In Oracle, is there a straightforward way to get the first day of the week given a week number?
For example, today's date is 12/4/2012. If I run:
select to_char(sysdate,'WW') from dual;
It returns 49 for the week number.
What I would like to do is somehow return 12/2/2012 for the first day...given week 49 (assuming Sunday as first day of the week).
Any ideas? Thanks in advance for any help!
try this:
select next_day(max(d), 'sun') requested_sun
from (select to_date('01-01-2012', 'dd-mm-yyyy') + (rownum-1) d from dual connect by level <= 366)
where to_char(d, 'ww') = 49-1;
just set your year to_date('01-01-2012' and week number-1 49-1 as applicable.
the sunday in the 49th week of 2008?
SQL> select next_day(max(d), 'sun') requested_sun
2 from (select to_date('01-01-2008', 'dd-mm-yyyy') + (rownum-1) d from dual connect by level <= 366)
3 where to_char(d, 'ww') = 49-1;
REQUESTED
---------
07-DEC-08
and 2012
SQL> select next_day(max(d), 'sun') requested_sun
2 from (select to_date('01-01-2012', 'dd-mm-yyyy') + (rownum-1) d from dual connect by level <= 366)
3 where to_char(d, 'ww') = 49-1;
REQUESTED
---------
02-DEC-12
Try this,
select
next_day(trunc(to_date(in_year,'yyyy'),'yyyy') -1,'Mon') + (7 * (in_week - 1))
from dual;
If you have the date, not just the week number, you can try this:
Get the day number of the week of your date with: to_char(theDate, 'D')
substract that number from your date plus 1, and you'll get the Sunday of that week.
Add 7 and you'll get the date of end of the week(Saturday).
Like this:
SELECT theDate - to_char(theDate, 'D') + 1 as BeginOfWeek,
theDate,
theDate - to_char(theDate, 'D') + 7 as EndOfWeek
FROM TableName
I can't comment on questions yet, so I'll add another one. But this is based on #Dazzals answer.
His solution doesn't work for week one and for ISO-weeks. Also it doesn't work, if the first day of the week is not sunday, which can be controlled by NLS_SETTINGS.
This one does:
SELECT MIN(D)
FROM (SELECT TO_DATE('01-01-2013', 'dd-mm-yyyy') + (ROWNUM-10) D, ROWNUM R
FROM DUAL
CONNECT BY LEVEL <= 376)
WHERE TO_CHAR(D,'IYYYIW') = '201301'
Because we are spanning more than one year, we need to check the year too.
Using the trunc function #Justin used, I think this is what you want:
select trunc(to_date('2012-01-01', 'YYYY-MM-DD') + (49 - 1) * 7, 'WW') from dual;
I ended up doing this:
function getFirstDayOfWeek(y in binary_integer, w in binary_integer) return date
is
td date;
begin
td:=TO_DATE(TO_CHAR(y)||'0101', 'YYYYMMDD');
for c in 0..52
loop
if TO_NUMBER(TO_CHAR(td, 'IW'))=w then
return TRUNC(td, 'IW');
end if;
td:=td+7;
end loop;
return null;
end;
Related
This question already has answers here:
How to get the previous working day from Oracle?
(4 answers)
Closed 1 year ago.
need help for some oracle stuff ..
I need to get Day-1 from sysdate, holiday and weekend will be excluded .
And for holiday, we need to get the range to get the last workday before holiday.
The start date and end date will coming from my holiday table.
ex :
Holiday Table
HolidayName
Start_date
End_Date
holiday1
5th Aug'21
6th Aug'21
condition :
this query run on 9th Aug 2021
expected result :
4th Aug'21
I've tried some query and function but I just can't get what I need.
Thanks a lot for help!
Here's one way to do it.
select max(d) as last_workday
from (select trunc(sysdate)-level as d from dual connect by level < 30) prior_month
where to_char(d, 'DY') not in ('SAT','SUN')
and not exists (select holidayname from holiday_table
where prior_month.d between start_date and end_date)
;
Without seeing your Holiday table, it's hard to say how many days back you would need to look to find the last workday. If you have a holiday that lasts for more than 30 days, you'll need to change the 30 to a larger number.
