I want to display result set of Years between From date - To date using oracle SQL on dual table
e.g.
if i pass - From date as 1/1/1900 and To Date as 1/1/2000
then it shoold display
Only Years
1900
1901
1902
-
-
2000
There are two parts to this question. Generating the range of dates is quite simple: just use the trick with CONNECT BY that I demonstrated here.
edit
Generating a list of first of New Year's Days is quite simple:
SQL> select add_months(to_date('01-jan-1900'), (level-1)*12) as year
2 from dual
3 connect by level <= 101
4 /
YEAR
---------
01-JAN-00
01-JAN-01
01-JAN-02
...
01-JAN-98
01-JAN-99
01-JAN-00
101 rows selected.
SQL>
You just want the years? Well either use to_char(... , 'YYYY') on that. Or cut to the chase and just generate a list of numbers from 1900 - 2000.
The trickiest part of your request is getting the number of years. It would be easier to be given a start date and an offset, rather than an end date. Anyway ...
SQL> select to_char(add_months(to_date('&&start_date'), (level-1)*12), 'YYYY') as year
2 from dual
3 connect by level <= ( to_number(to_char(to_date('&&end_date'), 'yyyy'))
4 -to_number(to_char(to_date('&&start_date'), 'yyyy')) ) + 1
5 /
Enter value for start_date: 01-jan-1900
old 1: select add_months(to_date('&&start_date'), (level-1)*12) as year
new 1: select add_months(to_date('01-jan-1900'), (level-1)*12) as year
Enter value for end_date: 01-jan-2000
old 3: connect by level <= ( to_number(to_char(to_date('&&end_date'), 'yyyy'))
new 3: connect by level <= ( to_number(to_char(to_date('01-jan-2000'), 'yyyy'))
old 4: -to_number(to_char(to_date('&&start_date'), 'yyyy')) ) - 1
new 4: -to_number(to_char(to_date('01-jan-1900'), 'yyyy')) ) - 1
YEAR
----
1900
1901
1902
...
1998
1999
2000
101 rows selected.
SQL>
Related
i have a query and I want to calculate the number of sat+sun total count in oracle, for example, I have a query pasted below there should be a total count of Saturday and Sunday, how can I achieve that please help, I really appreciate any help you can provide.
SELECT TO_DATE('01-12-2022','dd-mm-yyyy') start_date , TO_DATE(sysdate) end_date
FROM dual;
Don't use a row-generator to create a calendar (as it is very inefficient); just calculate the number by calculating the number of full weeks and then deal with the part weeks at the start and end of the range:
WITH range (start_date, end_date) AS (
SELECT DATE '2022-12-01', TRUNC(SYSDATE) FROM DUAL
)
SELECT -- Number of full weeks
(TRUNC(end_date, 'IW') - TRUNC(start_date, 'IW')) * 2/7
-- Number of weekend days in final week
+ GREATEST(end_date - TRUNC(end_date, 'IW') - 4, 0)
-- Number of weekend days in before first week
- GREATEST(start_date - TRUNC(start_date, 'IW') - 5, 0)
AS weekend_day_count
FROM range;
Which outputs:
WEEKEND_DAY_COUNT
8
fiddle
One option is to create a calendar between these two dates and then count number of Saturdays and Sundays:
SQL> with
2 test (start_date, end_date) as
3 -- period
4 (select date '2022-12-01', date '2022-12-29' from dual),
5 calendar as
6 -- calendar (all dates between START_DATE and END_DATE)
7 (select start_date + level - 1 as datum
8 from test
9 connect by level <= end_date - start_date + 1
10 )
11 -- number of Saturdays and Sundays
12 select count(*)
13 from calendar
14 where to_char(datum, 'dy', 'nls_date_language = english') in ('sat', 'sun');
COUNT(*)
----------
8
SQL>
You'd change dates at line #4.
P.S. If you look at code MT0 posted and their objection that row generator is inefficient, that's true. Although both queries return the same result, timing is different. For example:
Period 01.01.2022 - 31.12.2022 01.01.1900 - 31.12.2022 01.01.0001 - 31.12.2022
------ ----------------------- ----------------------- -----------------------
LF 00:00:00.00 00:00:00.17 00:00:03.72
MT0 00:00:00.02 00:00:00.02 00:00:00.05
It is obvious that my timing gets worse with period length. If you're looking at one year or a century, the difference is mostly irrelevant. For 2000 years, the difference is huge!
