I have a table and each record has a date. We can assume that a date range is contiguous if there's not a 3 month break. How can I find the start of the most recent contiguous date range?
For example, imagine if I had this data:
1990-5-1
1990-6-4
1990-10-28
1990-11-14
1990-12-19
1991-1-20
1991-4-30
1991-5-13
I'd like for it to return 1991-4-30 because it's the start of the most recent contiguous range of dates.
I think this does what you're looking for. Using my own table and column names as test data. This is on Oracle.
select * from (
select * from sm_ss_tickets t1 where exists (
select * from sm_ss_tickets t2 where t2.created_date between t1.created_date and t1.created_date+90 and t1.rowid <> t2.rowid
) order by created_date asc
) where rownum = 1;
Maybe something like the following would work:
WITH d1 AS (
SELECT date'1990-05-01' AS dt FROM dual
UNION ALL
SELECT date'1990-06-04' AS dt FROM dual
UNION ALL
SELECT date'1990-10-28' AS dt FROM dual
UNION ALL
SELECT date'1990-11-14' AS dt FROM dual
UNION ALL
SELECT date'1990-12-19' AS dt FROM dual
UNION ALL
SELECT date'1991-01-20' AS dt FROM dual
UNION ALL
SELECT date'1991-04-30' AS dt FROM dual
UNION ALL
SELECT date'1991-05-13' AS dt FROM dual
)
SELECT MAX(dt) FROM (
SELECT dt, LAG(dt) OVER ( ORDER BY dt ) AS prev_dt, LEAD(dt) OVER ( ORDER BY dt ) AS next_dt
FROM d1
) WHERE ( dt > ADD_MONTHS(prev_dt, 3) OR prev_dt IS NULL )
AND dt > ADD_MONTHS(next_dt, -3)
In the above, a date can only be the start of a contiguous sequence if there is no prior date within 3 months (either it is more than three months ago or it doesn't exist at all) and there is also a subsequent date within 3 months.
You can use LAG and LEAD. Find the query below. I think it works fine.
tmp_year is the table I have created. tdate is the column.
The records in the table are
28-JAN-15
27-JAN-15
26-JAN-15
25-JAN-15
12-JUL-14
11-JUL-14
10-JUL-14
09-JUL-14
24-DEC-13
23-DEC-13
22-DEC-13
21-DEC-13
15-SEP-13
07-JUN-13
27-FEB-13
19-NOV-12
11-AUG-12
Please find the query which returns 25th Jan 2015.
select max(d.tdate) from (
select c.tdate,c.next_date,c.date_diff,lag(date_diff) over( order by tdate) prev_diff from (
select b.tdate ,b.next_date,(next_date-tdate) date_diff from
(select a.tdate,lead(a.tdate) over(order by a.tdate) next_date from tmp_year a ) b ) c) d where d.date_diff<90 and d.prev_diff>=90;
Related
Ex:
Date - up,
1/2 - 1.1.127.0 ,
1/3 - 1.1.127.1,
1/3 - 1.1.127.0,
1/4 - 1.1.127.3,
1/4 - 1.1.127.5,
1/5 - 1.1.127.3,
Output:
Date-count,
1/2 - 1,
1/3 - 1,
1/4 - 2,
1/5 -0
New and unique ip logged in in each day
You want to count how many IPs exist for a date that have not occurred on a previous date. You want to use analytic functions for this.
The number of new IDs is the total number of distinct IDs on a date minus the number of the previous date. In order to get this, first select the running count per row. Then aggregate per date to get the distinct number of IDs per date. Then use LAG to get the difference per day.
select
date,
max(cnt) - lag(max(cnt)) over (order by date) as new_ips
from
(
select date, count(distinct ip) over (order by date) as cnt
from mytable
) running_counts
group by date
order by date;
The same without analytic functions, which is probably more readable:
select date, count(distinct ip) as cnt
from mytable
where not exists
(
select null
from mytable before
where before.date < mytable.date
and before.id = mytable.id
)
group by date
order by date;
The DISTINCT in this latter query is not necessary, if there can be no duplicates (two rows with the same date and IP) in the table.
