How to get count of consecutive dates - sql-server

Question

For example there is some table with dates:
2015-01-01
2015-01-02
2015-01-03
2015-01-06
2015-01-07
2015-01-11
I have to write ms sql query, which will return count of consecutive dates starting from every date in the table. So the result will be like:
2015-01-01 1
2015-01-02 2
2015-01-03 3
2015-01-06 1
2015-01-07 2
2015-01-11 1
It seems to me that I should use LAG and LEAD functions, but now I even can not imagine the way of thinking.

Answer 1

CREATE TABLE #T ( MyDate DATE) ;
INSERT #T VALUES ('2015-01-01'),('2015-01-02'),('2015-01-03'),('2015-01-06'),('2015-01-07'),('2015-01-11')
SELECT
RW=ROW_NUMBER() OVER( PARTITION BY GRP ORDER BY MyDate) ,MyDate
FROM
(
SELECT
MyDate, DATEDIFF(Day, '1900-01-01' , MyDate)- ROW_NUMBER() OVER( ORDER BY MyDate ) AS GRP
FROM #T
) A
DROP TABLE #T;

Answer 2

You can use this CTE:
;WITH CTE AS (
SELECT [Date],
ROW_NUMBER() OVER(ORDER BY [Date]) AS rn,
CASE WHEN DATEDIFF(Day, PrevDate, [Date]) IS NULL THEN 0
WHEN DATEDIFF(Day, PrevDate, [Date]) > 1 THEN 0
ELSE 1
END AS flag
FROM (
SELECT [Date], LAG([Date]) OVER (ORDER BY [Date]) AS PrevDate
FROM #Dates ) d
)
to produce the following result:
Date rn flag
===================
2015-01-01 1 0
2015-01-02 2 1
2015-01-03 3 1
2015-01-06 4 0
2015-01-07 5 1
2015-01-11 6 0
All you have to do now is to calculate a running total of flag up to the first occurrence of a preceding zero value:
;WITH CTE AS (
... cte statements here ...
)
SELECT [Date], b.cnt + 1
FROM CTE AS c
OUTER APPLY (
SELECT TOP 1 COALESCE(rn, 1) AS rn
FROM CTE
WHERE flag = 0 AND rn < c.rn
ORDER BY rn DESC
) a
CROSS APPLY (
SELECT COUNT(*) AS cnt
FROM CTE
WHERE c.flag <> 0 AND rn < c.rn AND rn >= a.rn
) b
OUTER APPLY calculates the rn value of the first zero-valued flag that comes before the current row. CROSS APPLY calculates the number of records preceding the current record up to the first occurrence of a preceding zero valued flag.

Answer 3

I'm assuming this table:
SELECT *
INTO #Dates
FROM (VALUES
(CAST('2015-01-01' AS DATE)),
(CAST('2015-01-02' AS DATE)),
(CAST('2015-01-03' AS DATE)),
(CAST('2015-01-06' AS DATE)),
(CAST('2015-01-07' AS DATE)),
(CAST('2015-01-11' AS DATE))) dates(d);
Here's a recursive solution with explanations:
WITH
dates AS (
SELECT
d,
-- This checks if the current row is the start of a new group by using LAG()
-- to see if the previous date is adjacent
CASE datediff(day, d, LAG(d, 1) OVER(ORDER BY d))
WHEN -1 THEN 0
ELSE 1 END new_group,
-- This will be used for recursion
row_number() OVER(ORDER BY d) rn
FROM #Dates
),
-- Here, the recursion happens
groups AS (
-- We initiate recursion with rows that start new groups, and calculate "GRP"
-- numbers
SELECT d, new_group, rn, row_number() OVER(ORDER BY d) grp
FROM dates
WHERE new_group = 1
UNION ALL
-- We then recurse by the previously calculated "RN" until we hit the next group
SELECT dates.d, dates.new_group, dates.rn, groups.grp
FROM dates JOIN groups ON dates.rn = groups.rn + 1
WHERE dates.new_group != 1
)
-- Finally, we enumerate rows within each group
SELECT d, row_number() OVER (PARTITION BY grp ORDER BY d)
FROM groups
ORDER BY d
SQLFiddle