how to execute jobs when start?

[iziyang]

Now cron job execute when next ticker coming. For example, I set @every 5 min，but when I start my program, it must wait for 5 min to execute, I want execute when start.

[robfig]

There is no way to do that with the schedule. I think the best workaround is to execute your job immediately, e.g.

cron.AddJob("@every 5m", myJob)
go myJob()

[tsarna]

Several cron implementations support @reboot for this. "reboot" seems a little weird in the context of a go program instead of system cron, so maybe an alias like "@startup" or "@once" would be good.

[flowchartsman]

The Add* methods return an opaque ID as per the docs, it might be handy to have a companion to Remove() that allows you to manually run a job if it's not already running. This seems like it could have a clean interaction with the scheduler, rather than needing to have a singleflight in myJob(), though I suppose the implications for that are more broad.

[freecells]

just run the func once

[zeewell]

#436

[alaingilbert]

@iziyang 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)