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comp.lang.ruby

Calculating Business Hours Left

Gavin Kistner

4/6/2007 5:51:00 AM

current_time = Time.new
due_time = # ... some Time instance in the future
work_hours_left_before_I_am_screwed = # ... calculate

I'm writing a little task-tracking app to help me put out the most
urgent fires. As above, given any two Time instances, I'd like to know
how many hours are left to perform the task. This should skip weekends
and hours outside of specified start- and end-hours for a work day.

Anyone got such code handy? Or feel like writing it for me? :)

Right now, my best idea involves looping over the range one day at a
time and manually accumulating the available hours. It feels
inelegant, but it's all I have right now.
2 Answers

Keith Fahlgren

4/6/2007 2:14:00 PM

0

On 4/5/07, Phrogz <gavin@refinery.com> wrote:
> current_time = Time.new
> due_time = # ... some Time instance in the future
> work_hours_left_before_I_am_screwed = # ... calculate
>
> I'm writing a little task-tracking app to help me put out the most
> urgent fires. As above, given any two Time instances, I'd like to know
> how many hours are left to perform the task. This should skip weekends
> and hours outside of specified start- and end-hours for a work day.
>
> Anyone got such code handy? Or feel like writing it for me? :)

Hmm... no idea if this works, but it might be a start:

#!/usr/bin/env ruby
class WorkHours
SECONDS_IN_HOUR = 3600
HOURS_I_WORK = (7..16)
DAYS_I_WORK = (1..5)
def self.diff_times(a, b)
hours = []
(a.to_i..b.to_i).step(SECONDS_IN_HOUR) {|t|
time = Time.at(t)
hours << t if DAYS_I_WORK.include?(time.wday) &&
HOURS_I_WORK.include?(time.hour)
}
return hours.length
end
end
nil

now = Time.now
=> Fri Apr 06 07:12:08 PDT 2007
twenty_days_from_now = now + 1728000
=> Thu Apr 26 07:12:08 PDT 2007
one_day_from_now = now + 86400
=> Sat Apr 07 07:12:08 PDT 2007
WorkHours.diff_times(now, twenty_days_from_now)
=> 141
WorkHours.diff_times(now, one_day_from_now)
=> 10


HTH,
Keith

Gavin Kistner

4/6/2007 2:24:00 PM

0

On Apr 6, 8:13 am, "Keith Fahlgren" <k...@audiobeta.com> wrote:
> Hmm... no idea if this works, but it might be a start:

That feels like it might be good, thanks! I'll have to investigate
more. It certainly looks cleaner than what I came up with:

class Time
def at( hrs, mins=0 )
self.class.parse strftime("%Y-%b-%d #{hrs}:#{mins}" )
end
end

# Fails for tasks one year or farther in the future
def work_hours_until( due_time, work_starts_at=9, work_ends_at=17 )
start_time = [
Time.now,
Time.parse( "#{work_starts_at}:00" )
].max
end_time = [
due_time,
due_time.at( work_ends_at )
].min

weekdays = 1..5
one_hour = 3600.0
one_day = one_hour * 24

if start_time.yday == end_time.yday
if weekdays.include?( start_time.wday )
( end_time - start_time ) / one_hour
else
0
end
else
business_hours = work_ends_at - work_starts_at

# Handle partial hours on the first day
hours = ( start_time.at( work_ends_at ) - start_time ) / one_hour

t = start_time + one_day
until t.yday == due_time.yday
hours += business_hours if weekdays.include?( t.wday )
t += one_day
end

# Handle partial hours on the last day
if weekdays.include?( t.wday )
hours += ( end_time - end_time.at( work_starts_at ) ) / one_hour
end

hours
end

end