Calculate the number of days per month between two dates - oracle

Using Oracle 12c, I need to run a script on an existing summary table of projects. The summary table has a project, a start date, and an end date. I need to break this data out into the number of days per month for each project.
An example is Project A has a start date of 2/10/2016 and an end date of 3/10/2016. My ending result for this example should be:
Project A, February, 19
Project A, March, 10
This was an easier one as some dates may span 2 or 3 months. This doesn't seem too difficult but for some reason I'm having trouble wrapping my head around it and overthinking it. Does someone have an quick and easy solution to this? I would like to run this as a SQL statement but a PL/SQL script would also work.

In this solution we don't assume any prior knowledge of the time period covered. Also, this solution does not use joins (which may be important for performance).
with
-- begin test data (this section can be deleted)
inputs ( project, start_date, end_date ) as (
select 'A', date '2014-10-03', date '2014-12-15' from dual union all
select 'B', date '2015-03-01', date '2015-03-31' from dual union all
select 'C', date '2015-11-30', date '2016-03-01' from dual
),
-- end test data; solution begins here (it includes the word "with" from the first line)
prep ( project, end_date, dt ) as (
select project, end_date, start_date from inputs union all
select project, end_date, end_date + 1 from inputs union all
select project, end_date, add_months( trunc(start_date, 'mm'), level )
from inputs
connect by add_months (trunc(start_date, 'mm'), level) <= end_date
and prior project = project
and prior sys_guid() is not null
),
computations ( project, dt, month, day_count ) as (
select project, dt, to_char(dt, 'Mon-yyyy'),
lead(dt) over (partition by project order by dt) - dt
from prep
where dt <= end_date + 1
)
select project, month, day_count
from computations
where day_count > 0
order by project, dt
;
OUTPUT:
PROJECT MONTH DAY_COUNT
------- -------- ---------
A Oct-2014 29
A Nov-2014 30
A Dec-2014 15
B Mar-2015 31
C Nov-2015 1
C Dec-2015 31
C Jan-2016 31
C Feb-2016 29
C Mar-2016 1
9 rows selected

If you can do an assumption on the maximum number of days for a project (1000 in my example), you can use the following:
with yourTable(project, startDate, endDate) as
(
select 'Project a' as project,
date '2016-02-10' as startDate,
date '2016-03-10' as endDate
from dual
UNION ALL
select 'Project XX',
date '2016-01-01',
date '2016-01-10'
from dual
)
select project, to_char(startDate + n, 'MONTH'), count(1)
from yourTable
inner join (
select level n
from dual
connect by level <= 1000
)
on (startDate + n <= endDate)
group by project, to_char(startDate + n, 'MONTH')
The part with the CONNECT BY is used as a date generator, assuming that every project is at maximum 1000 days long; the external query uses the date generator to split the row of a project in many rows, one for each day between start and end date, and then aggregates by month and project to build the output.
A slightly different way, based on months and not days, could be:
select project, to_char(add_months(startDate, n ), 'MONTH'),
case
when trunc(add_months(startDate, n ), 'MONTH') = trunc(add_months(endDate, n ), 'MONTH')
then endDate - startDate +1
when trunc(add_months(startDate, n ), 'MONTH') <= startDate
then last_day(add_months(startDate, n)) - startDate
when last_day(add_months(startDate, n )) >= endDate
then endDate - trunc(add_months(startDate, n ), 'MONTH') +1
else
last_day(add_months(startDate, n )) - trunc(last_day(add_months(startDate, n )), 'MONTH')
end as numOfDays
from yourTable
inner join (
select level -1 n
from dual
connect by level <= 1000
)
on trunc(add_months(startDate, n ), 'MONTH') <= trunc(endDate, 'MONTH')
This is a bit more complicated, to handle the different cases, but more efficient, given that it works at month level, not day level

