Scheduler doesn't execute function within class (is it because it's not a first-class function)? about schedule HOT 3 CLOSED

It looks like your problem is caused by reusing the schedule returned by parse_work_schedule, rather than where the functions are defined; you can see this by swapping the order in which you call sched -- only the last one called will actually be scheduled (this is expected, since do will just overwrite the job function each time it's called).

Instead, you could try something like changing all to a lambda so that it returns a new schedule every time:

def parse_work_schedule(self, work):
    options = {
        'all': lambda: self.schedule.every().minute.do,
    }
    return options['all']

and then using it as such:

for work in self.works:
    if sched and sched():
        sched()(pp, work['name'])
        sched()(work['func'], work['name'])
        print 'Scheduled: {0}'.format(sched)
    else:
        print 'Function pointer {0} returned None value.'.format(sched)

from schedule.

Hi dylwhich,

I tried your suggestion but it didn't work.

Perhaps I'm not clear with my question (and code).

If you uncomment and comment out:

sched(pp, work['name'])
# sched(work['func'], work['name'])

The code, as it is, works. However, when the comments are swapped, i.e.

# sched(pp, work['name'])
sched(work['func'], work['name'])

the code doesn't work. Why is that? Both are functions.

from schedule.

I found the answer. Both are not functions. pp is a function hence it works correctly. But work['func'] is an instancemethod type. So the solution is to have a method within the class that returns a function object. Then I can provide that function object to the schedule.

Thanks for your help.

from schedule.