英文:
Whether to create connection every time when amqp.Dial is threadsafe or not in go lang
问题
根据RabbitMQ文档中提到的,建立TCP连接是昂贵的。因此,引入了通道的概念。现在我遇到了这个示例。在main()
函数中,每次发布消息时都会创建连接。
conn, err := amqp.Dial("amqp://guest:guest@localhost:5672/")
。
它不应该在全局声明一次,并且应该有故障转移机制,以防连接关闭,就像单例对象一样。如果amqp.Dial是线程安全的,我认为它应该是线程安全的。
编辑后的问题:
我以以下方式处理连接错误。在错误时,我监听一个通道并创建一个新的连接。但是当我关闭现有连接并尝试发布消息时,我会收到以下错误。
错误:
2016/03/30 19:20:08 Failed to open a channel: write tcp 172.16.5.48:51085->172.16.0.20:5672: use of closed network connection
exit status 1
7:25 PM
代码:
func main() {
Conn, err := amqp.Dial("amqp://guest:guest@172.16.0.20:5672/")
failOnError(err, "Failed to connect to RabbitMQ")
context := &appContext{queueName: "QUEUENAME", exchangeName: "ExchangeName", exchangeType: "direct", routingKey: "RoutingKey", conn: Conn}
c := make(chan *amqp.Error)
go func() {
error := <-c
if(error != nil){
Conn, err = amqp.Dial("amqp://guest:guest@172.16.0.20:5672/")
failOnError(err, "Failed to connect to RabbitMQ")
Conn.NotifyClose(c)
}
}()
Conn.NotifyClose(c)
r := web.New()
// We pass an instance to our context pointer, and our handler.
r.Get("/", appHandler{context, IndexHandler})
graceful.ListenAndServe(":8086", r)
}
英文:
As it is mentioned in the RabbitMQ docs that tcp connections are expensive to make. So, for that concept of channel was introduced. Now i came across this example. In the main()
it creates the connection everytime a message is publised.
conn, err := amqp.Dial("amqp://guest:guest@localhost:5672/")
.
Shouldn't it be declared globally once and there should be failover mechanism in case connection get closed like singleton object. If amqp.Dial is thread-safe, which i suppose it should be
Edited question :
I am handling the connection error in the following manner. In which i listen on a channel and create a new connection on error. But when i kill the existing connection and try to publish message. I get the following error.
error :
2016/03/30 19:20:08 Failed to open a channel: write tcp 172.16.5.48:51085->172.16.0.20:5672: use of closed network connection
exit status 1
7:25 PM
Code :
func main() {
Conn, err := amqp.Dial("amqp://guest:guest@172.16.0.20:5672/")
failOnError(err, "Failed to connect to RabbitMQ")
context := &appContext{queueName: "QUEUENAME",exchangeName: "ExchangeName",exchangeType: "direct",routingKey: "RoutingKey",conn: Conn}
c := make(chan *amqp.Error)
go func() {
error := <-c
if(error != nil){
Conn, err = amqp.Dial("amqp://guest:guest@172.16.0.20:5672/")
failOnError(err, "Failed to connect to RabbitMQ")
Conn.NotifyClose(c)
}
}()
Conn.NotifyClose(c)
r := web.New()
// We pass an instance to our context pointer, and our handler.
r.Get("/", appHandler{context, IndexHandler})
graceful.ListenAndServe(":8086", r)
}
答案1
得分: 11
当然,你不应该为每个请求创建一个连接。将其设置为全局变量,或者更好地将其作为应用程序上下文的一部分,在启动时初始化一次。
您可以通过使用Connection.NotifyClose
注册一个通道来处理连接错误:
func initialize() {
c := make(chan *amqp.Error)
go func() {
err := <-c
log.Println("reconnect: " + err.Error())
initialize()
}()
conn, err := amqp.Dial("amqp://guest:guest@localhost:5672/")
if err != nil {
panic("cannot connect")
}
conn.NotifyClose(c)
// create topology
}
英文:
Of course, you shouldn't create a connection for each request. Make it a global variable or better part of an application context which you initialize once at startup.
You can handle connection errors by registering a channel using Connection.NotifyClose
:
func initialize() {
c := make(chan *amqp.Error)
go func() {
err := <-c
log.Println("reconnect: " + err.Error())
initialize()
}()
conn, err := amqp.Dial("amqp://guest:guest@localhost:5672/")
if err != nil {
panic("cannot connect")
}
conn.NotifyClose(c)
// create topology
}
通过集体智慧和协作来改善编程学习和解决问题的方式。致力于成为全球开发者共同参与的知识库,让每个人都能够通过互相帮助和分享经验来进步。
评论