Find the last weekday for a given month in PostgreSQL - sql

Question

Find the last weekday for a given month in PostgreSQL
Usage: If month end falls on a Saturday or a Sunday, return the previous Friday, else use month end
Examples:
3/31/2013 falls on a Sunday, so return 3/29/2013
11/30/2013 falls on a Saturday, so return 11/29/2013
How to write this in PostgreSQL SQL?
What I have so far is this (returns only Month Ends, but month ends don't exist when they fall on a Saturday or Sunday):
SELECT as_of_dt, sum(bank_shr_bal) as bank_shr_bal
FROM hm_101.vw_gl_bal
WHERE as_of_dt = (date_trunc('MONTH', as_of_dt) + INTERVAL '1 MONTH - 1 day')::date
GROUP BY 1
Thanks

Answer 1

A calendar table really simplifies the SQL for queries like these. (The table "weekdays" is actually a view based on the calendar table. The structure of it should be obvious.)
select max(cal_date)
from weekdays
where cal_date < '2013-05-01'
or
select max(cal_date)
from weekdays
where cal_date between '2013-04-01' and '2013-04-30'

Answer 2

with s as (
select *, (date_trunc('MONTH', as_of_dt) + INTERVAL '1 MONTH - 1 day')::date last_day
from
hm_101.vw_gl_bal
)
SELECT
as_of_dt,
gl_acct_nbr,
cc_nbr,
sum(bank_shr_bal) as bank_shr_bal
FROM s
WHERE as_of_dt = (
last_day
-
(extract(dow from last_day) = 5)::int
-
2 * (extract(dow from last_day) = 6)::int
)
GROUP BY 1,2,3

Answer 3

One solution is to use a CTE, find the last day by month in the data and the actual last day for each month
WITH s1
as
(
SELECT
date_part('YEAR', as_of_dt) AOD_Year
,date_part('MONTH', as_of_dt) AOD_Month
,(date_trunc('MONTH', as_of_dt) + INTERVAL '1 MONTH - 1 day')::date AOD_MonthEnd
,max(as_of_dt) AOD_LastFound
FROM hm_101.vw_gl_bal
where (date_trunc('MONTH', as_of_dt) + INTERVAL '1 MONTH - 1 day')::date = '2013-03-31'
group by 1, 2, 3
)
SELECT
s1.AOD_MonthEnd
,s1.AOD_LastFound
,sum(v.bank_shr_bal) as bank_shr_bal
FROM hm_101.vw_gl_bal v
INNER JOIN s1
on v.as_of_dt = s1.AOD_LastFound
WHERE v.as_of_dt = '2013-03-29'
GROUP BY 1, 2

Answer 4

What you want to do is remove between 0 and 2 days from the last day of the month (which you have).
By extracting the Day Of Week (DOW) and check if it's 0 (Sunday) or 6 (Saturday), we know how many days to remove.
You can do it like this:
... - INTERVAL '1 day' * CASE date_part('DOW', last_day_of_month)
WHEN 0 THEN 2 -- Sunday, remove 2 days.
WHEN 6 THEN 1 -- Saturday, remove 1 day.
ELSE 0 -- Don't remove any days.
END
For the sake of readability I didn't include the complete last_day_of_month-calculation in there.

Answer 5

You can actually do this without CTEs or stored procedures.
select
case
when extract(dow from last_day_of_month) = 0
then last_day_of_month - 2
when extract(dow from last_day_of_month) = 6
then last_day_of_month - 1
else
last_day_of_month
end as last_weekday_of_month
from(
SELECT (date_trunc('MONTH', as_of_dt)
+ INTERVAL '1 MONTH - 1 day')::date as last_day_of_month
from hm_101.vw_gl_bal
)subquery;

Answer 6

select
case
when extract(dow from first_day_of_month) = 0 then first_day_of_month
when extract(dow from first_day_of_month) = 1 then first_day_of_month - 1
when extract(dow from first_day_of_month) = 2 then first_day_of_month - 2
when extract(dow from first_day_of_month) = 3 then first_day_of_month - 3
when extract(dow from first_day_of_month) = 4 then first_day_of_month - 4
when extract(dow from first_day_of_month) = 5 then first_day_of_month - 5
when extract(dow from first_day_of_month) = 6 then first_day_of_month - 6
end as first_weekday_of_month,
case
when extract(dow from last_day_of_month) = 6 then last_day_of_month
when extract(dow from last_day_of_month) = 5 then last_day_of_month - 6
when extract(dow from last_day_of_month) = 4 then last_day_of_month - 5
when extract(dow from last_day_of_month) = 3 then last_day_of_month - 4
when extract(dow from last_day_of_month) = 2 then last_day_of_month - 3
when extract(dow from last_day_of_month) = 1 then last_day_of_month - 2
when extract(dow from last_day_of_month) = 0 then last_day_of_month - 1
end as last_weekday_of_month
from(
SELECT
(date_trunc('month', current_date) -'7day'::interval)::date first_day_of_month,
(date_trunc('month', current_date) -'1day'::interval)::date as last_day_of_month
)subquery;