标签:str 抛出异常 password 处理过程 info icon ESS 服务 span
默认情况下,EasyNetQ的消息处理过程中,如果throw exception,那么,依然是认为消息已经送达,不会再次推送,为了让RabbitMQ再次推送,可以这么实现:
public sealed class AlwaysRequeueErrorStrategy : IConsumerErrorStrategy
{
public void Dispose()
{
}
public AckStrategy HandleConsumerError(ConsumerExecutionContext context, Exception exception)
{
return AckStrategies.NackWithRequeue;
}
public AckStrategy HandleConsumerCancelled(ConsumerExecutionContext context)
{
return AckStrategies.NackWithRequeue;
}
}
[TestMethod]
public void test()
{
var hostName = _configuration["RabbitMQ:HostName"];
var mqport = _configuration["RabbitMQ:Port"];
var userName = _configuration["RabbitMQ:UserName"];
var password = _configuration["RabbitMQ:Password"];
var connectionConfiguration = new ConnectionConfiguration
{
Hosts = new List<HostConfiguration>
{
new HostConfiguration
{
Host = hostName,
Port = Convert.ToUInt16(mqport)
}
},
Port = Convert.ToUInt16(mqport),
VirtualHost = "/",
UserName = userName,
Password = password
};
var _bus = RabbitHutch.CreateBus(connectionConfiguration , x=> {
x.Register<IConsumerErrorStrategy, AlwaysRequeueErrorStrategy>();
});
var exchange = _bus.Advanced.ExchangeDeclare("ExchangeNAME_test", ExchangeType.Topic);
var nickNameQu = _bus.Advanced.QueueDeclare("queuename2222");
int count = 0;
_bus.Advanced.Bind(exchange, nickNameQu, "RouterKey1");
_bus.Advanced.Consume(nickNameQu, (data, properties, info) =>
{
count++;
if (count < 2)
throw new Exception("my error");
var msg = Encoding.UTF8.GetString(data);
});
Task.Run(() => {
_bus.PublishAsync(new UpdateAvaterMessage() { Avater = "abcc22" }, "RouterKey1");
});
Thread.Sleep(100000);
}
原理是重写IConsumerErrorStrategy
(如果不是使用EasyNetQ,传统的RabbitMQ库是用手动ack实现)
另外,如果消息处理过程中,服务器死机,这种情况消息是会重发的,不需要担心
EasyNetQ(RabbitMQ)在处理消息时,如果抛出异常,继续不断发送到订阅队列,不断处理(也就是不自动确认消息已到达)
标签:str 抛出异常 password 处理过程 info icon ESS 服务 span
原文地址:https://www.cnblogs.com/IWings/p/14681745.html