2

I had a problem in a script using this package:

    "github.com/jasonlvhit/gocron"

I wrote this little testscript after I couldn't find a mistake and it resulted that there were twice as many cronjobs executed as intended:

func main() {
    for i := 0; i < 3; i++ {
        channel := make(chan string)
        go taskCron(channel, i)
    }

    time.Sleep(time.Second * 5)
    gocron.Clear()
    fmt.Println("stop this shit")
}

func task(i int) {
    fmt.Println("still running...", i)
}

func taskCron(channel chan string, i int) {
    gocron.Every(4).Seconds().Do(task, i)
    <-gocron.Start()
}

running it gave me this output:

——▶go run *.go
still running... 0
still running... 0
still running... 1
still running... 1
still running... 2
still running... 2
still running... 0
still running... 1
still running... 2
stop this

Does anyone know how I can create a dynamic amount of gocron jobs without duplicating them?

Thanks :)

Hendrik Evert
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1 Answers1

3

Ok apparently 

<-gocron.Start()

will start jobs that were already started again so to fix my issue, I had to change the script to this:

func main() {
    for i := 0; i < 3; i++ {
        taskCron(i)
    }
    channel2 := make(chan int)
    go startCron(channel2)

    time.Sleep(time.Second * 5)
    gocron.Clear()
    fmt.Println("stop this")
}

func task(i int) {
    fmt.Println("still running...", i)
}

func taskCron(i int) {
    gocron.Every(4).Seconds().Do(task, i)
}

func startCron(channel chan int) {
    <-gocron.Start()
}

I hope this helps anybody who had the same Problem!

Hendrik Evert
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