I think you're after something like:
WITH sample_data AS (SELECT 'A' PROJECT, to_date('10/02/2016', 'dd/mm/yyyy') start_date, to_date('10/03/2016', 'dd/mm/yyyy') end_date FROM dual UNION ALL
SELECT 'B' PROJECT, to_date('10/02/2016', 'dd/mm/yyyy') start_date, to_date('10/06/2016', 'dd/mm/yyyy') end_date FROM dual UNION ALL
SELECT 'C' PROJECT, to_date('10/02/2016', 'dd/mm/yyyy') start_date, to_date('18/02/2016', 'dd/mm/yyyy') end_date FROM dual)
SELECT PROJECT,
to_char(add_months(trunc(start_date, 'mm'), LEVEL -1), 'fmMonth yyyy', 'nls_date_language=english') mnth,
CASE WHEN trunc(end_date, 'mm') = add_months(trunc(start_date, 'mm'), LEVEL -1)
THEN end_date
ELSE add_months(trunc(start_date, 'mm'), LEVEL) -1
END - CASE WHEN trunc(start_date, 'mm') = add_months(trunc(start_date, 'mm'), LEVEL -1)
THEN start_date + 1
ELSE add_months(trunc(start_date, 'mm'), LEVEL -1)
END + 1 num_days
FROM sample_data
CONNECT BY PRIOR PROJECT = PROJECT
AND PRIOR sys_guid() IS NOT NULL
AND add_months(trunc(start_date, 'mm'), LEVEL -1) <= TRUNC(end_date, 'mm');
PROJECT MNTH NUM_DAYS
------- -------------- ----------
A February 2016 19
A March 2016 10
B February 2016 19
B March 2016 31
B April 2016 30
B May 2016 31
B June 2016 10
C February 2016 8
This uses the multi-row connect-by-level technique (the presence of the and prior sys_guid() is not null enables the connect by to loop through each row separately) to loop through each project row in the sample_data table (you presumably have the project information in a table already, so you wouldn't need to have the sample_data subquery at all; you could just reference your table directly in the main SQL).
We then compare the month of the start date with the month of the row being generated by the connect by, and if it's the same month, then we know we need to use the start date, otherwise we use the first of the month of the generated row; we do similarly for the end date.
That way, we can now subtract one from the other and make adjustments to make the calculation correct. You may need to tweak this yourself if you need a start and end date of the same day to count as 1 day, rather than 0 - it'll probably need an extra case statement to take account of when the start and end date are in the same month.
Using this approach won't limit your project length; it could be as long as you liked.
ETA: Looks like Mathguy posted an answer whilst I was typing out my answer, and whilst our basic methods are the same, mine doesn't use an analytic function to determine the difference in the number of days. You may or may not find their answer more performant than mine - you should test both to see which one works best with your data.

Related

Stop condition for recursive CTE on Oracle (ORA-32044)

I have the following recursive CTE which splits each element coming from base per month:
with
base (id, start_date, end_date) as (
select 1, date '2022-01-15', date '2022-03-15' from dual
union
select 2, date '2022-09-15', date '2022-12-31' from dual
union
select 3, date '2023-09-15', date '2023-09-25' from dual
),
split (id, start_date, end_date) as (
select base.id, base.start_date, least(last_day(base.start_date), base.end_date) from base
union all
select base.id, split.end_date + 1, least(last_day(split.end_date + 1), base.end_date) from base join split on base.id = split.id and split.end_date < base.end_date
)
select * from split order by id, start_date, end_date;
It works on Oracle and gives the following result:
id
start_date
end_date
1
2022-01-15
2022-01-31
1
2022-02-01
2022-02-28
1
2022-03-01
2022-03-15
2
2022-09-15
2022-09-30
2
2022-10-01
2022-10-31
2
2022-11-01
2022-11-30
2
2022-12-01
2022-12-31
3
2023-09-15
2023-09-25
The two following stop conditions work correctly:
... from base join split on base.id = split.id and split.end_date < base.end_date
... from base, split where base.id = split.id and split.end_date < base.end_date
The following one fails with the message ORA-32044: cycle detected while executing recursive WITH query:
... from base join split on base.id = split.id where split.end_date < base.end_date
I fail to understand how the last one is different from the two others.
It looks like a bug as all your queries should result in identical explain plans.
However, you can rewrite the recursive sub-query without the join (and using a SEARCH clause so you may not have to re-order the query later):
WITH split (id, start_date, month_end, end_date) AS (
SELECT id,
start_date,
LEAST(
ADD_MONTHS(TRUNC(start_date, 'MM'), 1) - INTERVAL '1' SECOND,
end_date
),
end_date
FROM base
UNION ALL
SELECT id,
month_end + INTERVAL '1' SECOND,
LEAST(
ADD_MONTHS(month_end, 1),
end_date
),
end_date
FROM split
WHERE month_end < end_date
) SEARCH DEPTH FIRST BY id, start_date SET order_id
SELECT id,
start_date,
month_end AS end_date
FROM split;
Note: if you want to just use values at midnight rather than the entire month then use INTERVAL '1' DAY rather than 1 second.
Which, for the sample data:
CREATE TABLE base (id, start_date, end_date) as
select 1, date '2022-01-15', date '2022-04-15' from dual union all
select 2, date '2022-09-15', date '2022-12-31' from dual union all
select 3, date '2023-09-15', date '2023-09-25' from dual;
Outputs:
ID
START_DATE
END_DATE
1
2022-01-15T00:00:00Z
2022-01-31T23:59:59Z
1
2022-02-01T00:00:00Z
2022-02-28T23:59:59Z
1
2022-03-01T00:00:00Z
2022-03-31T23:59:59Z
1
2022-04-01T00:00:00Z
2022-04-15T00:00:00Z
2
2022-09-15T00:00:00Z
2022-09-30T23:59:59Z
2
2022-10-01T00:00:00Z
2022-10-31T23:59:59Z
2
2022-11-01T00:00:00Z
2022-11-30T23:59:59Z
2
2022-12-01T00:00:00Z
2022-12-31T00:00:00Z
3
2023-09-15T00:00:00Z
2023-09-25T00:00:00Z
fiddle
It's because WHERE and ON conditions are not evaluated at the same level:
when the condition is in the ON clause it's limiting the rows concerned by the JOIN, where it's in the WHERE it's filtering the results after the JOIN has been applied, and since a recursive CTE see all rows selected up to now...

How to get last workday before holiday in Oracle [duplicate]

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

How to convert this code from oracle to redshift?

I am trying to implement the same in redshift and i am finding it little difficult to do that. Since redshift is in top of postgresql engine, if any one can do it in postgresql it would be really helpfull. Basically the code gets the count for previous two month at column level. If there is no count for exact previous month then it gives 0.
This is my code:
with abc(dateval,cnt) as(
select 201908, 100 from dual union
select 201907, 200 from dual union
select 201906, 300 from dual union
select 201904, 600 from dual)
select dateval, cnt,
last_value(cnt) over (order by dateval
range between interval '1' month preceding
and interval '1' month preceding ) m1,
last_value(cnt) over (order by dateval
range between interval '2' month preceding
and interval '2' month preceding ) m2
from (select to_date(dateval, 'yyyymm') dateval, cnt from abc)
I get error in over by clause. I tried to give cast('1 month' as interval) but still its failing. Can someone please help me with this windows function.
expected output:
Regards
This is how I would do it. In Redshift there's no easy way to generate sequences, do I select row_number() from an arbitrary table to create a sequence:
with abc(dateval,cnt) as(
select 201908, 100 union
select 201907, 200 union
select 201906, 300 union
select 201904, 600),
cal(date) as (
select
add_months(
'20190101'::date,
row_number() over () - 1
) as date
from <an arbitrary table to generate a sequence of rows> limit 10
),
with_lag as (
select
dateval,
cnt,
lag(cnt, 1) over (order by date) as m1,
lag(cnt, 2) over (order by date) as m2
from abc right join cal on to_date(dateval, 'YYYYMM') = date
)
select * from with_lag
where dateval is not null
order by dateval

Just the month january not working

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.

Retrieve rows with 0 count from Oracle

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;

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