How to get last month's date from specific date in postgresql - sql

Question

I have a max date in my query and If the max date is current month then I want always get previous month’s date. How can I do that?
For example, today is 20160825. If the max date is 20160801 then I want to get 20160701. But, if the max date is 20160501 then I just want to get without changes - 20160501.
SELECT
DEFN_DK,
MAX(SNAPSHOT_MTH)
FROM myTable
WHERE TOT_AMT >0
GROUP BY DEFN_DK

Answer 1

Since your SNAPSHOT_MTH column is an integer (why?) you can not use any of the otherwise very useful timestamp functions. So it's back to integer math, creating a "month" from your snapshot "date" through integer division by 100. This can be compared to CURRENT_DATE by converting that to a string and then casting it to an integer. Going back 1 month similarly requires some math. Not entirely efficient, but here goes:
SELECT DEFN_DK,
CASE max(SNAPSHOT_MTH) / 100
WHEN tochar(CURRENT_DATE, 'YYYYMM')::int THEN
((max(SNAPSHOT_MTH) / 100) - 1) * 100 + 1
-- or max(SNAPSHOT_MTH) - 100, if you know it always ends in 01
ELSE max(SNAPSHOT_MTH)
END AS SNAPSHOT_MTH
FROM myTable
WHERE TOT_AMT > 0
GROUP BY DEFN_DK;

Answer 2

select
defn_dk,
case max(snapshot_mth)
when date_trunc('month', current_date) then max(snapshot_mth) - interval '1 month'
else max(snapshot_mth)
end
from mytable
where tot_amt >0
group by defn_dk

Answer 3

show this, two examples:
select
case when date_trunc('month',dd) = date_trunc('month',now())
then dd - interval'1 month' else date_trunc('day',dd) end
from
(
select '2016-08-04'::date as dd
) d;
and
select
case when date_trunc('month',dd) = date_trunc('month',now())
then dd - interval'1 month' else date_trunc('day',dd) end
from
(
select '2016-05-04'::date as dd
) d;