You can use a simple case expression to determine what day of the week the start of your holiday is, then subtract a number of days based on that.
WITH
holiday (holidayname, start_date, end_date)
AS
(SELECT 'holiday1', DATE '2021-8-5', DATE '2021-8-6' FROM DUAL
UNION ALL
SELECT 'Christmas', DATE '2021-12-25', DATE '2021-12-26' FROM DUAL
UNION ALL
SELECT 'July 4th', DATE '2021-7-4', DATE '2021-7-5' FROM DUAL)
SELECT holidayname,
start_date,
end_date,
start_date - CASE TO_CHAR (start_date, 'Dy') WHEN 'Mon' THEN 3 WHEN 'Sun' THEN 2 ELSE 1 END AS prior_business_day
FROM holiday;
HOLIDAYNAME START_DATE END_DATE PRIOR_BUSINESS_DAY
______________ _____________ ____________ _____________________
holiday1 05-AUG-21 06-AUG-21 04-AUG-21
Christmas 25-DEC-21 26-DEC-21 24-DEC-21
July 4th 04-JUL-21 05-JUL-21 02-JUL-21
You can use a recursive sub-query factoring clause from this answer:
WITH start_date (dt) AS (
SELECT DATE '2021-05-02' FROM DUAL
),
days ( dt, day, found ) AS (
SELECT dt,
TRUNC(dt) - TRUNC(dt, 'IW'),
0
FROM start_date
UNION ALL
SELECT dt - CASE day WHEN 0 THEN 3 WHEN 6 THEN 2 ELSE 1 END,
CASE WHEN day IN (0, 6, 5) THEN 4 ELSE day - 1 END,
CASE WHEN h.start_date IS NULL THEN 1 ELSE 0 END
FROM days d
LEFT OUTER JOIN holidays h
ON ( dt - CASE day WHEN 0 THEN 3 WHEN 6 THEN 2 ELSE 1 END
BETWEEN h.start_date AND h.end_date )
WHERE found = 0
)
SELECT dt
FROM days
WHERE found = 1;
Which, for the sample data:
CREATE TABLE holidays (HolidayName, Start_date, End_Date) AS
SELECT 'holiday1', DATE '2021-08-05', DATE '2021-08-06' FROM DUAL;
Outputs:
DT
2021-08-04 00:00:00
db<>fiddle here
Don't know if it's very efficient. Did it just for fun
create table holidays (
holiday_name varchar2(100) primary key,
start_date date not null,
end_date date not null
)
/
Table created
insert into holidays (holiday_name, start_date, end_date)
values ('holiday1', date '2021-08-05', date '2021-08-06');
1 row inserted
with days_before(day, wrk_day) as
(select trunc(sysdate - 1) d,
case
when h.holiday_name is not null then 0
when to_char(trunc(sysdate - 1), 'D') in ('6', '7') then 0
else 1
end work_day
from dual
left join holidays h
on trunc(sysdate - 1) between h.start_date and h.end_date
union all
select db.day - 1,
case
when h.holiday_name is not null then 0
when to_char(db.day - 1, 'D') in ('6', '7') then 0
else 1
end work_day
from days_before db
left join holidays h
on db.day - 1 between h.start_date and h.end_date
where db.wrk_day = 0) search depth first by day set order_no
select day from days_before where wrk_day = 1;
DAY
-----------
04.08.2021
In this query I found the weeks of month, for example november begin 1th Wednesday and ends sunday 5.
These are the first week on november.
SELECT * FROM (
WITH days AS
(SELECT to_date('01012017','ddmmyyyy') + level-1 date_in
FROM dual
CONNECT BY level < 32)
SELECT date_in,
TO_CHAR(date_in,'IW') - TO_CHAR(TRUNC(date_in,'MM'),'IW') + 1 week_number
FROM days) where week_number = 1;
The week_number can change depending of the weeks of the month, but in January is not working.
