Keep a MQTT Go client running - go

Question

I think it is a silly question, I need a MQTT Client to keep running after connection and subscription. I never encountered the problem because my MQTT clients are always coupled with an HTTP server, and when launching a HTTP server, the code don't stop running.
But in the present use case I only need a MQTT Client to subscribe to some topic and stay alive.
Here is what I do (the function just connect to a broker and subcribe to one topic.)
func main() {
godotenv.Load("./.env")
_initMqttConnection()
}
I need the client to stay connected and not stop just after the subscription is done.
How to perform that simple thing ?
Edit 1 : Complete Code
package main
import (
"encoding/json"
"fmt"
"log"
"net/http"
"os"
"path/filepath"
"strings"
"github.com/yosssi/gmq/mqtt"
"github.com/yosssi/gmq/mqtt/client"
"github.com/joho/godotenv"
"github.com/skratchdot/open-golang/open"
)
var cli *client.Client
func _initMqttConnection() {
cli = client.New(&client.Options{
ErrorHandler: func(err error) {
fmt.Println(err)
},
})
defer cli.Terminate()
log.Println("Connecting to " + os.Getenv("mqtt_host"))
err := cli.Connect(&client.ConnectOptions{
Network: "tcp",
Address: os.Getenv("mqtt_host"),
UserName: []byte(os.Getenv("mqtt_user")),
Password: []byte(os.Getenv("mqtt_password")),
ClientID: []byte("mqtt_video_launcher"),
})
if err != nil {
log.Println("Error 1")
panic(err)
}
log.Println("Connected to MQTT")
topic_to_sub := []byte("/" + os.Getenv("video_topic"))
err = cli.Subscribe(&client.SubscribeOptions{
SubReqs: []*client.SubReq{
&client.SubReq{
TopicFilter: topic_to_sub,
QoS: mqtt.QoS0,
Handler: func(topicName, message []byte) {
//do struff with message
fmt.Println(string(topicName), string(message))
},
},
},
})
if err != nil {
panic(err)
}
log.Println("Subscription OK : " + string(topic_to_sub[:len(topic_to_sub)]))
}
func main() {
godotenv.Load("./.env")
_initMqttConnection()
}
The temporary solution I use is adding :
http.ListenAndServe(":", nil)
at the end.

Answer 1

You have to make the program run infinitely or unless you want to explicitly end it (Cntrl c). One good solution that worked for me is to wait for a channel before exiting the main function and that channel can keep listening for an interrupt.
Eg:
func main() {
keepAlive := make(chan os.Signal)
signal.Notify(keepAlive, os.Interrupt, syscall.SIGTERM)
// All your code
<-keepAlive
}