You can also use below solution using left join.
with t (dt, ip) as (
select to_date( '1/2', 'MM/DD' ), '1.1.127.0' from dual union all
select to_date( '1/3', 'MM/DD' ), '1.1.127.1' from dual union all
select to_date( '1/3', 'MM/DD' ), '1.1.127.0' from dual union all
select to_date( '1/4', 'MM/DD' ), '1.1.127.3' from dual union all
select to_date( '1/4', 'MM/DD' ), '1.1.127.5' from dual union all
select to_date( '1/5', 'MM/DD' ), '1.1.127.3' from dual
)
select t.DT, count( decode(t2.IP, null, 1, null) ) cnt
from t
left join t t2
on ( t2.DT < t.DT and t2.IP = t.IP )
group by t.DT
order by 1
;
demo
I have three tabels, each of them has a date column (the date column is an INT field and needs to stay that way). I need a UNION accross all three tables so that I get the list of unique dates in accending order like this:
20040602
20051215
20060628
20100224
20100228
20100422
20100512
20100615
Then I need to add a column to the result of the query where I subtract one from each date and place it one row above as the end date. Basically I need to generate the end date from the start date somehow and this is what I got so far (not working):
With Query1 As (
Select date_one As StartDate
From table_one
Union
Select date_two As StartDate
From table_two
Union
Select date_three e As StartDate
From table_three
Order By Date Asc
)
Select Query1.StartDate - 1 As EndDate
From Query1
Thanks a lot for your help!
Building on your existing union cte, we can use lead() in the outer query to get the start_date of the next record, and withdraw 1 from it.
with q as (
select date_one start_date from table_one
union select date_two from table_two
union select date_three from table_three
)
select
start_date,
dateadd(day, -1, lead(start_date) over(order by start_date)) end_date
from q
order by start_date
If the datatype the original columns are numeric, then you need to do some casting before applying date functions:
with q as (
select cast(cast(date_one as varchar(8)) as date) start_date from table_one
union select cast(cast(date_two as varchar(8)) as date) from table_two
union select cast(cast(date_three as varchar(8)) as date) from table_three
)
select
start_date,
dateadd(day, -1, lead(start_date) over(order by start_date)) end_date
from q
order by start_date
I want to fetch customers balances at the maximum date of every month, in every year in database. The Balance table has balances at the end of everyday when customer does transaction.
I just want to pick the balance at the maximum date of every month.Any help??
Below is a snip of My dataset.
You can try using window function - row_number()
select * from
(
SELECT *,row_number() over(partition by extract(YEAR FROM Date), extract(MONTH FROM Date) order by date desc) as rn
FROM t
)rn=1
You can do it also without a sub-query:
WITH b(ID, "date",bal) AS
(
SELECT 'CUST_I',DATE '2013-07-27', 14777.44 FROM dual UNION ALL
SELECT 'CUST_H',DATE '2013-07-26', 71085.13 FROM dual UNION ALL
SELECT 'CUST_I',DATE '2013-08-27', 66431.35656 FROM dual UNION ALL
SELECT 'CUST_H',DATE '2013-08-26', 63102.68622 FROM dual UNION ALL
SELECT 'CUST_H',DATE '2013-08-20', 6310.68622 FROM dual UNION ALL
SELECT 'CUST_H',DATE '2013-08-10', 630.68622 FROM dual UNION ALL
SELECT 'CUST_G',DATE '2013-09-25', 89732.04889 FROM dual UNION ALL
SELECT 'CUST_E',DATE '2013-09-23', 83074.70822 FROM dual
)
SELECT ID,
MAX("date") KEEP (DENSE_RANK FIRST ORDER BY "date" desc) AS MAX_DATE,
MAX(bal) KEEP (DENSE_RANK FIRST ORDER BY "date" desc) AS MAX_BAL
FROM b
GROUP BY ID, TRUNC("date", 'MM');
+-----------------------------+
|ID |MAX_DATE |MAX_BAL |
+-----------------------------+
|CUST_E|23.09.2013|83074.70822|
|CUST_G|25.09.2013|89732.04889|
|CUST_H|26.07.2013|71085.13 |
|CUST_H|26.08.2013|63102.68622|
|CUST_I|27.07.2013|14777.44 |
|CUST_I|27.08.2013|66431.35656|
+-----------------------------+
You may use a self join for your table call cust_balances :
select c1.*
from cust_balances c1
join
(
select max("date") max_date
from cust_balances
group by to_char("date",'yyyymm')
) c2 on ( c1."date" = c2.max_date );
SQL Fiddle Demo
I am working on a "counting days" problem almost identical to this one. I have a list of date(s), and need to count how many days used excluding duplicate, and handling the gaps. Same input and output.
