rabbitmq-tutorials/go/receive_logs.go
Simon Alling bd2c06f28f Clarify forever channel declarations
The current idiom

    forever := make(chan bool)

gives the impression that booleans will be sent through `forever`.  But
the channel never has anything sent through it; its purpose is just to
block indefinitely.

Go doesn't have any bottom type (empty type; ⊥), but it does have a unit
type, namely `struct{}`.  Because not even such values will be sent, the
channel need not even exist at all, leaving no doubts as to its purpose.
2022-03-09 16:12:15 +01:00

75 lines
1.4 KiB
Go

package main
import (
"log"
amqp "github.com/rabbitmq/amqp091-go"
)
func failOnError(err error, msg string) {
if err != nil {
log.Panicf("%s: %s", msg, err)
}
}
func main() {
conn, err := amqp.Dial("amqp://guest:guest@localhost:5672/")
failOnError(err, "Failed to connect to RabbitMQ")
defer conn.Close()
ch, err := conn.Channel()
failOnError(err, "Failed to open a channel")
defer ch.Close()
err = ch.ExchangeDeclare(
"logs", // name
"fanout", // type
true, // durable
false, // auto-deleted
false, // internal
false, // no-wait
nil, // arguments
)
failOnError(err, "Failed to declare an exchange")
q, err := ch.QueueDeclare(
"", // name
false, // durable
false, // delete when unused
true, // exclusive
false, // no-wait
nil, // arguments
)
failOnError(err, "Failed to declare a queue")
err = ch.QueueBind(
q.Name, // queue name
"", // routing key
"logs", // exchange
false,
nil)
failOnError(err, "Failed to bind a queue")
msgs, err := ch.Consume(
q.Name, // queue
"", // consumer
true, // auto-ack
false, // exclusive
false, // no-local
false, // no-wait
nil, // args
)
failOnError(err, "Failed to register a consumer")
var forever chan struct{}
go func() {
for d := range msgs {
log.Printf(" [x] %s", d.Body)
}
}()
log.Printf(" [*] Waiting for logs. To exit press CTRL+C")
<-forever
}