Counting working days


Counting working days


I have a table with a list of delivery dates for orders that have failed to go out. I need to report on how many working days these are passed that delivery date.

For example I have a table like this:

OrderNo, RequestedDeliveryDate

Order1  , 2012-08-27

Order2  , 2012-08-31

So if I was to run this today (2012-09-03) I need an output of:

OrderNo, RequestedDeliveryDate,DaysLate

Order1  , 2012-08-27                   , 5

Order2  , 2012-08-31                   , 1

Any help or direction on how to solve this will be a fantastic.



Re: Counting working days

How do you define "working days"?

Always Mo - Fr, Mo-Sa, exclude Holidays?

If you need to consider Holidays you would need a kalendar table to maintain these.

Re: Counting working days

Hi ulrich,

Working days a Monday to Friday. I do not need to consider holidays.



Re: Counting working days

so let's assume the following table 

create volatile table vt_dates
select c1.calendar_Date from_dt, c2.calendar_date to_dt
from sys_calendar.calendar c1
cross join
sys_calendar.calendar c2
where c1.calendar_Date between current_date - 30 and current_Date
and c2.calendar_Date between current_date - 30 and current_Date
and c2.calendar_date > c1.calendar_date
) with data primary index (from_dt)
on commit preserve rows;

You have two choices:

1. simple logic but join on calendar table

select from_dt, 
sum(case when c.day_of_week in (1,7) then 0 else 1 end) - (case when sum(case when c.day_of_week in (1,7) then 0 else 1 end) = 0 then 0 else 1 end) as count_of_weekdays
from vt_dates v
join sys_calendar.calendar c
on c.calendar_Date between v.from_dt and v.to_dt
group by 1,2
order by 1,2

2. direct calculation with not too obvious formula

select from_dt, 
case when day_of_week(from_dt) = 7 then 2
when day_of_week(from_dt) = 1 then 1
else 0
end + from_dt as next_work_dt,
case when day_of_week(to_dt) = 7 then -1
when day_of_week(to_dt) = 1 then -2
else 0
end + to_dt as prev_work_dt,
(prev_work_dt - next_work_dt)+1 as diff,
(case when diff mod 7 = 0 then 1 else 0 end) as correct_fw,
(diff - diff mod 7) / 7 as full_week,

case when diff < 0
then 0
else diff - (full_week*2) - 1
end +
case when day_of_week(prev_work_dt) < day_of_week(next_work_dt)
then -2
else 0
end +
case when diff mod 7 = 0 then 2 else 0 end
as num_of_weekdays
from vt_dates v
order by 1,2

If your table is big and the number of workdays can also be big it might be worth to consider solution two.

A SQL UDF might be usefull to have a central logic for it - even if the UDF code becomes more unreadable due to repeating cases...

Re: Counting working days

Hi ulrich,

Solution 1 works wonderfully.

Thanks so much this has solved somthing that has stumpt me for a while now.