From: Markus Jarderot
Input
ID d1 d2
1 2011-08-01 2011-08-08
1 2011-08-02 2011-08-06
1 2011-08-03 2011-08-10
1 2011-08-12 2011-08-14
2 2011-08-01 2011-08-03
2 2011-08-02 2011-08-06
2 2011-08-05 2011-08-09
Output
ID hold_days
1 11
2 8
SQL to find time elapsed from multiple overlapping intervals
But for the life of me I couldn't understand Markus Jarderot's solution.
SELECT DISTINCT
t1.ID,
t1.d1 AS date,
-DATEDIFF(DAY, (SELECT MIN(d1) FROM Orders), t1.d1) AS n
FROM Orders t1
LEFT JOIN Orders t2 -- Join for any events occurring while this
ON t2.ID = t1.ID -- is starting. If this is a start point,
AND t2.d1 <> t1.d1 -- it won't match anything, which is what
AND t1.d1 BETWEEN t2.d1 AND t2.d2 -- we want.
GROUP BY t1.ID, t1.d1, t1.d2
HAVING COUNT(t2.ID) = 0
Why is DATEDIFF(DAY, (SELECT MIN(d1) FROM Orders), t1.d1) picking from the min(d1) from the entire list? Is that regardless of ID.
And what does t1.d1 BETWEEN t2.d1 AND t2.d2 do? Is that to ensure only overlapped interval are calculated?
Same thing with group by, I think because if in the event the same identical period will be discarded? I tried to trace the solution by hand but getting more confused.
This is mostly a duplicate of my answer here (including explanation) but with the inclusion of grouping on an id column. It should use a single table scan and does not require a recursive sub-query factoring clause (CTE) or self joins.
SQL Fiddle
Oracle 11g R2 Schema Setup:
CREATE TABLE your_table ( id, usr, start_date, end_date ) AS
SELECT 1, 'A', DATE '2017-06-01', DATE '2017-06-03' FROM DUAL UNION ALL
SELECT 1, 'B', DATE '2017-06-02', DATE '2017-06-04' FROM DUAL UNION ALL -- Overlaps previous
SELECT 1, 'C', DATE '2017-06-06', DATE '2017-06-06' FROM DUAL UNION ALL
SELECT 1, 'D', DATE '2017-06-07', DATE '2017-06-07' FROM DUAL UNION ALL -- Adjacent to previous
SELECT 1, 'E', DATE '2017-06-11', DATE '2017-06-20' FROM DUAL UNION ALL
SELECT 1, 'F', DATE '2017-06-14', DATE '2017-06-15' FROM DUAL UNION ALL -- Within previous
SELECT 1, 'G', DATE '2017-06-22', DATE '2017-06-25' FROM DUAL UNION ALL
SELECT 1, 'H', DATE '2017-06-24', DATE '2017-06-28' FROM DUAL UNION ALL -- Overlaps previous and next
SELECT 1, 'I', DATE '2017-06-27', DATE '2017-06-30' FROM DUAL UNION ALL
SELECT 1, 'J', DATE '2017-06-27', DATE '2017-06-28' FROM DUAL UNION ALL -- Within H and I
SELECT 2, 'K', DATE '2011-08-01', DATE '2011-08-08' FROM DUAL UNION ALL -- Your data below
SELECT 2, 'L', DATE '2011-08-02', DATE '2011-08-06' FROM DUAL UNION ALL
SELECT 2, 'M', DATE '2011-08-03', DATE '2011-08-10' FROM DUAL UNION ALL
SELECT 2, 'N', DATE '2011-08-12', DATE '2011-08-14' FROM DUAL UNION ALL
SELECT 3, 'O', DATE '2011-08-01', DATE '2011-08-03' FROM DUAL UNION ALL
SELECT 3, 'P', DATE '2011-08-02', DATE '2011-08-06' FROM DUAL UNION ALL
SELECT 3, 'Q', DATE '2011-08-05', DATE '2011-08-09' FROM DUAL;
Query 1:
SELECT id,
SUM( days ) AS total_days
FROM (
SELECT id,
dt - LAG( dt ) OVER ( PARTITION BY id
