Oracle SQL counting the days of the week of the same month - sql

Question

How can I count the days of each week for a specific month that the user will give to my SQL query? For example, if the user gives April 2021, the result will be:
If the user gives May 2021 the result will be:

Answer 1

Here is a direct computation of the same. The input is given as a string, such as May 2021; you can use a bind variable in its place. Just keep in mind the possibility that the user may be in a non-English-speaking locale; as long as they use their local language in passing the month to the query, everything should work fine.
with
inputs (mth) as (select 'May 2021' from dual)
, first_day (dt) as (select to_date(mth, 'fmMonth yyyy') from inputs)
, mondays (dt, ord, lst) as (
select trunc(dt, 'iw') + 7 * (level - 1), level, max(level) over ()
from first_day
connect by level <= 1 + (trunc(add_months(dt, 1), 'iw') - trunc(dt, 'iw')) / 7
)
select to_number(to_char(dt, 'iw')) as week_number,
case ord when 1 then dt + 7 - trunc(dt + 7, 'mm')
when lst then last_day(dt) + 1 - dt
else 7 end as week_days
from mondays
order by week_number
;
WEEK_NUMBER WEEK_DAYS
----------- ----------
17 2
18 7
19 7
20 7
21 7
22 1

Answer 2

I think this is the query you are looking for:
SELECT WEEK AS WEEK_NUMBER, COUNT(*) AS WEEK_DAYS
FROM (SELECT TO_CHAR(FIRST_DAY + (LEVEL-1), 'IW') AS WEEK
FROM (SELECT supplied_date AS FIRST_DAY, LAST_DAY(supplied_date) - SUPPLIED_DATE+1 AS DAYS
FROM (SELECT TO_DATE('05/2021', 'MM/YYYY') AS SUPPLIED_DATE
FROM DUAL))
CONNECT BY LEVEL <= DAYS)
GROUP BY WEEK
ORDER BY WEEK;
Just replace the inner TO_DATE with your date.
Edit 1: Additional Column (See Comments)
SELECT MONTH_NAME, WEEK AS WEEK_NUMBER, COUNT(*) AS WEEK_DAYS
FROM (SELECT TO_CHAR(FIRST_DAY, 'MONTH') AS MONTH_NAME, TO_CHAR(FIRST_DAY + (LEVEL-1), 'IW') AS WEEK
FROM (SELECT supplied_date AS FIRST_DAY, TO_CHAR(LAST_DAY(supplied_date), 'DD') AS DAYS
FROM (SELECT TO_DATE('05/2021', 'MM/YYYY') AS SUPPLIED_DATE
FROM DUAL))
CONNECT BY LEVEL <= DAYS)
GROUP BY MONTH_NAME, WEEK
ORDER BY WEEK;
Note: I also changed the way I calculate the number of days in the month.

Answer 3

You can generate the weeks (without needing to use any aggregation) using:
WITH input ( month ) AS (
SELECT DATE '2021-04-01' FROM DUAL
),
weeks_of_month ( month_start, week_start, end_day ) AS (
SELECT TRUNC( month, 'MM' ),
TRUNC( month, 'IW' ),
LAST_DAY( TRUNC( month ) )
FROM input
UNION ALL
SELECT month_start,
week_start + INTERVAL '7' DAY,
end_day
FROM weeks_of_month
WHERE week_start + INTERVAL '7' DAY <= end_day
)
SELECT TO_CHAR( week_start, 'IW' ) AS iso_week,
GREATEST( month_start, week_start ) AS first_day_of_week,
LEAST( end_day, week_start + INTERVAL '6' DAY ) AS last_day_of_week,
LEAST( end_day + 1, week_start + INTERVAL '7' DAY )
- GREATEST( month_start, week_start ) AS days
FROM weeks_of_month;
Which outputs (with the NLS_DATE_FORMAT set to YYYY-MM-DD (DY)):
ISO_WEEK
FIRST_DAY_OF_WEEK
LAST_DAY_OF_WEEK
DAYS
13
2021-04-01 (THU)
2021-04-04 (SUN)
4
14
2021-04-05 (MON)
2021-04-11 (SUN)
7
15
2021-04-12 (MON)
2021-04-18 (SUN)
7
16
2021-04-19 (MON)
2021-04-25 (SUN)
7
17
2021-04-26 (MON)
2021-04-30 (FRI)
5
db<>fiddle here