Detect consecutive dates ranges using SQL

I want to fill the calendar object which requires start and end date information. I have one column which contains a sequence of dates. Some of the dates are consecutive (have one day difference) and some are not.

InfoDate  

2013-12-04  consecutive date [StartDate]
2013-12-05  consecutive date
2013-12-06  consecutive date [EndDate]

2013-12-09                   [startDate]
2013-12-10                   [EndDate]

2014-01-01                   [startDate]
2014-01-02 
2014-01-03                   [EndDate]

2014-01-06                   [startDate]
2014-01-07                   [EndDate]

2014-01-29                   [startDate]
2014-01-30 
2014-01-31                   [EndDate]

2014-02-03                   [startDate]
2014-02-04                   [EndDate]

I want to pick each consecutive dates range’s start and end date (the first one and the last one in the block).

StartDate     EndDate

2013-12-04    2013-12-06
2013-12-09    2013-12-10
2014-01-01    2014-01-03
2014-01-06    2014-01-07
2014-01-29    2014-01-31
2014-02-03    2014-02-04

I want to solve the problem using SQL only.


No joins or recursive CTEs needed. The standard gaps-and-island solution is to group by (value minus row_number), since that is invariant within a consecutive sequence. The start and end dates are just the MIN() and MAX() of the group.

WITH t AS (
  SELECT InfoDate d,ROW_NUMBER() OVER(ORDER BY InfoDate) i
  FROM @d
  GROUP BY InfoDate
)
SELECT MIN(d),MAX(d)
FROM t
GROUP BY DATEDIFF(day,i,d)

Here you go..

;WITH CTEDATES
AS
(
    SELECT ROW_NUMBER() OVER (ORDER BY Infodate asc ) AS ROWNUMBER,infodate FROM YourTableName  

),
 CTEDATES1
AS
(
   SELECT ROWNUMBER, infodate, 1 as groupid FROM CTEDATES WHERE ROWNUMBER=1
   UNION ALL
   SELECT a.ROWNUMBER, a.infodate,case datediff(d, b.infodate,a.infodate) when 1 then b.groupid else b.groupid+1 end as gap FROM CTEDATES A INNER JOIN CTEDATES1 B ON A.ROWNUMBER-1 = B.ROWNUMBER
)

select min(mydate) as startdate, max(infodate) as enddate from CTEDATES1 group by groupid

please don't forget to mark it as answer, if this answers your question.