How to Select continuous date in sql

Question

Is there any function to check for continuous date. I'm having problem on working with this issue below:

My table has a datetime column with the following data:

----------
2015-03-11
2015-03-12
2015-03-13
2015-03-16

Given start date as 2015-3-11 and end date as 2015-3-17. I want the result as:

----------
2015-03-11
2015-03-12
2015-03-13

Can anyone suggest anything ?

Answer 1

I'm thinking this is somewhat a variation of Grouping Islands of Contiguous Dates problem. This can be done using ROW_NUMBER():

SQL Fiddle

CREATE TABLE Test(
    tDate   DATETIME
)
INSERT INTO Test VALUES
('20150311'), ('20150312'), ('20150313'), ('20150316');

DECLARE @startDate  DATE = '20150311'
DECLARE @endDate    DATE = '20150317'

;WITH Cte AS(
    SELECT
        *,
        RN = DATEADD(DD, - (ROW_NUMBER() OVER(ORDER BY tDATE) - 1), tDate)
    FROM Test
    WHERE 
        tDate >= @startDate
        AND tDate < DATEADD(DAY, 1, @endDate)
)
SELECT CAST(tDate AS DATE)
FROM CTE
WHERE RN = @startDate

RESULT

|------------|
| 2015-03-11 |
| 2015-03-12 |
| 2015-03-13 |

---

Here is the SQL Server 2005 version:

SQL Fiddle

DECLARE @startDate  DATETIME
DECLARE @endDate    DATETIME

SET @startDate  = '20150311'
SET @endDate    = '20150317'

;WITH Cte AS(
    SELECT
        *,
        RN = DATEADD(DD, -(ROW_NUMBER() OVER(ORDER BY tDATE)-1), tDate)
    FROM Test
    WHERE 
        tDate >= @startDate
        AND tDate < DATEADD(DAY, 1, @endDate)
)
SELECT CONVERT(VARCHAR(10), tDate, 121)
FROM CTE
WHERE RN = @startDate

Answer 2

For MSSQL 2012. This will return MAX continuous groups:

DECLARE @t TABLE(d DATE)

INSERT INTO @t VALUES
('20150311'),
('20150312'),
('20150313'),
('20150316')


;WITH
c1 AS(SELECT d, IIF(DATEDIFF(dd,LAG(d, 1, DATEADD(dd, -1, d)) OVER(ORDER BY d), d) = 1, 0, 1) AS n FROM @t),
c2 AS(SELECT d, SUM(n) OVER(ORDER BY d) AS n FROM c1) 

SELECT TOP 1 WITH TIES MIN(d) AS StartDate, MAX(d) AS EndDate, COUNT(*) AS DayCount
FROM c2
GROUP BY n
ORDER BY DayCount desc

Output:

StartDate   EndDate     DayCount
2015-03-11  2015-03-13  3

For

('20150311'),
('20150312'),
('20150313'),
('20150316'),
('20150317'),
('20150318'),
('20150319'),
('20150320')

Output:

StartDate   EndDate     DayCount
2015-03-16  2015-03-20  5

Apply filtering in c1 CTE:

c1 AS(SELECT d, IIF(DATEDIFF(dd,LAG(d, 1, DATEADD(dd, -1, d)) OVER(ORDER BY d), d) = 1, 0, 1) AS n FROM @t WHERE d BETWEEN '20150311' AND '20150320'),

For MSSQL 2008:

;WITH
c1 AS(SELECT d, (SELECT MAX(d) FROM @t it WHERE it.d < ot.d) AS pd FROM @t ot),
c2 AS(SELECT d, CASE WHEN DATEDIFF(dd,ISNULL(pd, DATEADD(dd, -1, d)),  d) = 1 THEN  0 ELSE 1 END AS n FROM c1),
c3 AS(SELECT d, (SELECT SUM(n) FROM c2 ci WHERE ci.d <= co.d)  AS n FROM c2 co) 

SELECT TOP 1 WITH TIES MIN(d) AS StartDate, MAX(d) AS EndDate, COUNT(*) AS DayCount
FROM c3
GROUP BY n
ORDER BY DayCount desc

Answer 3

you don't need to declare any start date or end date as other answers says, you need a row_num with datediff function:

create table DateFragTest (cDate date);
insert into DateFragTest 
       values ('2015-3-11'),
              ('2015-3-12'),
              ('2015-3-13'),
              ('2015-3-16')

   with cte as 
      (select 
        cDate,
        row_number() over (order by cDate ) as rn
       from
        DateFragTest)
    select cDate 
    from cte t1  
    where datediff(day,
                   (select cDate from cte t2 where t2.rn=t1.rn+1),
                   t1.cDate)<>1

Output:

cDate
2015-03-11
2015-03-12
2015-03-13

SQLFIDDLE DEMO

Answer 4

For sql server 2012-

WITH cte
AS
(
    SELECT [datex]
    ,      lead([datex]) OVER ( ORDER BY [datex]) lead_datex
    ,      datediff(dd,[datex],lead([datex]) OVER ( ORDER BY [datex]) ) AS diff
    FROM [dbo].[datex]
)
SELECT c.[datex]
FROM [cte] AS c
WHERE diff >=1

Answer 5

Use BETWEEN

The query will go like this:

SELECT * 
FROM your_table_name
WHERE your_date_column_name BETWEEN '2015-3-11' AND '2015-3-13'

Answer 6

(dt between x and y) or just (dt >= x and dt <= y).