# Calculating Business Hours Left **URL:** \ **Category:** ruby-talk\ **Created:** [6 April 2007 05:55 UTC](https://rubytalk.org/t/calculating-business-hours-left/36732 "2007-04-06T05:55:05Z")\ **Posts on this page:** 3\ **Page:** 1
**Author:** ![Gavin\_Kistner2](https://avatars.discourse-cdn.com/v4/letter/g/b9e5f3/32.png) [@Gavin\_Kistner2](https://rubytalk.org/u/Gavin_Kistner2)\ **Post date:** [6 April 2007 05:55 UTC](https://rubytalk.org/t/calculating-business-hours-left/36732/1 "2007-04-06T05:55:05Z")
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? 馃檪 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. ---
**Author:** ![Keith\_Fahlgren](https://avatars.discourse-cdn.com/v4/letter/k/e9c0ed/32.png) [@Keith\_Fahlgren](https://rubytalk.org/u/Keith_Fahlgren)\ **Post date:** [6 April 2007 14:13 UTC](https://rubytalk.org/t/calculating-business-hours-left/36732/2 "2007-04-06T14:13:31Z")
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 > **路路路** > > On 4/5/07, Phrogz \ 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? 馃檪 ---
**Author:** ![Gavin\_Kistner2](https://avatars.discourse-cdn.com/v4/letter/g/b9e5f3/32.png) [@Gavin\_Kistner2](https://rubytalk.org/u/Gavin_Kistner2)\ **Post date:** [6 April 2007 14:25 UTC](https://rubytalk.org/t/calculating-business-hours-left/36732/3 "2007-04-06T14:25:09Z")
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 > **路路路** > > On Apr 6, 8:13 am, "Keith Fahlgren" \ wrote: > > > Hmm... no idea if this works, but it might be a start: