I have a table 'Holiday' which lists a set of holiday details.If i specify a date,I should obtain a result date after 5 days of specified date.If there is holiday in between it should exclude them and display the non holiday date.I have table named holiday which includes holiday date,holiday type|(weekly off,local holiday).Now i have used nested decode for continuous holiday checking.Tell me how this can be changed in case function.
DECODE
(date,
holidaydate, DECODE
(date + 1,
holidaydate + 1, DECODE
(date + 2,
holidaydate + 2, DECODE
(date + 3,holidaydate+3,date+4,date+3),date+2),date+1),date);
This can be achieved with a simple subquery which counts the number of holiday dates between a specified date and date+5. The following will return a date that is five non-holiday days in the future:
testdate+(select 5+count(1)
from holiday
where holidaydate between testdate
and testdate + 5)
Simply change both "5"s so another number to change the evaluation period.
SQLFiddle here
Edit - based on comment below, my code doesn't evaluate any days after the fifth day. This would probably be much easier with a function, but the following cte-based code will work also:
with cte as ( (select alldate,holidaydate
from (select to_date('20130101','yyyymmdd')+level alldate
from dual
connect by level < 10000 -- adjust for period to evaluate
) alldates
left join holiday on alldate=holidaydate) )
select
testdate,test_plus_five
from (
select
alldate test_plus_five,testdate,
sum(case when holidaydate is null
then 1
else 0 end) over (partition by testdate order by alldate) lastday
from
cte,
testdates
where
alldate >= testdate
group by
alldate,holidaydate,testdate)
where
lastday = 6
This script builds a calendar table so it can evaluate each day (holiday or non-holiday); then we get a running count of non-holiday days, and use the sixth one.
SQLFiddle here
AFAIK, You can use CASE alternative to DECODE in Oracle
CASE [ expression ]
WHEN condition_1 THEN result_1
WHEN condition_2 THEN result_2
...
WHEN condition_n THEN result_n
ELSE result
END
Finally i found the optimal solution.Thanks for ur response guys. SELECT dt FROM
(SELECT dt FROM (SELECT TO_DATE('15-AUG-2013','dd-mon-yyyy')+LEVEL dt FROM DUAL
CONNECT BY LEVEL < 30)
WHERE
(SELECT COUNT (*) FROM mst_holiday WHERE holidaydate = dt) = 0 )
where rownum=1
Related
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...
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
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.
I have an account where interest is debited corresponding to each account as below
amount Date
2 01-01-2012
5 02-01-2012
2 05-01-2012
1 07-01-2012
If the total credit in the account is 8. Ineed a query to find till what dates interest the credit amount can adjust.
Here the query should give output as 02-01-2012(2+5 < 8). I know this can be handled through cursor. But is there any method to write this as a single query in ORACLE.
SELECT pdate
FROM (
SELECT t.*,
LAG(date) OVER (ORDER BY date) AS pdate
8 - SUM(amount) OVER (ORDER BY date ROWS BETWEEN UNBOUNDED PRECEDING AND CURRENT ROW) AS diff
FROM mytable t
ORDER BY
date
)
WHERE diff < 0
AND rownum = 1
Not knowing the structure of your table, here's a guess:
SELECT date from your_table
GROUP BY AMOUNT
HAVING SUM(AMOUNT) < 8
Note: this is LESS THAN 8. Change the conditional as appropriate.
Doesn't do the (2+5)<8 thing yet:
select max(cum_sum), max(date)
from (
select date,
sum(amount) over (order by date) cum_sum
) where cum_sum < 8
The question I need to answer is this "What is the maximum number of page requests we have ever received in a 60 minute period?"
I have a table that looks similar to this:
date_page_requested date;
page varchar(80);
I'm looking for the MAX count of rows in any 60 minute timeslice.
I thought analytic functions might get me there but so far I'm drawing a blank.
I would love a pointer in the right direction.
You have some options in the answer that will work, here is one that uses Oracle's "Windowing Functions with Logical Offset" feature instead of joins or correlated subqueries.
First the test table:
Wrote file afiedt.buf
1 create table t pctfree 0 nologging as
2 select date '2011-09-15' + level / (24 * 4) as date_page_requested
3 from dual
4* connect by level <= (24 * 4)
SQL> /
Table created.
SQL> insert into t values (to_date('2011-09-15 11:11:11', 'YYYY-MM-DD HH24:Mi:SS'));
1 row created.
SQL> commit;
Commit complete.
T now contains a row every quarter hour for a day with one additional row at 11:11:11 AM. The query preceeds in three steps. Step 1 is to, for every row, get the number of rows that come within the next hour after the time of the row:
1 with x as (select date_page_requested
2 , count(*) over (order by date_page_requested
3 range between current row
4 and interval '1' hour following) as hour_count
5 from t)
Then assign the ordering by hour_count:
6 , y as (select date_page_requested
7 , hour_count
8 , row_number() over (order by hour_count desc, date_page_requested asc) as rn
9 from x)
And finally select the earliest row that has the greatest number of following rows.
10 select to_char(date_page_requested, 'YYYY-MM-DD HH24:Mi:SS')
11 , hour_count
12 from y
13* where rn = 1
If multiple 60 minute windows tie in hour count, the above will only give you the first window.
This should give you what you need, the first row returned should have
the hour with the highest number of pages.
select number_of_pages
,hour_requested
from (select to_char(date_page_requested,'dd/mm/yyyy hh') hour_requested
,count(*) number_of_pages
from pages
group by to_char(date_page_requested,'dd/mm/yyyy hh')) p
order by number_of_pages
How about something like this?
SELECT TOP 1
ranges.date_start,
COUNT(data.page) AS Tally
FROM (SELECT DISTINCT
date_page_requested AS date_start,
DATEADD(HOUR,1,date_page_requested) AS date_end
FROM #Table) ranges
JOIN #Table data
ON data.date_page_requested >= ranges.date_start
AND data.date_page_requested < ranges.date_end
GROUP BY ranges.date_start
ORDER BY Tally DESC
For PostgreSQL, I'd first probably write something like this for a "window" aligned on the minute. You don't need OLAP windowing functions for this.
select w.ts,
date_trunc('minute', w.ts) as hour_start,
date_trunc('minute', w.ts) + interval '1' hour as hour_end,
(select count(*)
from weblog
where ts between date_trunc('minute', w.ts) and
(date_trunc('minute', w.ts) + interval '1' hour) ) as num_pages
from weblog w
group by ts, hour_start, hour_end
order by num_pages desc
Oracle also has a trunc() function, but I'm not sure of the format. I'll either look it up in a minute, or leave to see a friend's burlesque show.
WITH ranges AS
( SELECT
date_page_requested AS StartDate,
date_page_requested + (1/24) AS EndDate,
ROWNUMBER() OVER(ORDER BY date_page_requested) AS RowNo
FROM
#Table
)
SELECT
a.StartDate AS StartDate,
MAX(b.RowNo) - a.RowNo + 1 AS Tally
FROM
ranges a
JOIN
ranges b
ON a.StartDate <= b.StartDate
AND b.StartDate < a.EndDate
GROUP BY a.StartDate
, a.RowNo
ORDER BY Tally DESC
or:
WITH ranges AS
( SELECT
date_page_requested AS StartDate,
date_page_requested + (1/24) AS EndDate,
ROWNUMBER() OVER(ORDER BY date_page_requested) AS RowNo
FROM
#Table
)
SELECT
a.StartDate AS StartDate,
( SELECT MIN(b.RowNo) - a.RowNo
FROM ranges b
WHERE b.StartDate > a.EndDate
) AS Tally
FROM
ranges a
ORDER BY Tally DESC