However, if you consider debugging, from my own point of view, my code is easier to read: "select number of rows from the calendar where date is either saturday or sunday" - plain English.
On the other hand, the other code isn't that straightforward; truncate date to week, subtract them, multiply by 2/7 (why "2/7" and not 4/9?), add result returned by the GREATEST function minus 4 (why 4? Why not 7?), subtract GREATEST of something minus 5 (why 5? Why not 2?) - as I said, that's NOT easy to read nor understand.
Therefore, it depends on what you actually need, timing vs. readability. Pick one :)
I have a table with business days BUSINESS_DAYS which has all the dates
I have another table with payment information and DUE_DATES
I want to return in my query the next business day IF the DUE_DATE is not a business day
SELECT SQ1.DUE_DATE, SQ2.DATE FROM
(SELECT * FROM
PAYMENTS
ORDER BY
DUE_DATE) SQ1,
(SELECT MIN(DATE) DATE FROM BUSINESS_DAYS WHERE SQ1.DUE_DATE <= DATE GROUP BY DATE) SQ2
Anyone can shed some light?
The way I see it, code you posted doesn't do what you wanted anyway (otherwise, you won't be asking a question at all). Therefore, I'd suggest another approach:
Altering the session (you don't have to do it; my database speaks Croatian so I'm switching to English; also, setting date format to display day name):
SQL> alter session set nls_date_language = 'english';
Session altered.
SQL> alter session set nls_date_format = 'dd.mm.yyyy, dy';
Session altered.
Two CTEs contain
business_days: as commented, only this year's July, weekends excluded, there are no holidays)
payments: two rows, one whose due date is a working day and another whose isn't
Sample data end at line #15, query you might be interested in begins at line #16. Its CASE expression check whether due_date is one of weekend days; if not, due date to be returned is exactly what it is. Otherwise, another SELECT statement returns the first (MIN) business day larger than due_date.
SQL> with
2 business_days (datum) as
3 -- for simplicity, only all dates in this year's July,
4 -- weekends excluded (as they aren't business days), no holidays
5 (select date '2021-07-01' + level - 1
6 from dual
7 where to_char(date '2021-07-01' + level - 1, 'dy')
8 not in ('sat', 'sun')
9 connect by level <= 31
10 ),
11 payments (id, due_date) as
12 (select 1, date '2021-07-14' from dual -- Wednesday, business day
13 union all
14 select 2, date '2021-07-25' from dual -- Sunday, non-business day
15 )
16 select p.id,
17 p.due_date current_due_date,
18 --
19 case when to_char(p.due_date, 'dy') not in ('sat', 'sun') then
20 p.due_date
21 else (select min(b.datum)
22 from business_days b
23 where b.datum > p.due_date
24 )
25 end new_due_date
26 from payments p
27 order by id;
ID CURRENT_DUE_DAT NEW_DUE_DATE
---------- --------------- ---------------
1 14.07.2021, wed 14.07.2021, wed --> Wednesday remains "as is"
2 25.07.2021, sun 26.07.2021, mon --> Sunday switched to Monday
SQL>
I have a schedule of my training, three times a week, for example -
MON, WED,FRI. I need to generate records for my schedule table with dates till the end of the current year when I have training.
The schedule table is:
CREATE TABLE trainingSchedule (
id NUMBER,
training_date DATE
);
If training date already exist - don’t insert a record.
Here's one option. Read comments within code.
SQL> CREATE TABLE trainingSchedule
2 (id NUMBER,
3 training_date DATE
4 );
Table created.
SQL> create sequence seq_tra;
Sequence created.
SQL> -- initial insert (just to show that MERGE will skip it
SQL> insert into trainingschedule values (seq_tra.nextval, date '2021-03-22');
1 row created.
MERGE will skip rows that are already inserted. I understood that you want to insert only dates that follow today's date; if that's not so, just remove the last condition.
SQL> merge into trainingschedule t
2 using (-- this is a calendar for current year
3 select trunc(sysdate, 'yyyy') + level - 1 datum
4 from dual
5 connect by level <= add_months(trunc(sysdate, 'yyyy'), 12) - trunc(sysdate, 'yyyy')
6 ) c
7 on (c.datum = t.training_date)
8 when not matched then insert (id, training_date) values (seq_tra.nextval, c.datum)
9 -- insert only Mondays, Wednesdays and Fridays
10 where to_char(c.datum, 'dy', 'nls_date_language = english') in ('mon', 'wed', 'fri')
11 -- insert only dates that follow today's date ("till the end of the current year")
12 and datum >= trunc(sysdate);
122 rows merged.
SQL>
What's in there?
SQL> select id,
2 to_char(training_date, 'dd.mm.yyyy, dy', 'nls_date_language = english') tr_date
3 from trainingschedule
4 order by training_date;
ID TR_DATE
---------- ------------------------
1 22.03.2021, mon --> see? No duplicates
311 24.03.2021, wed
309 26.03.2021, fri
207 29.03.2021, mon
354 31.03.2021, wed
321 02.04.2021, fri
<skip>
I have afile where i recieve Birthdates and insert them into my Database.
the format is like the following
03-JUN-52
I use the following script to insert the date
update data."PersonBDates" set BIRTHDATE = to_date('13-SEP-47', 'DD-MON-YY');
and i also used
update data."PersonBDates" set BIRTHDATE = to_date('13-SEP-47', 'DD-MON-RR');
but when i check if find it 2074 not 1947.
How to insert this date into my oracle database?
Generally speaking, RR should work, but - not in all cases. You'll have to fix data first because RR will return different values:
for years from 00 to 49 you'll get this century, 20xx, while
50 to 99 will return previous century, 19xx
Here's an example:
SQL> alter session set nls_date_format = 'dd.mm.yyyy';
Session altered.
SQL> select
2 to_date('03-07-52', 'dd-mm-rr') rr1,
3 to_date('03-07-52', 'dd-mm-yy') yy1 ,
4 --
5 to_date('03-07-47', 'dd-mm-rr') rr2,
6 to_date('03-07-47', 'dd-mm-yy') yy2
7 from dual;
RR1 YY1 RR2 YY2
---------- ---------- ---------- ----------
03.07.1952 03.07.2052 03.07.2047 03.07.2047
SQL>
As you can see, both RR and YY format mask for year 47 return 2047.
What to do? Concatenate 19 to all years, e.g.
SQL> with test (col) as
2 (select '03-07-52' from dual union all
3 select '03-07-47' from dual
4 )
5 select col,
6 to_date(substr(col, 1, 6) || '19' || substr(col, -2), 'dd-mm-rrrr') result
7 ---------------- ---------------
8 -- this is "03-07-" "19" the last 2 digits
9 --
10 from test;
COL RESULT
-------- ----------
03-07-52 03.07.1952
03-07-47 03.07.1947
SQL>
[EDIT]
If your current inserting script works OK - which I doubt, regarding error code you mentioned in a comment:
ORA-01858: a non-numeric character was found where a numeric was expected
which means that not all input data have the same, expected & correct format of DD-MON-YY, then a simple way to fix birthdates might be this:
subtract 100 years from all dates whose year is larger than 2000
Here's how:
SQL> create table test (birthdate date);
Table created.
SQL> insert into test
2 select to_date('03-07-52', 'dd-mm-rr') from dual union all
3 select to_date('03-07-47', 'dd-mm-rr') from dual;
2 rows created.
SQL> select * from test;
BIRTHDATE
----------
03.07.1952
03.07.2047
SQL> update test set
2 birthdate = add_months(birthdate, -100 * 12)
3 where extract (year from birthdate) > 2000;
1 row updated.
SQL> select * from test;
BIRTHDATE
----------
03.07.1952
03.07.1947
SQL>
You can modify that, of course, if there's someone who actually was born in 2000 or later.
As of error you got (ORA-01858), well, fixing it depends on how exactly you're entering those values into a table.
if it was a SQL*Loader, invalid values would be rejected and stored into the .bad file and you could fix them and reprocess them later
if it was using an external tables, you could use a where clause and omit invalid rows; for example, use regexp_like
Or, your best option is to make sure that all input values are valid dates. Then any option you choose (I mentioned previously) would work without ORA-xxxxx errors.
Alternate way of concatenating 19 to all years, as Littlefoot suggested.
to_date(regexp_replace('13-SEP-47', '([0-9]+$)', '19\1'), 'DD-MON-YYYY')
I would suggest to implement the solution where 01 is not considered as 1901 but 2001 or something similar (I assume that birthday year is not 1901 for any person in your system).
Case when substr(col, -2) < to_char(sysdate,'YY')
then to_date(col, 'DD-MON-YY')
else to_date(substr(col, 1, 6) || '19' || substr(col, -2), 'dd-mm-rrrr'
End
Cheers!!
Ok, I've had a query that has been working fine that calculates the week number from December 1 (the start of our Sales Fiscal Year).
Now the requirements have changed. I still need to calculate the week number based on the field (Invoice_Date). However, instead of starting to count from December 1 (Dec 1-7, Week 1, etc.) now I need to start counting on the nearest Monday to December 1st. As I understand it, the ISO week is kind of what I'm looking for but it starts January 1. How do I modify this to work from December 1?
Any help would be greatly appreciated.
select next_day(to_date('0112' || to_char(sysdate, 'YYYY'),'ddmmyyyy') - 1, 'MONDAY') dec_mon from dual; gives you first Monday of December current year
Number of week is just ceil((sysdate - dec_mon)/7).
If you want last Monday before 1st Dec you can get it by:
select next_day(to_date('2511' || to_char(sysdate, 'YYYY'),'ddmmyyyy') - 1, 'MONDAY')
from dual;
In this proposed solution, I build a "helper table" first, showing the Monday_from and Monday_to for each fiscal year (in the third CTE, named ranges). Then I build a few test dates - I was lazy, I should have used to_date() so I can include time-of-day component as well. The join condition in the actual solution (at the end of the code) is written so it works without modification for dates with non-zero "time-of-day" component.
I used the nice feature of Oracle 11.2 which allows us to give column aliases in the declaration of CTEs - otherwise the column aliases would need to be moved inside the respective SELECTs. Otherwise the solution should work at least for Oracle 9 and above (I think).
with
y ( dt ) as (
select add_months(date '2000-12-01', 12 * level )
from dual
connect by level <= 30
),
m ( dt ) as (
select trunc(dt, 'iw') + case when dt - trunc(dt, 'iw') <= 3 then 0 else 7 end
from y
),
ranges ( monday_from, monday_to ) as (
select dt, lead(dt) over (order by dt) - 1
from m
),
test_dates ( t_date ) as (
select date '2013-02-23' from dual union all
select date '2008-12-01' from dual union all
select date '2008-04-28' from dual union all
select date '2016-11-29' from dual
)
select t_date, monday_from, 1 + trunc((t_date - monday_from)/7) as week_no
from test_dates t inner join ranges r
on t.t_date >= r.monday_from and t.t_date < r.monday_to
;
T_DATE MONDAY_FROM WEEK_NO
------------------- ------------------- ----------
2008-04-28 00:00:00 2007-12-03 00:00:00 22
2008-12-01 00:00:00 2008-12-01 00:00:00 1
2013-02-23 00:00:00 2012-12-03 00:00:00 12
2016-11-29 00:00:00 2016-11-28 00:00:00 1
The nearest Monday to any any given date is returned with the following function:
NEXT_DAY(some_date-4,'Monday')
as shown by this query:
with dts(some_date) as (
select date '2006-12-1' from dual
union all
select add_months(some_date,12)
from dts
where some_date <= date '2014-12-1'
)
select some_date
, next_day(some_date-4,'monday') nearest
, some_date - next_day(some_date-4,'monday') dist
from dts;
SOME_DATE NEAREST DIST
----------- ----------- ----------
01-DEC-2006 04-DEC-2006 -3
01-DEC-2007 03-DEC-2007 -2
01-DEC-2008 01-DEC-2008 0
01-DEC-2009 30-NOV-2009 1
01-DEC-2010 29-NOV-2010 2
01-DEC-2011 28-NOV-2011 3
01-DEC-2012 03-DEC-2012 -2
01-DEC-2013 02-DEC-2013 -1
01-DEC-2014 01-DEC-2014 0
01-DEC-2015 30-NOV-2015 1
10 rows selected