{
Execute(_action, _times, _interval);
}
private void Execute(Action action, int? times, TimeSpan? interval)
{
action();
if (times.HasValue && times.Value <= 1) return;
if (times.HasValue && interval.HasValue) Thread.Sleep(interval.Value);
Execute(action, times - 1, interval);
}
public IRetryConfiguration Times(int times)
{
return new ActionRetryConfiguration(_action, times, _interval);
}
public IRetryConfiguration Interval(TimeSpan interval)
{
return new ActionRetryConfiguration(_action, _times, interval);
}
}
public class FunctionRetryConfiguration<TResult> : IRetryConfiguration<TResult>
{
private readonly Func<TResult> _function;
private readonly int? _times;
private readonly TimeSpan? _interval;
private readonly Func<TResult, bool> _condition;
public FunctionRetryConfiguration(Func<TResult> function, int? times, TimeSpan? interval, Func<TResult, bool> condition)
{
_function = function;
_times = times;
_interval = interval;
_condition = condition;
}
public TResult Execute()
{
return Execute(_function, _times, _interval, _condition);
}
private TResult Execute(Func<TResult> function, int? times, TimeSpan? interval, Func<TResult, bool> condition)
{
TResult result = function();
if (condition != null && condition(result)) return result;
if (times.HasValue && times.Value <= 1) return result;
if ((times.HasValue || condition != null) && interval.HasValue) Thread.Sleep(interval.Value);
return Execute(function, times - 1, interval, condition);
}
public IRetryConfiguration<TResult> Times(int times)
{
return new FunctionRetryConfiguration<TResult>(_function, times, _interval, _condition);
}
public IRetryConfiguration<TResult> Interval(TimeSpan interval)
{
return new FunctionRetryConfiguration<TResult>(_function, _times, interval, _condition);
}
public IRetryConfiguration<TResult> Until(Func<TResult, bool> condition)
{
return new FunctionRetryConfiguration<TResult>(_function, _times, _interval, condition);
}
}
最后，Retry静态类，入口点：
public static class Retry
{
public static IRetryConfiguration Do(Action action)
{
return new ActionRetryConfiguration(action, 1, null);
}
public static IRetryConfiguration<TResult> Do<TResult>(Func<TResult> action)
{
return new FunctionRetryConfiguration<TResult>(action, 1, null, null);
}
}
我认为这种方法不那么多，而且更清洁。
此外，它允许你做这样的事情：
int result = Retry.Do(SomeIntMethod).Interval(TimeSpan.FromSeconds(1)).Until(n => n > 20).Execute();
Retry.Do(SomeVoidMethod).Times(4).Execute();
答案 1 :(得分：1)
微软的Reactive Framework（NuGet“Rx-Main”）已经构建了所有已经开发的运营商来开箱即用。
试试这个：
IObservable<int> query =
Observable
.Defer(() =>
Observable.Start(() => GetSomeValue()))
.Where(x => x == 1)
.Timeout(TimeSpan.FromSeconds(0.1))
.Retry()
.Take(1);
query
.Subscribe(x =>
{
// Can only be a `1` if produced in less than 0.1 seconds
Console.WriteLine(x);
});
答案 2 :(得分：0)
好吧，如果我理解了一切，这样的事情可以解决你的问题：
public static T Do<T>(Func<T> action, TimeSpan retryInterval, Predicate<T> predicate)
{
var exceptions = new List<Exception>();
try
{
bool succeeded;
T result;
do
{
result = action();
succeeded = predicate(result);
} while (!succeeded);
return result;
}
catch (Exception ex)
{
exceptions.Add(ex);
}
throw new AggregateException(exceptions);
}
将此方法添加到您的重试类。
我已尝试使用示例ConsoleApplication，并使用以下代码：
class Program
{
static void Main(string[] args)
{
var _random = new Random();
Func<int> func = () =>
{
var result = _random.Next(10);
Console.WriteLine(result);
return result;
};
Retry.Do(func, TimeSpan.FromSeconds(1), i => i == 5);
Console.ReadLine();
}
}
事实上，它在randoms 5时会停止。
答案 3 :(得分：-1)
好像你在想这个：
int returnValue = -1;
while (returnValue != 3)
{
returnValue = DoStuff();
// DoStuff should include a step to avoid maxing out cpu
}
return returnValue;
当然，＆＃34; 3＆＃34;可以是传递给函数的变量。