SQL server to do like Group By task - sql

Question

I have a table with SQL server as below,
Date Value
---------------------------------------------------
08-01-2016 1
08-02-2016 1
08-03-2016 1
08-04-2016 1
08-05-2016 1
08-06-2016 2
08-07-2016 2
08-08-2016 2
08-09-2016 2.5
08-10-2016 1
08-11-2016 1
Since the original table is too large, even I used 'Results to file', it still raise the exception 'System.OutOfMemoryException'. That's why I want to organize the table into this kind.
But I don't have a good logic to deal with. Therefore, I want to change the table into this kind as below.
Date_from Date_to Value
-------------------------------------------------
08-01-2016 08-05-2016 1
08-06-2016 08-08-2016 2
08-09-2016 08-09-2016 2.5
08-10-2016 08-11-2016 1
I appreciate your ideas!

Answer 1

Commonly called as Groups and Island problem. Here is one trick to do this
;WITH data
AS (SELECT *,Lag(Value, 1)OVER(ORDER BY Dates) [pVal]
FROM (VALUES ('08-01-2016',1 ),
('08-02-2016',1 ),
('08-03-2016',1 ),
('08-04-2016',1 ),
('08-05-2016',1 ),
('08-06-2016',2 ),
('08-07-2016',2 ),
('08-08-2016',2 ),
('08-09-2016',2.5 ),
('08-10-2016',1 ),
('08-11-2016',1 )) tc (Dates, Value)),
intr
AS (SELECT Dates,
Value,
Sum(Iif(pVal = Value, 0, 1)) OVER(ORDER BY Dates) AS [Counter]
FROM data)
SELECT Min(Dates) AS Dates_from,
Max(Dates) AS Dates_to,
Value
FROM intr
GROUP BY [Counter],
Value

Answer 2

The cumulative sum/lag approach is one method. In this case, a simpler method is:
select min(date) as date_from, max(date) as date_to, value
from (select t.*,
dateadd(day, - row_number() over (partition by value order by date),date) as grp
from t
) t
group by value, grp;
This uses the observation that the dates are consecutive with no gaps. Hence, subtracting a sequence from the date will yield a constant -- when the values are the same.

Answer 3

Here is an example:
DECLARE #T TABLE (
[Date] DATE,
[Value] DECIMAL(9,2)
)
INSERT #T VALUES
( '08-01-2016', 1 ),
( '08-02-2016', 1 ),
( '08-03-2016', 1 ),
( '08-04-2016', 1 ),
( '08-05-2016', 1 ),
( '08-06-2016', 2 ),
( '08-07-2016', 2 ),
( '08-08-2016', 2 ),
( '08-09-2016', 2.5 ),
( '08-10-2016', 1 ),
( '08-11-2016', 1 )
SELECT * FROM #T
SELECT A.[Date] StartDate, B.[Date] EndDate, A.[Value] FROM (
SELECT A.*, ROW_NUMBER() OVER (ORDER BY A.[Date], A.[Value]) O FROM #T A
LEFT JOIN #T B ON B.[Value] = A.[Value] AND B.[Date] = DATEADD(d, -1, A.[Date])
WHERE B.[Date] IS NULL
) A
JOIN (
SELECT A.*, ROW_NUMBER() OVER (ORDER BY A.[Date], A.[Value]) O FROM #T A
LEFT JOIN #T B ON B.[Value] = A.[Value] AND B.[Date] = DATEADD(d, 1, A.[Date])
WHERE B.[Date] IS NULL
) B ON B.O = A.O

Answer 4

Prdp's solution is great but just in case if anyone is still using SQL Server 2008 where LAG() and The Parallel Data Warehouse (PDW) features are not available here is an alternative:
SAMPLE DATA:
IF OBJECT_ID('tempdb..#Temp') IS NOT NULL
DROP TABLE #Temp;
CREATE TABLE #Temp([Dates] DATE
, [Value] FLOAT);
INSERT INTO #Temp([Dates]
, [Value])
VALUES
('08-01-2016'
, 1),
('08-02-2016'
, 1),
('08-03-2016'
, 1),
('08-04-2016'
, 1),
('08-05-2016'
, 1),
('08-06-2016'
, 2),
('08-07-2016'
, 2),
('08-08-2016'
, 2),
('08-09-2016'
, 2.5),
('08-10-2016'
, 1),
('08-11-2016'
, 1);
QUERY:
;WITH Seq
AS (SELECT SeqNo = ROW_NUMBER() OVER(ORDER BY [Dates]
, [Value])
, t.Dates
, t.[Value]
FROM #Temp t)
SELECT StartDate = MIN([Dates])
, EndDate = MAX([Dates])
, [Value]
FROM
(SELECT [Value]
, [Dates]
, SeqNo
, rn = SeqNo - ROW_NUMBER() OVER(PARTITION BY [Value] ORDER BY SeqNo)
FROM Seq s) a
GROUP BY [Value]
, rn
ORDER BY StartDate;
RESULTS: