2015-11-21 27 views

回答

0
select 
    s.BusDepartureStation, s.BusArrivalStation, s.DepartureTime, 
    dateadd(
     mi, 
     min(case 
      when d.DepartureTime >= s.DepartureTime 
      then datediff(mi, s.DepartureTime, d.DepartureTime), s.DepartureTime) 
      else null 
     end), 
     s.DepartureTime 
    ) as NextNearestDepartureTime 
from Scheduled s inner join Departed d 
    on  d.BusDepartureStation = s.BusDepartureStation 
     and d.ArrivalStation = s.ArrivalStation 
group by 
    s.BusDepartureStation, s.BusArrivalStation, s.DepartureTime 

爲此,我假設你的出發時間不早,並且早先的預定出發總是在後面的公共汽車之前離開。如果我對SQL Server的假設不正確,你應該能夠調整時間差異邏輯來改用MySQL函數。

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