Your requirement is not clear for me but perhaps you are looking for this:
SELECT TRUNC (next_day(TRUNC(TO_DATE('20022017','ddmmyyyy'), 'mm'),'monday') - 1), 'IW') AS first_week
FROM dual;
I'm still a bit unclear as to what you're looking for, but perhaps the following query will help you on your way:
WITH YEAR_START AS (SELECT TO_DATE('01012017', 'DDMMYYYY') AS FIRST_DAY_OF_YEAR
FROM DUAL),
MONTH_START AS (SELECT FIRST_DAY_OF_YEAR AS FIRST_DAY_OF_MONTH
FROM YEAR_START
UNION ALL
SELECT ADD_MONTHS(FIRST_DAY_OF_YEAR, LEVEL) AS FIRST_DAY_OF_MONTH
FROM YEAR_START
CONNECT BY LEVEL <= 11),
MONTH_START_AND_END AS (SELECT FIRST_DAY_OF_MONTH,
ADD_MONTHS(FIRST_DAY_OF_MONTH, 1) - INTERVAL '1' DAY AS LAST_DAY_OF_MONTH
FROM MONTH_START),
ABS_MONTH_WEEKS AS (SELECT FIRST_DAY_OF_MONTH,
LAST_DAY_OF_MONTH,
TO_NUMBER(TO_CHAR(FIRST_DAY_OF_MONTH, 'IW')) AS ABS_FIRST_WEEK_OF_MONTH,
TO_NUMBER(TO_CHAR(LAST_DAY_OF_MONTH, 'IW')) AS ABS_LAST_WEEK_OF_MONTH
FROM MONTH_START_AND_END),
REL_MONTH_WEEKS AS (SELECT a.*,
1 AS REL_FIRST_WEEK_OF_MONTH,
ABS_LAST_WEEK_OF_MONTH - CASE
WHEN ABS_FIRST_WEEK_OF_MONTH > ABS_LAST_WEEK_OF_MONTH THEN 0
ELSE ABS_FIRST_WEEK_OF_MONTH-1
END AS REL_LAST_WEEK_OF_MONTH
FROM ABS_MONTH_WEEKS a)
SELECT *
FROM REL_MONTH_WEEKS;
Best of luck.
In Oracle we get week number from following query:
select to_char(TO_DATE(SYSDATE,'DD-MM-YY'),'IW') from dual
I want to get date range of given week number, for example for week no:1 date range is 01-01-2017 to 08-01-2017.
is there any way to get the date range for given week number?
"week no:1 date range is 01-01-2017 to 08-01-2017"
No it isn't. You're confusing 'IW' (which runs MON - SUN) with 'WW' which runs from the first day of the year:
SQL> with dts as (
2 select date '2017-01-01' + (level-1) as dt
3 from dual
4 connect by level <= 8
5 )
6 select dt
7 , to_char(dt, 'DY') as dy_dt
8 , to_char(dt, 'IW') as iw_dt
9 , to_char(dt, 'WW') as ww_dt
10 from dts
11 order by 1;
DT DY_DT IW WW
--------- ------------ -- --
01-JAN-17 SUN 52 01
02-JAN-17 MON 01 01
03-JAN-17 TUE 01 01
04-JAN-17 WED 01 01
05-JAN-17 THU 01 01
06-JAN-17 FRI 01 01
07-JAN-17 SAT 01 01
08-JAN-17 SUN 01 02
8 rows selected.
SQL>
However, it's easy enough to generate a range for the the IW week number. You need to multiple the IW number by 7 which you can convert to a date with the day of year mask. Then you can use next_day() function to get the previous Monday and the next Sunday relative to that date:
SQL> with tgt as (
2 select to_date( &iw *7, 'DDD') as dt from dual
3 )
4 select next_day(dt-8, 'mon') as start_date
5 , next_day(dt, 'sun') as end_date
6* from tgt;
Enter value for iw: 23
old 2: select to_date( &iw *7, 'DDD') as dt from dual
new 2: select to_date( 23 *7, 'DDD') as dt from dual
START_DAT END_DATE
--------- ---------
05-JUN-17 11-JUN-17
SQL>
Obvious this solution uses my NLS Settings (English): you may need to tweak the solution if you use different settings.
These kinds of problems are easy to solve with calendar tables.
