Daily average for the month (needs number of days in month) - sql

Question

I have a table as follow:
CREATE TABLE counts
(
T TIMESTAMP NOT NULL,
C INTEGER NOT NULL
);
I create the following views from it:
CREATE VIEW micounts AS
SELECT DATE_TRUNC('minute',t) AS t,SUM(c) AS c FROM counts GROUP BY 1;
CREATE VIEW hrcounts AS
SELECT DATE_TRUNC('hour',t) AS t,SUM(c) AS c,SUM(c)/60 AS a
FROM micounts GROUP BY 1;
CREATE VIEW dycounts AS
SELECT DATE_TRUNC('day',t) AS t,SUM(c) AS c,SUM(c)/24 AS a
FROM hrcounts GROUP BY 1;
The problem now comes in when I want to create the monthly counts to know what to divide the daily sums by to get the average column a i.e. the number of days in the specific month.
I know to get the days in PostgreSQL you can do:
SELECT DATE_PART('days',DATE_TRUNC('month',now())+'1 MONTH'::INTERVAL-DATE_TRUNC('month',now()))
But I can't use now(), I have to somehow let it know what the month is when the grouping gets done. Any suggestions i.e. what should replace ??? in this view:
CREATE VIEW mocounts AS
SELECT DATE_TRUNC('month',t) AS t,SUM(c) AS c,SUM(c)/(???) AS a
FROM dycounts
GROUP BY 1;

Answer 1

A bit shorter and faster and you get the number of days instead of an interval:
SELECT EXTRACT(day FROM date_trunc('month', now()) + interval '1 month'
- interval '1 day')
It's possible to combine multiple units in a single interval value . So we can use '1 mon - 1 day':
SELECT EXTRACT(day FROM date_trunc('month', now()) + interval '1 mon - 1 day')
(mon, month or months work all the same for month units.)
To divide the daily sum by the number of days in the current month (orig. question):
SELECT t::date AS the_date
, SUM(c) AS c
, SUM(c) / EXTRACT(day FROM date_trunc('month', t::date)
+ interval '1 mon - 1 day') AS a
FROM dycounts
GROUP BY 1;
To divide monthly sum by the number of days in the current month (updated question):
SELECT DATE_TRUNC('month', t)::date AS t
,SUM(c) AS c
,SUM(c) / EXTRACT(day FROM date_trunc('month', t)::date
+ interval '1 mon - 1 day') AS a
FROM dycounts
GROUP BY 1;
You have to repeat the GROUP BY expression if you want to use a single query level.
Or use a subquery:
SELECT *, c / EXTRACT(day FROM t + interval '1 mon - 1 day') AS a
FROM (
SELECT date_trunc('month', t)::date AS t, SUM(c) AS c
FROM dycounts
GROUP BY 1
) sub;