ORDER BY dt ) + 1 AS days,
start_end
FROM (
SELECT id,
dt,
CASE SUM( value ) OVER ( PARTITION BY id
ORDER BY dt ASC, value DESC, ROWNUM ) * value
WHEN 1 THEN 'start'
WHEN 0 THEN 'end'
END AS start_end
FROM your_table
UNPIVOT ( dt FOR value IN ( start_date AS 1, end_date AS -1 ) )
)
WHERE start_end IS NOT NULL
)
WHERE start_end = 'end'
GROUP BY id
Results:
| ID | TOTAL_DAYS |
|----|------------|
| 1 | 25 |
| 2 | 13 |
| 3 | 9 |
The brute force method is to create all days (in a recursive query) and then count:
with dates(id, day, d2) as
(
select id, d1 as day, d2 from mytable
union all
select id, day + 1, d2 from dates where day < d2
)
select id, count(distinct day)
from dates
group by id
order by id;
Unfortunately there is a bug in some Oracle versions and recursive queries with dates don't work there. So try this code and see whether it works in your system. (I have Oracle 11.2 and the bug still exists there; so I guess you need Oracle 12c.)
I guess Markus' idea is to find all starting points that are not within other ranges and all ending points that aren't. Then just take the first starting point till the first ending point, then the next starting point till the next ending point, etc. As Markus isn't using a window function to number starting and ending points, he must find a more complicated way to achieve this. Here is the query with ROW_NUMBER. Maybe this gives you a start what to look for in Markus' query.
select startpoint.id, sum(endpoint.day - startpoint.day)
from
(
select id, d1 as day, row_number() over (partition by id order by d1) as rn
from mytable m1
where not exists
(
select *
from mytable m2
where m1.id = m2.id
and m1.d1 > m2.d1 and m1.d1 <= m2.d2
)
) startpoint
join
(
select id, d2 as day, row_number() over (partition by id order by d1) as rn
from mytable m1
where not exists
(
select *
from mytable m2
where m1.id = m2.id
and m1.d2 >= m2.d1 and m1.d2 < m2.d2
)
) endpoint on endpoint.id = startpoint.id and endpoint.rn = startpoint.rn
group by startpoint.id
order by startpoint.id;
If all your intervals start at different dates, consider them in ascending order by d1 counting how many days are from d1 to the next interval.
You can discard an interval of it is contained in another one.
The last interval won't have a follower.
This query should give you how many days each interval give
select a.id, a.d1,nvl(min(b.d1), a.d2) - a.d1
from orders a
left join orders b
on a.id = b.id and a.d1 < b.d1 and a.d2 between b.d1 and b.d2
group by a.id, a.d1
Then group by id and sum days
I am arriving at the total number of days a service has been used in a month.
(Start_Date and End_Date are - both inclusive)
Sample Data 1:
User Start_Date End_Date
A 01-Jun-2017 30-Jun-2017
B 06-Jun-2017 30-Jun-2017
Ans: Service used days = 30 days.
Sample Data 2:
User Start_Date End_Date
C 06-Jun-2017 10-Jun-2017
D 02-Jun-2017 02-Jun-2017
Ans: Service used days = 6 days.
How do I write a code to find the same, preferable in SQL to PLSQL.
