一个连接能否在go API中为Rabbitmq支持多个通道？

Question

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)
    }
}

I thought there is only one connection between the program and mq-server,

but there are two connection,one connection can only support one channel,why?

can't the two goroutine share the same tcp connection?

Socket descriptor can share in all threads of a process in the theory.

Why the two goroutine don't share one socket but have their own channel?

The model by hand:

The real model in rabbitmq:

Answer 1

Looking at the source for the library it appears as though you can call conn.Channel() as many times as you like and it creates a new stream of communication over the same connection.

Ok, I tried it, here's a working example... One goroutine, one connection, two channels I setup the receiver, then send a message, then read from the receiver channel

if you wanted multiple queue's bound in one goroutine, you would call rec.Consume twice and then select across the queues.

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)
    }
}

Output on my box:

$ go run rmq.go 
Connected
Setup receiver
Setup sender
Send message
Received from: {go-test 0 0} msg: This is a test