CODEWITHSUNDEEP
c#exceptionpolly
user2197446
user2197446

Reputation: 1146

TimeoutException unhandled by the code, do I need a try catch when using Polly?

I've been working with a remote web service over an unreliably and/or slow VPN that brought to light a point of failure in my code when I call a web service I will get timeout exceptions.

I did a little googling and found Polly and seemed to be exactly what I needed, but I'm still getting an unhandled TimeoutException and would like to know what I'm doing wrong and how to update the code so that the TimeoutException is handled, preferably using Polly.

        var networkPolicy = Policy
              .Handle<TimeoutException>()
              .Or<CommunicationException>()
              .WaitAndRetry(
                   5,
                   retryAttempt => TimeSpan.FromSeconds(Math.Pow(2, retryAttempt)),
                   (exception, timeSpan, context) =>
                   {
                        System.Diagnostics.Debug.WriteLine("Exception being retried" + exception );
                   });
        // The following line is giving me the exception.
        response = networkPolicy.Execute(() => soapClient.WebServiceFunction(request));

I'd also like to know if it is best practice to define the policy as a static readonly variable?

Upvotes: 1

Views: 1811

Answers (1)

Bartosz Czerwonka
Bartosz Czerwonka

Reputation: 1651

Polly works so that repeated operations if an error occurs, but only a certain number of times. If the error occurs more times Polly throws it.

If you want to repeat it until when it is not thrown expetion then use RetryForever.

Usually I use Polly with a try catch - like here:

    try
    {
        return
            await
            Policy.Handle<MongoConnectionException>()
                  .RetryAsync(3,
                      (exception, i) =>
                          {
                              logger.Warn(exception,
                                  string.Format("Mongo Connection Exception - Retry Count : {0}", i));
                          })
                  .ExecuteAsync(async () => await operation());
    }
    catch (MongoConnectionException ex)
    {
        logger.Error(ex);
        return null;
    }

Upvotes: 1

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