How to determine number of days in a month in Presto? - sql

Question

I have data with date, userid, and amount. I want to calculate sum(amount) divided by total day for each month. the final will be presented in monthly basis.
The table I have is looks like this
date userid amount
2019-01-01 111 10
2019-01-15 112 20
2019-01-20 113 10
2019-02-01 114 30
2019-02-15 111 20
2019-03-01 115 40
2019-03-23 155 50
desired result is like this
date avg_qty_sol
Jan-19 1.29
Feb-19 1.79
Mar-19 2.90
avg_qty_sold is coming from sum(amount) / total day for respective month
e.g for jan 2019 sum amount is 40 and total days in jan is 31. so the avg_qty_sold is 40/31
Currently Im using case when for this solution. is there any better approach to this?

Answer 1

Since Presto 318, you this is as easy as:
SELECT day(last_day_of_month(some_date))
See https://trino.io/docs/current/functions/datetime.html#last_day_of_month
Before Presto 318, You can combine date_trunc with EXTRACT:
date_trunc('month', date_value)) gives beginning of the month, while date_add('month', 1, date_trunc('month', date_value)) gives beginning of the next month
subtracting date values returns an interval day to second
EXTRACT(DAY FROM interval) returns day-portion of the interval. You can also use day convenience function instead of EXTRACT(DAY FROM ...). The EXTRACT syntax is more verbose and more standard.
presto:default> SELECT
-> date_value,
-> EXTRACT(DAY FROM (
-> date_add('month', 1, date_trunc('month', date_value)) - date_trunc('month', date_value)))
-> FROM (VALUES DATE '2019-01-15', DATE '2019-02-01') t(date_value);
date_value | _col1
------------+-------
2019-01-15 | 31
2019-02-01 | 28
(2 rows)
Less natural, but a bit shorter alternative would be to get day number for the last day of given month with day(date_add('day', -1, date_add('month', 1, date_trunc('month', date_value)))):
presto:default> SELECT
-> date_value,
-> day(date_add('day', -1, date_add('month', 1, date_trunc('month', date_value))))
-> FROM (VALUES DATE '2019-01-15', DATE '2019-02-01') t(date_value);
date_value | _col1
------------+-------
2019-01-15 | 31
2019-02-01 | 28
(2 rows)