Get sequence of days from days - sql

Question

We have a table of days like:
ID Date
1 2015-07-29
2 2015-07-30
3 2015-07-31
4 2015-08-01
5 2015-08-03
7 2015-08-04
8 2015-08-05
9 2015-08-06
10 2015-08-07
11 2015-08-10
And we want to find all the sequences (day+1). The result should be something like this:
Start End
2015-07-29 2015-08-01
2015-08-03 2015-08-07
2015-08-10 2015-08-10
1. Update
First I modified Deepanshu Kalra answer to use RowNumber instead of the Id (Id is autoincrement, so its possible that ids are missing)
DECLARE #P TABLE(DATE DATE)
INSERT INTO #P
SELECT MIN([DATE])
FROM MietvertragsArtikelDays
UNION
SELECT T1.[DATE]
FROM (SELECT ROW_NUMBER() OVER (ORDER BY [DATE]) RowNumber, [DATE] FROM MietvertragsArtikelDays) AS T1 INNER JOIN (SELECT ROW_NUMBER() OVER (ORDER BY [DATE]) RowNumber, [DATE] FROM MietvertragsArtikelDays) AS T2 ON T1.RowNumber=T2.RowNumber+1
WHERE DATEDIFF(DAY,T2.[DATE],T1.[DATE]) <>1
UNION
SELECT T2.[DATE]
FROM (SELECT ROW_NUMBER() OVER (ORDER BY [DATE]) RowNumber, [DATE] FROM MietvertragsArtikelDays) AS T1 INNER JOIN (SELECT ROW_NUMBER() OVER (ORDER BY [DATE]) RowNumber, [DATE] FROM MietvertragsArtikelDays) AS T2 ON T1.RowNumber=T2.RowNumber+1
WHERE DATEDIFF(DAY,T2.[DATE],T1.[DATE]) <>1
DECLARE #X TABLE(DATE DATE, RN INT)
INSERT INTO #X
SELECT *, ROW_NUMBER() OVER(ORDER BY [DATE]) AS X FROM #P
SELECT A.[DATE] Start, ISNULL(B.[DATE],A.[DATE]) [End] FROM #X A
LEFT JOIN (SELECT [DATE], RN-1 AS RN FROM #X) B
ON A.RN=B.RN
WHERE A.RN%2=1
2. Update
The most elegant solution is Ughais
;WITH CTE as
(
SELECT *,DATEDIFF(D,0,[Date]) - ROW_NUMBER()OVER(ORDER BY ID ASC) grp
FROM MietvertragsArtikelDays
)
SELECT MIN([Date]),MAX([Date])
FROM CTE
GROUP BY grp

Answer 1

This is an Islands and Gap problem. You can use ROW_NUMBER and DATEDIFF. Something like this.
SQL Fiddle
Sample Data
DECLARE #Dates TABLE
([ID] int, [Date] datetime);
INSERT INTO #Dates
([ID], [Date])
VALUES
(1, '2015-07-29 00:00:00'),
(2, '2015-07-30 00:00:00'),
(3, '2015-07-31 00:00:00'),
(4, '2015-08-01 00:00:00'),
(5, '2015-08-03 00:00:00'),
(7, '2015-08-04 00:00:00'),
(8, '2015-08-05 00:00:00'),
(9, '2015-08-06 00:00:00'),
(10, '2015-08-07 00:00:00'),
(11, '2015-08-10 00:00:00');
Query
;WITH CTE as
(
SELECT *,DATEDIFF(D,0,[Date]) - ROW_NUMBER()OVER(ORDER BY ID ASC) grp
FROM #Dates
)
SELECT MIN([Date]),MAX([Date])
FROM CTE
GROUP BY grp
Output
2015-07-29 00:00:00.000 2015-08-01 00:00:00.000
2015-08-03 00:00:00.000 2015-08-07 00:00:00.000
2015-08-10 00:00:00.000 2015-08-10 00:00:00.000

Answer 2

As I started from an answer which was already posted, maybe I made it very complex. But it works.
DECLARE #T TABLE(ID INT, DATE DATE)
INSERT INTO #T
SELECT 1, '2015-07-29' UNION ALL
SELECT 2, '2015-07-30' UNION ALL
SELECT 3, '2015-07-31' UNION ALL
SELECT 4, '2015-08-01' UNION ALL
SELECT 5, '2015-08-03' UNION ALL
SELECT 7, '2015-08-04' UNION ALL
SELECT 8, '2015-08-05' UNION ALL
SELECT 9, '2015-08-06' UNION ALL
SELECT 10, '2015-08-07' UNION ALL
SELECT 11, '2015-08-10'
DECLARE #P TABLE(DATE DATE)
INSERT INTO #P
SELECT MIN([DATE])
FROM #T
UNION
SELECT T1.[DATE]
FROM #T AS T1 INNER JOIN #T AS T2 ON T1.ID=T2.ID+1
WHERE DATEDIFF(DAY,T2.[DATE],T1.[DATE]) <>1
UNION
SELECT T2.[DATE]
FROM #T AS T1 INNER JOIN #T AS T2 ON T1.ID=T2.ID+1
WHERE DATEDIFF(DAY,T2.[DATE],T1.[DATE]) <>1
DECLARE #X TABLE(DATE DATE, RN INT)
INSERT INTO #X
SELECT *, ROW_NUMBER() OVER(ORDER BY [DATE]) AS X FROM #P
SELECT A.[DATE], B.[DATE] FROM #X A
LEFT JOIN (SELECT [DATE], RN-1 AS RN FROM #X) B
ON A.RN=B.RN
WHERE A.RN%2=1
Please excuse the standards and all. Will edit later in the day. Sorry for that.

Answer 3

This will work whatever the order and value of Id is:
Declare #dates table(ID int, D datetime)
Insert Into #dates(ID, D)
values (1, '2015-07-29')
, (2, '2015-07-30')
, (3, '2015-07-31')
, (4, '2015-08-01')
, (5, '2015-08-03')
, (7, '2015-08-04')
, (8, '2015-08-05')
, (9, '2015-08-06')
, (10, '2015-08-07')
, (11, '2015-08-10')
; With start(ID, D) as (
-- Get 1st Dates
Select d1.ID, d1.D From #dates as d1
Left Join #dates as d2 On d1.D = DATEADD(DAY, 1, d2.D)
Where d2.ID is NULL
), loop(startD, endD) as (
-- Loop through consecutives dates
Select D, D From start
Union All
Select l.startD, s.D From loop as l
Inner Join #dates as s On s.D = DATEADD(DAY, 1, l.endD)
)
-- Get max end date for each start date
Select startD as [Start], max(endD) as [End] From loop group by startD
Output:
Start End
2015-07-29 2015-08-01
2015-08-03 2015-08-07
2015-08-10 2015-08-10