How to get the number of days in a month? - postgresql

Question

I am trying to get the following in Postgres:
select day_in_month(2);
Expected output:
28
Is there any built-in way in Postgres to do that?

Answer 1

SELECT
DATE_PART('days',
DATE_TRUNC('month', NOW())
+ '1 MONTH'::INTERVAL
- '1 DAY'::INTERVAL
)
Substitute NOW() with any other date.

Answer 2

Using the smart "trick" to extract the day part from the last date of the month, as demonstrated by Quassnoi. But it can be a bit simpler / faster:
SELECT extract(days FROM date_trunc('month', now()) + interval '1 month - 1 day');
Rationale
extract is standard SQL, so maybe preferable, but it resolves to the same function internally as date_part(). The manual:
The date_part function is modeled on the traditional Ingres equivalent to the SQL-standard function extract:
But we only need to add a single interval. Postgres allows multiple time units at once. The manual:
interval values can be written using the following verbose syntax:
[#] quantity unit[quantity unit...] [direction]
where quantity is a number (possibly signed); unit is microsecond,
millisecond, second, minute, hour, day, week, month, year, decade,
century, millennium, or abbreviations or plurals of these units;
ISO 8601 or standard SQL format are also accepted. Either way, the manual again:
Internally interval values are stored as months, days, and seconds.
This is done because the number of days in a month varies, and a day
can have 23 or 25 hours if a daylight savings time adjustment is
involved. The months and days fields are integers while the seconds
field can store fractions.
(Output / display depends on the setting of IntervalStyle.)
The above example uses default Postgres format: interval '1 month - 1 day'. These are also valid (while less readable):
interval '1 mon - 1 d' -- unambiguous abbreviations of time units are allowed
IS0 8601 format:
interval '0-1 -1 0:0'
Standard SQL format:
interval 'P1M-1D';
All the same.

Answer 3

Note that expected output for day_in_month(2) can be 29 because of leap years. You might want to pass a date instead of an int.
Also, beware of daylight saving : remove the timezone or else some monthes calculations could be wrong (next example in CET / CEST) :
SELECT DATE_TRUNC('month', '2016-03-12'::timestamptz) + '1 MONTH'::INTERVAL
- DATE_TRUNC('month', '2016-03-12'::timestamptz) ;
------------------
30 days 23:00:00
SELECT DATE_TRUNC('month', '2016-03-12'::timestamp) + '1 MONTH'::INTERVAL
- DATE_TRUNC('month', '2016-03-12'::timestamp) ;
----------
31 days

Answer 4

This works as well.
WITH date_ AS (SELECT your_date AS d)
SELECT d + INTERVAL '1 month' - d FROM date_;
Or just:
SELECT your_date + INTERVAL '1 month' - your_date;
These two return interval, not integer.

Answer 5

SELECT cnt_dayofmonth(2016, 2); -- 29
create or replace function cnt_dayofmonth(_year int, _month int)
returns int2 as
$BODY$
-- ZU 2017.09.15, returns the count of days in mounth, inputs are year and month
declare
datetime_start date := ('01.01.'||_year::char(4))::date;
datetime_month date := ('01.'||_month||'.'||_year)::date;
cnt int2;
begin
select extract(day from (select (datetime_month + INTERVAL '1 month -1 day'))) into cnt;
return cnt;
end;
$BODY$
language plpgsql;

Answer 6

You can write a function:
CREATE OR REPLACE FUNCTION get_total_days_in_month(timestamp)
RETURNS decimal
IMMUTABLE
AS $$
select cast(datediff(day, date_trunc('mon', $1), last_day($1) + 1) as decimal)
$$ LANGUAGE sql;