package main
import (
"fmt"
"github.com/streadway/amqp"
"time"
)
// Every connection should declare the topology they expect
func setup(url, queue string) (*amqp.Connection, *amqp.Channel, error) {
//setup connection
conn, err := amqp.Dial(url)
if err != nil {
return nil, nil, err
}
//build channel in the connection
ch, err := conn.Channel()
if err != nil {
return nil, nil, err
}
//queue declare
if _, err := ch.QueueDeclare(queue, false, true, false, false, nil); err != nil {
return nil, nil, err
}
return conn, ch, nil
}
func main() {
//amqp url
url := "amqp://guest:guest@127.0.0.1:5672";
for i := 1; i <= 2; i++ {
fmt.Println("connect ", i)
//two goroutine
go func() {
//queue name
queue := fmt.Sprintf("example.reconnect.%d", i)
//setup channel in the tcp connection
_, pub, err := setup(url, queue)
if err != nil {
fmt.Println("err publisher setup:", err)
return
}
// Purge the queue from the publisher side to establish initial state
if _, err := pub.QueuePurge(queue, false); err != nil {
fmt.Println("err purge:", err)
return
}
//publish msg
if err := pub.Publish("", queue, false, false, amqp.Publishing{
Body: []byte(fmt.Sprintf("%d", i)),
}); err != nil {
fmt.Println("err publish:", err)
return
}
//keep running
for{
time.Sleep(time.Second * 20)
}
}()
}
//keep running
for {
time.Sleep(time.Second * 20)
}
}
我认为程序和mq-server之间只有一个连接,
但有两个连接,一个连接只能支持一个通道,为什么?
两个goroutine可以共享相同的tcp连接吗?
套接字描述符可以在理论中共享进程的所有线程。
为什么两个goroutine不共享一个插槽但有自己的频道?
手工模型:
rabbitmq中的真实模型:
答案 0 :(得分:5)
看source for the library,好像你可以多次调用conn.Channel(),它会在同一个连接上创建一个新的通信流。
好的,我试过了,这是一个有效的例子......一个goroutine,一个连接,两个通道 我设置接收器,然后发送消息,然后从接收器通道
读取如果你想在一个goroutine中绑定多个队列,你可以调用rec.Consume两次,然后在队列中选择。
package main
import (
"fmt"
"github.com/streadway/amqp"
"os"
)
func main() {
conn, err := amqp.Dial("amqp://localhost")
e(err)
defer conn.Close()
fmt.Println("Connected")
rec, err := conn.Channel()
e(err)
fmt.Println("Setup receiver")
rq, err := rec.QueueDeclare("go-test", false, false, false, false, nil)
e(err)
msgs, err := rec.Consume(rq.Name, "", true, false, false, false, nil)
e(err)
fmt.Println("Setup sender")
send, err := conn.Channel()
e(err)
sq, err := send.QueueDeclare("go-test", false, false, false, false, nil)
e(err)
fmt.Println("Send message")
err = send.Publish("", sq.Name, false, false, amqp.Publishing{
ContentType: "text/plain",
Body: []byte("This is a test"),
})
e(err)
msg := <-msgs
fmt.Println("Received from:", rq, "msg:", string(msg.Body))
}
func e(err error) {
if err != nil {
fmt.Println(err)
os.Exit(1)
}
}
我的盒子上的输出:
$ go run rmq.go
Connected
Setup receiver
Setup sender
Send message
Received from: {go-test 0 0} msg: This is a test