Execute the function in the scheduled task immediately when the program starts。

[wuruipeng404]

For example, I have a task that is executed every minute. When I start the program, cron will start to execute my task in the next minute. Can I add an option such as "Execute once now" task。

If I manually call the task once in the program, it will look very ugly, and some situations cannot be satisfied

[windzhu0514]

you can write a custom schedule,but this only support stand parser.

type StartNowSchedule struct {
	firstTime int32
	schedule  cron.Schedule
}

func NewStartNow(spec string) (*StartNowSchedule, error) {
	schedule, err := cron.ParseStandard(spec)
	if err != nil {
		return nil, err
	}

	return &StartNowSchedule{schedule: schedule}, nil
}

func (schedule *StartNowSchedule) Next(t time.Time) time.Time {
	if schedule.firstTime == 0 {
		schedule.firstTime = 1
		return t
	}

	return schedule.schedule.Next(t)
}

[ivanjaros]

this is quite needed indeed. the AddJob(spec string, cmd Job) could be simply refactored to AddJob(spec string, cmd Job, runOnStart ...bool) to keep the backwards compatibility and the scheduler would fire the tasks one after another in the order they were added.

[alaingilbert]

@wuruipeng404 I made a fork that addresses almost all issues of this repo.
https://github.com/alaingilbert/cron

This is how you could do what you want to do ->

_, _ = c.AddJob("*/10 * * * * *", func() {
    fmt.Println("this job runs as soon as cron is started, then at every 10th seconds", time.Now())
}, cron.RunOnStart)