ORACLE CONVERT CHAR FORMAT "YYYY-WW" TO DATE FORMAT - sql

Question

I am trying to convert char date format "YYYY-WW" in ORACLE to date for calculating by week in year but have a trouble with error message format code cannot appear in date input format
If i write
TO_DATE(TO_CHAR(TO_DATE('1970-01-01 07:00:00', 'YYYY-MM-DD HH:MI:SS'),'YYYY-MM'),'YYYY-MM')
It will be normal
But if i write
TO_DATE(TO_CHAR(TO_DATE('1970-01-01 07:00:00', 'YYYY-MM-DD HH:MI:SS'),'YYYY-WW'),'YYYY-WW')
The message error format code cannot appear in date input format appear. I don't have no idea how to convert it right
Thanks for any advice

Answer 1

You can use:
TO_DATE( year || '-01-01', 'YYYY-MM-DD' ) + INTERVAL '7' DAY * (week - 1)
For your code:
SELECT TO_DATE( TO_CHAR(dt, 'YYYY') || '-01-01', 'YYYY-MM-DD' )
+ INTERVAL '7' DAY * (TO_CHAR(dt, 'WW') - 1) AS week_start
FROM (
SELECT TO_DATE('1970-01-01 07:00:00', 'YYYY-MM-DD HH24:MI:SS') AS dt
FROM DUAL
);
If you want to use ISO weeks (which always start on a Monday, rather than counting from the 1st January as WW does) then:
TRUNC( TO_DATE( iso_year || '-01-04', 'YYYY-MM-DD' ), 'IW')
+ INTERVAL '7' DAY * (iso_week - 1)
db<>fiddle here

Answer 2

You can have a look at this for guidance. https://asktom.oracle.com/pls/apex/asktom.search?tag=week-of-year-in-sql-confusing

Answer 3

There isn't a built-in way to convert a week number back to a date, as the result would be a bit arbitrary - which of the (up to) seven days in the week should it return?
If you want the first day of the week, as would be found by truncating the original date to WW precision (which is based on the day of the week of the first day of the year - docs), then you can reverse that process by getting the first day of the year and then adding the number of days in the specified number of weeks.
To do that you need to split your string into the two component parts:
substr('1970-01', 1, 4)
substr('1970-01', 6, 2)
and convert the first part to a date:
to_date(substr('1970-01', 1, 4) || '-01-01', 'YYYY-MM-DD')
and the second part to a number of weeks:
to_number(substr('1970-01', 6, 2))
then subtract one from that number of weeks, multiply by seven, and add it to the base date:
select to_date(substr('1970-01', 1, 4) || '-01-01', 'YYYY-MM-DD')
+ (7 * (to_number(substr('1970-01', 6, 2)) - 1))
from dual
1970-01-01
It's bit hard to tell that has been adjusted, so trying with today's date, which is week 30 of 2022:
select to_date(substr('2022-30', 1, 4) || '-01-01', 'YYYY-MM-DD')
+ (7 * (to_number(substr('2022-30', 6, 2)) - 1))
from dual
2022-07-23
which matches what trunc(sysdate, 'WW') gives.
db<>fiddle
If you want a specific day of the week then you'll need to figure out how much to adjust that, based on what day of the week the first day of the year was.

Answer 4

Select
Extract (year from TO_DATE('1970-01-01 07:00:00', 'YYYY-MM-DD HH:MI:SS')) || '-' ||
TO_CHAR(TO_DATE('1970-01-01 07:00:00', 'YYYY-MM-DD HH:MI:SS'),'WW')
From dual

Answer 5

If you are starting from a string like '1970-01-01 07:00:00' and want the first day of the corresponding week, you don't need to convert to and from strings and dates repeatedly; you just need to convert that string to a date once, and truncate it:
trunc(to_date('1970-01-01 07:00:00'), 'WW')
or if you want the ISO week:
trunc(to_date('1970-01-01 07:00:00'), 'IW')
For your example that would give:
select to_date('1970-01-01 07:00:00') as dt,
trunc(to_date('1970-01-01 07:00:00'), 'WW') as ww,
trunc(to_date('1970-01-01 07:00:00'), 'IW') as iw
from dual;
DT
WW
IW
1970-01-01 07:00:00
1970-01-01 00:00:00
1969-12-29 00:00:00
And for today it would give:
select to_date('2022-07-28 07:00:00') as dt,
trunc(to_date('2022-07-28 07:00:00'), 'WW') as ww,
trunc(to_date('2022-07-28 07:00:00'), 'IW') as iw
from dual;
DT
WW
IW
2022-07-28 07:00:00
2022-07-23 00:00:00
2022-07-25 00:00:00
db<>fiddle