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Golang - RabbitMq : channel/connection is not open

Tags:

go

rabbitmq

I'm new to golang, and I would like to refactorate my code so that the rabbitmq initialization is in another function that main. So I use a struct pointer (containing all the rabbitmq infos initilized) and pass it to the send function, but it tells me : Failed to publish a message: Exception (504) Reason: "channel/connection is not open"

struct :

type RbmqConfig struct {
    q amqp.Queue
    ch *amqp.Channel
    conn *amqp.Connection
    rbmqErr error
}

the init function :

func initRabbitMq() *RbmqConfig {

    config := &RbmqConfig{}

    config.conn, config.rbmqErr = amqp.Dial("amqp://guest:guest@localhost:5672/")
    failOnError(config.rbmqErr, "Failed to connect to RabbitMQ")
    defer config.conn.Close()

    config.ch, config.rbmqErr = config.conn.Channel()
    failOnError(config.rbmqErr, "Failed to open a channel")
    defer config.ch.Close()

    config.q, config.rbmqErr = config.ch.QueueDeclare(
        "<my_queue_name>",
        true,   // durable
        false,   // delete when unused
        false,   // exclusive
        false,   // no-wait
        nil,     // arguments
    )
    failOnError(config.rbmqErr, "Failed to declare a queue")

    return config
}

main :

config := initRabbitMq()

fmt.Println("queue name : ", config.q.Name)

sendMessage(config, <message_to_send>)

in send message :

func sendMessage(config *RbmqConfig, <message_to_send>) {

    config.rbmqErr = config.ch.Publish(
        "",           // exchange
        config.q.Name,       // routing key
        false,        // mandatory
        false,
        amqp.Publishing{
            DeliveryMode: amqp.Persistent,
            ContentType:  "text/plain",
            Body:         []byte(<message_to_send>),
        })
    failOnError(config.rbmqErr, "Failed to publish a message")

If someone has any idea, that would be very helpful. Thank you in advance

like image 224
Xys Avatar asked Apr 12 '16 17:04

Xys


1 Answers

Inside your init, you wrote defer config.conn.Close(), which will be executed when the function return. That is to say, whenever init finished, your connection will be closed, which causes unopen connection.

You need to defer the connection closing in main, or somewhere you want it to be closed.

like image 137
nevets Avatar answered Sep 27 '22 15:09

nevets