Oracle: Return the specific records based on one column date - oracle

Question

I have a database structure as below.
period
month
start_date
1
April
2022-04-01
2
May
2022-05-07
3
June
2022-06-04
4
July
2022-07-02
5
August
2022-08-06
6
September
2022-09-03
7
October
2022-10-01
8
November
2022-11-05
9
December
2022-12-03
10
January
2023-01-01
11
February
2023-02-04
12
March
2023-03-04
End date of the year is 2023-03-31.
Based on current_date, how do I select the query to return where the current date falls under Period 6.
My current query as below.
SELECT period FROM table1 as a
WHERE
a.start_date = (SELECT MAX(start_date) FROM table1 as b WHERE
b.start_date <=current_date) and ROWNUM <= 1
Is there anyway to improve the current query which to avoid using subquery?

Answer 1

Today is September 22nd, so - would this do?
Some sample data:
SQL> with test (period, month, start_date) as
2 (select 1, 'april' , date '2022-04-01' from dual union all
3 select 5, 'august' , date '2022-08-06' from dual union all
4 select 6, 'september', date '2022-09-03' from dual union all
5 select 7, 'october' , date '2022-10-01' from dual union all
6 select 10, 'january' , date '2023-01-01' from dual union all
7 select 12, 'march' , date '2023-03-04' from dual
8 ),
Query begins here:
9 temp as
10 (select period, month, start_date,
11 row_number() over (order by start_date desc) rn
12 from test
13 where start_date <= sysdate
14 )
15 select period
16 from temp
17 where rn = 1
18 /
PERIOD
----------
6
SQL>
It still uses a subquery (or a CTE, as in my example), but - as opposed to your approach, it selects from the source table only once, so performance should be improved.
A few more tests: instead of sysdate (line #13), presume that today is September 2nd (which means that it is in period #5):
9 temp as
10 (select period, month, start_date,
11 row_number() over (order by start_date desc) rn
12 from test
13 where start_date <= date '2022-09-02'
14 )
15 select period
16 from temp
17 where rn = 1;
PERIOD
----------
5
SQL>
Or, if today were August 7th:
9 temp as
10 (select period, month, start_date,
11 row_number() over (order by start_date desc) rn
12 from test
13 where start_date <= date '2022-08-07'
14 )
15 select period
16 from temp
17 where rn = 1;
PERIOD
----------
5
SQL>

Answer 2

Your rule for the start_date appears to be:
If the month is January (first month of the calendar year) or April (typically, first month of the financial year) then use the 1st of that month;
Otherwise use the 1st Saturday of the month.
If that is the case then you can calculate the start date of the next month and use the query:
SELECT *
FROM table1
WHERE start_date <= SYSDATE
AND SYSDATE < CASE
WHEN EXTRACT(MONTH FROM ADD_MONTHS(start_date, 1))
IN (1, 4) -- 1st month of calendar or financial year
THEN TRUNC(ADD_MONTHS(start_date, 1), 'MM')
ELSE NEXT_DAY(TRUNC(ADD_MONTHS(start_date, 1), 'MM') - 1, 'SATURDAY')
END
Then, for your sample data:
CREATE TABLE table1 (
period NUMBER(2,0),
month VARCHAR2(9)
GENERATED ALWAYS AS (
CAST(
TO_CHAR(start_date, 'FXMonth', 'NLS_DATE_LANGUAGE=English')
AS VARCHAR2(9)
)
),
start_date DATE
);
INSERT INTO table1 (period, start_date)
SELECT LEVEL,
CASE
WHEN EXTRACT(MONTH FROM ADD_MONTHS(DATE '2022-04-01', LEVEL - 1))
IN (1, 4) -- 1st month of calendar or financial year
THEN ADD_MONTHS(DATE '2022-04-01', LEVEL - 1)
ELSE NEXT_DAY(ADD_MONTHS(DATE '2022-04-01', LEVEL - 1) - 1, 'SATURDAY')
END
FROM DUAL
CONNECT BY LEVEL <= 12;
Outputs:
PERIOD
MONTH
START_DATE
6
September
2022-09-03 00:00:00
fiddle