The following query builds on the assumption (ISO 8601) that the 4th of January is present in the first week in a year. Therefore I can generate a valid date in the first week of any year by constructing the 4th of January like: to_date(year || '-01-04', 'yyyy-mm-dd'). Oracle will tell me the day of week (sun=1, sat=7) for any date using to_char(date, 'D'). The 4th of JAN 2017 happens to be a wednesday (day 4). Subtracting 3 days will give me the first day (sunday) of the first week of the year.
Now it is easy to find the start day in any given week in the year by simply adding 7 days for each week (not counting the first week).
with weeks as(
select 2017 as year, 39 as week from dual union all
select 2017 as year, 40 as week from dual union all
select 2018 as year, 35 as week from dual
)
select a.*
,to_date(year || '-01-04', 'yyyy-mm-dd') - to_number(to_char(to_date(year || '-01-04', 'yyyy-mm-dd'), 'D')) + 1 + (7 * (week-1)) as start_day
,to_date(year || '-01-04', 'yyyy-mm-dd') + 7 - to_number(to_char(to_date(year || '-01-04', 'yyyy-mm-dd'), 'D')) + (7 * (week-1)) as end_day
from weeks a;
Edit: These are the "convert" expressions you need to convert from week to date range. Note that 2017 and 39 are variable...
start date = to_date(2017 || '-01-04', 'yyyy-mm-dd') - to_number(to_char(to_date(2017 || '-01-04', 'yyyy-mm-dd'), 'D')) + 1 + (7 * (39-1))
end date = to_date(2017 || '-01-04', 'yyyy-mm-dd') + 7 - to_number(to_char(to_date(2017 || '-01-04', 'yyyy-mm-dd'), 'D')) + (7 * (39-1))
Here's a query to list all ISO weeks from 2001 to 2099
SELECT TO_CHAR(TRUNC(dt, 'IW') + 6, 'IYYY-IW') AS week,
TRUNC(dt, 'IW') AS start_date,
TRUNC(dt, 'IW') + 6 AS end_date
FROM (SELECT DATE '2001-01-01' + ((LEVEL - 1) * 7) dt
FROM DUAL
CONNECT BY LEVEL <= 5165);
For the first and last week of year this query needs some CASE logic, but for other weeks works good. This solution use current NLS settings.
select to_char( start_of_week, 'day dd.mm.yyyy' ) start_of_week,
to_char( start_of_week + 6, 'day dd.mm.yyyy' ) end_of_week
from
(
select trunc( date '2017-01-01' + 38*7 , 'day') start_of_week
from dual
)
1) date '2017-01-01' - in what year we look for weeks
or it may be trunc (sysdate, 'YEAR') to take first day of current year
2) date '2017-01-01' + 38*7 - jump to 38th week
3) trunc ( ... , 'day' ) - gives date of first day of the week
https://docs.oracle.com/cd/B19306_01/server.102/b14200/functions201.htm
https://docs.oracle.com/cd/B19306_01/server.102/b14200/functions230.htm
I use this function:
FUNCTION ISOWeekDate(WEEK INTEGER, YEAR INTEGER) RETURN DATE DETERMINISTIC IS
res DATE;
BEGIN
IF WEEK > 53 OR WEEK < 1 THEN
RAISE VALUE_ERROR;
END IF;
res := NEXT_DAY(TO_DATE( YEAR || '0104', 'YYYYMMDD' ) - 7, 'MONDAY') + ( WEEK - 1 ) * 7;
IF TO_CHAR(res, 'fmIYYY') = YEAR THEN
RETURN res;
ELSE
RAISE VALUE_ERROR;
END IF;
END ISOWeekDate;
Please note, according to my comment it is ambiguous if you only provide a week number without a year. The function returns the first day of given ISO Week.
I am woking on a query which can give back the count divided by month about the offices that will be closed this summer.
SELECT
qa.tmonth,
COUNT(qa.tmonth) AS qtn
FROM
(
SELECT TO_CHAR(CLOSURE_DATE, 'yyyymm') AS tmonth
FROM Holidays
WHERE CLOSURE_DATE >= TO_DATE('20160501', 'YYYY-MM-DD') AND
CLOSURE_DATE <= TO_DATE('20160901', 'YYYY-MM-DD')
) qa
GROUP BY qa.tmonth;
Since the months: May, June, August and September no office will be closed the output is the following:
TMONTH|QTN
201607|80
But I need a thing like this
TMONTH|QTN
201605|0
201606|0
201607|80
201608|0
201609|0
How could I achieve that?
