calculate date difference between two dates in the same column - sql

Question

Hi I have a input table format like.
ID1 ID2 date
1002 9648 2011-01-02
1003 9648 2011-06-06
1004 9648 2012-08-08
1005 9648 2016-01-06
1006 9648 2016-09-12
1007 9648 2018-01-22
1009 9744 2009-10-03
1010 9744 2012-01-10
1011 9744 2016-09-23
1012 9744 2017-10-25
1013 9923 2006-10-10
1014 10124 2017-10-11
1015 10124 2018-01-24
am looking for an output table as shown below.
Could you please help me with a sql query or how it can be achieved in talend.
if no of days between two dates exceed 1096 which is approximately 3 years I want to take it as zero and update the type to new.
ID1 ID2 date daysdifference type
1002 9648 2011-01-02 0 new
1003 9648 2011-06-06 156 old
1004 9648 2012-08-08 429 old
1005 9648 2016-01-06 0 new
1006 9648 2016-09-12 250 old
1007 9648 2018-01-22 497 old
1009 9744 2009-10-03 0 new
1010 9744 2012-01-10 829 old
1011 9744 2016-09-23 0 new
1012 9744 2017-10-25 397 old
1013 9923 2006-10-10 0 new
1014 10124 2017-10-11 0 new
1015 10124 2018-01-24 91 old
Thanks,
Ankush Reddy.

Answer 1

You can use LAG function to get the desired result you want. Basically, LAG function access the data of the previous row. From there, you can now calculate the difference between the previous date and the current date using DATEDIFF function. However, this function exist starting version 2012.
SELECT ID1, ID2, Date,
DaysDifference = CASE WHEN PreviousDate IS NULL THEN 0 ELSE
CASE WHEN Date_Diff < 1096 THEN Date_Diff ELSE 0 END END,
CASE WHEN PreviousDate IS NULL THEN 'NEW' ELSE
CASE WHEN Date_Diff < 1096 THEN 'OLD' ELSE 'NEW' END END
FROM
(
SELECT *,
LAG(date, 1,NULL) OVER (PARTITION BY ID2 ORDER BY ID1) AS PreviousDate ,
DATEDIFF(DAY, LAG(date, 1,0) OVER (PARTITION BY ID2 ORDER BY ID1), date) Date_Diff
FROM TableA
) a
Here's a Demo.
However, if you are using version below 2012, you can still have the result you want using ROW_NUMBER()
;WITH rows AS
(
SELECT *,
RN = ROW_NUMBER() OVER (PARTITION BY ID2 ORDER BY ID1)
FROM TableA
)
SELECT ID1, ID2, Date,
DaysDifference = CASE WHEN PreviousDate IS NULL THEN 0 ELSE
CASE WHEN Date_Diff < 1096 THEN Date_Diff ELSE 0 END END,
CASE WHEN PreviousDate IS NULL THEN 'NEW' ELSE
CASE WHEN Date_Diff < 1096 THEN 'OLD' ELSE 'NEW' END END
FROM
(
SELECT a.ID1, a.ID2, a.Date, b.Date as PreviousDate,
DATEDIFF(DAY, b.date, a.date) Date_Diff
FROM rows a
LEFT JOIN rows b
ON a.RN = b.RN + 1
AND a.ID2 = b.ID2
) a
ORDER BY ID1, ID2
Here's a Demo.

Answer 2

Try this:
select [id1], [id2], [date],
case when [daysdifference] > 1096 then 0 else [daysdifference] end [daysdifference],
case when [daysdifference] > 1096 then 'new' else 'old' end [type]
from (
select *, abs(isnull(datediff(day, LAG([date], 1) over (order by id2, id1),
[date]), 0)) [daysdifference] from #x
) [a]