Test Data:
CREATE TABLE your_table ( usr, start_date, end_date ) AS (
SELECT 'A', DATE '2017-06-01', DATE '2017-06-03' FROM DUAL UNION ALL
SELECT 'B', DATE '2017-06-02', DATE '2017-06-04' FROM DUAL UNION ALL -- Overlaps previous
SELECT 'C', DATE '2017-06-06', DATE '2017-06-06' FROM DUAL UNION ALL
SELECT 'D', DATE '2017-06-07', DATE '2017-06-07' FROM DUAL UNION ALL -- Adjacent to previous
SELECT 'E', DATE '2017-06-11', DATE '2017-06-20' FROM DUAL UNION ALL
SELECT 'F', DATE '2017-06-14', DATE '2017-06-15' FROM DUAL UNION ALL -- Within previous
SELECT 'G', DATE '2017-06-22', DATE '2017-06-25' FROM DUAL UNION ALL
SELECT 'H', DATE '2017-06-24', DATE '2017-06-28' FROM DUAL UNION ALL -- Overlaps previous and next
SELECT 'I', DATE '2017-06-27', DATE '2017-06-30' FROM DUAL UNION ALL
SELECT 'J', DATE '2017-06-27', DATE '2017-06-28' FROM DUAL; -- Within H and I
Query:
SELECT SUM( days ) AS total_days
FROM (
SELECT dt - LAG( dt ) OVER ( ORDER BY dt ) + 1 AS days,
start_end
FROM (
SELECT dt,
CASE SUM( value ) OVER ( ORDER BY dt ASC, value DESC, ROWNUM ) * value
WHEN 1 THEN 'start'
WHEN 0 THEN 'end'
END AS start_end
FROM your_table
UNPIVOT ( dt FOR value IN ( start_date AS 1, end_date AS -1 ) )
)
WHERE start_end IS NOT NULL
)
WHERE start_end = 'end';
Output:
TOTAL_DAYS
----------
25
Explanation:
SELECT dt, value
FROM your_table
UNPIVOT ( dt FOR value IN ( start_date AS 1, end_date AS -1 ) )
This will UNPIVOT the table so that the start and end dates are in the same column (dt) and are given a corresponding value of +1 for a start and -1 for an end date.
SELECT dt,
SUM( value ) OVER ( ORDER BY dt ASC, value DESC, ROWNUM ) AS total,
value
FROM your_table
UNPIVOT ( dt FOR value IN ( start_date AS 1, end_date AS -1 ) )
Will give the start and end dates and the cumulative sum of those generated values. The start of a range will always have value=1 and total=1 and the end of a range will always have total=0. If a date is mid-way through a range then it will either have total>1 or value=-1 and total=1. Using this, if you multiply value and total then the start of a range is when value*total=1 and the end of a range is when value*total=0 and any other value indicates a date that is midway through a range.
Which is what this gives:
SELECT dt,
CASE SUM( value ) OVER ( ORDER BY dt ASC, value DESC, ROWNUM ) * value
WHEN 1 THEN 'start'
WHEN 0 THEN 'end'
END AS start_end
FROM your_table
UNPIVOT ( dt FOR value IN ( start_date AS 1, end_date AS -1 ) )
You can then filter out the dates when the start_end is NULL which will leave you with a table with alternating start and end rows which you can use LAG to calculate the number of days difference:
SELECT dt - LAG( dt ) OVER ( ORDER BY dt ) + 1 AS days,
start_end
FROM (
SELECT dt,
CASE SUM( value ) OVER ( ORDER BY dt ASC, value DESC, ROWNUM ) * value
WHEN 1 THEN 'start'
WHEN 0 THEN 'end'
END AS start_end
FROM your_table
UNPIVOT ( dt FOR value IN ( start_date AS 1, end_date AS -1 ) )
)
WHERE start_end IS NOT NULL
All you need to do then is to SUM all the differences for the end - start; which gives the query above.
As #Pravin Satav addressed, your requirement it's not very clear, something like this is what I understood from your explanation:
SELECT sum(CASE WHEN end_date=start_date THEN 1 ELSE (end_date-start_date)+1 END) as total_days
FROM my_table
WHERE <conditions that determine your "sample data">;