Thanks to all!
You can try with something like this:
SQL> with holidays(closure_date) as
2 (
3 select date '2016-07-01' from dual union all
4 select date '2016-07-02' from dual union all
5 select date '2016-07-03' from dual union all
6 select date '2016-07-04' from dual union all
7 select date '2016-07-05' from dual
8 )
9 select count(closure_date) as closure_days, to_char(day, 'yyyymm') as month
10 from (
11 select date '2016-05-01' + level -1 as day
12 from dual
13 connect by date '2016-05-01' + level -1 <= date '2016-09-30'
14 ) days
15 left outer join holidays
16 on (day = closure_date)
17 group by to_char(day, 'yyyymm') ;
CLOSURE_DAYS MONTH
------------ ------
0 201608
5 201607
0 201606
0 201605
0 201609
SQL>
This uses a query to build the list of all the days between a starting and an ending date; I used 01/05 and 30/09 and called it days.
Then it queries days with the holidays table in outer join; this way you can count only the days for which there is a corrensponding value in the closure days list, thus counting the closure days for each day, month year; the aggregation for year and month completes the job
A similar approach like above. Tip: You can execute the two sub-queries separately, to analyse the logic.
select to_char (m.month, 'yyyymm') as TMONTH, m.month,
nvl (h.qtn, 0) as QTN
from
(
SELECT add_months(trunc (SYSDATE, 'MONTH'), -(LEVEL-1)) as MONTH
FROM dual
CONNECT BY LEVEL <= 12 -- generate a list of the last 12 month
) m
left join
(
SELECT trunc (closure_date, 'MONTH') as MONTH,
count (*) as QTN
FROM Holidays
group by trunc (closure_date, 'MONTH')
) h
on m.MONTH = h.MONTH
where m.month between DATE '2016-01-01' and sysdate
order by TMONTH desc;
I have a date field
DATE = 10/10/2010
sum = 4 (this are number of years by calculation)
is there a way to add four years to 10/10/2010 and make it
10/10/2014?
Try adding months (12 * number of years) instead. Like this-
add_months(date'2010-10-10', 48)
Use add_months
Example:
SELECT add_months( to_date('10-OCT-2010'), 48 ) FROM DUAL;
Warning
add_months, returns the last day of the resulting month if you input the last day of a month to begin with.
So add_months(to_date('28-feb-2011'),12) will return 29-feb-2012 as a result.
I believe you could use the ADD_MONTHS() function. 4 years is 48 months, so:
add_months(DATE,48)
Here is some information on using the function:
http://www.techonthenet.com/oracle/functions/add_months.php
http://asktom.oracle.com/pls/asktom/f?p=100:11:0::::P11_QUESTION_ID:1157035034361
You can try this:
someDate + interval '4' year
INTERVAL
I am not sure, if I understood Your question correctly, but
select add_months(someDate, numberOfYears * 12) from dual
might do the trick
One more option apart from ADD_MONTHS
SELECT
SYSDATE,
SYSDATE
+ TO_YMINTERVAL ( '1-0' )
FROM
DUAL;
SYSDATE SYSDATE+TO_YMINTERVAL('1-0')
--------- ----------------------------
29-OCT-13 29-OCT-14
1 row selected.
SELECT
SYSDATE,
SYSDATE
+ TO_YMINTERVAL ( '2-0' )
FROM
DUAL;
SYSDATE SYSDATE+TO_YMINTERVAL('2-0')
--------- ----------------------------
29-OCT-13 29-OCT-15
1 row selected.
SELECT
TO_DATE ( '29-FEB-2004',
'DD-MON-YYYY' )
+ TO_YMINTERVAL ( '1-0' )
FROM
DUAL
*
Error at line 4
ORA-01839: date not valid for month specified
But the last one is illegal since there is no 29th day of February in 2005, hence it fails on leap year cases (Feb 29)
Read the documentation for the same
SELECT TO_CHAR(SYSDATE,'YYYY')-2 ANO FROM DUAL