Detect consecutive dates ranges using SQL

Question

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.

Answer 1

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)

Answer 2

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

Answer 3

--MS SQL

with cte as (
select start_date, end_date,
    dateadd(d, -row_number() over (order by start_date), start_date) as GRN
from projects)
select min(start_date), max(end_date) from cte group by grn order by grn;

--Oracle

with cte as(
select start_date, end_date, 
    start_date - row_number() over (order by start_date) as GRN 
    from projects)
select min(start_date), max(end_date) from cte  group by grn order by grn;

Answer 4

I have inserted these values into a table called #consec and then perforemed the following:

select t1.*
,t2.infodate as binfod
into #temp1
from #consec t1
left join #consec t2 on dateadd(DAY,1,t1.infodate)=t2.infodate

select t1.*
,t2.infodate as binfod
into #temp2
from #consec t1
left join #consec t2 on dateadd(DAY,1,t2.infodate)=t1.infodate
;with cte as(
select infodate,  ROW_NUMBER() over(order by infodate asc) as seq from #temp1
where binfod is null
),
cte2 as(
select infodate, ROW_NUMBER() over(order by infodate asc) as seq from #temp2
where binfod is null
)

select t2.infodate as [start_date]
,t1.infodate as [end_date] from cte t1
left join cte2 t2 on t1.seq=t2.seq 

As long as your date periods are not overlapping, that should do the job for you.

Answer 5

Here it is my sample with test data:

--required output
-- 01 - 03
-- 08 - 09
-- 12 - 14

DECLARE @maxRN int;
WITH #tmp AS (
                SELECT CAST('2013-01-01' AS date) DT
    UNION ALL   SELECT CAST('2013-01-02' AS date)
    UNION ALL   SELECT CAST('2013-01-03' AS date)
    UNION ALL   SELECT CAST('2013-01-05' AS date)
    UNION ALL   SELECT CAST('2013-01-08' AS date)
    UNION ALL   SELECT CAST('2013-01-09' AS date)
    UNION ALL   SELECT CAST('2013-01-12' AS date)
    UNION ALL   SELECT CAST('2013-01-13' AS date)
    UNION ALL   SELECT CAST('2013-01-14' AS date)
),
#numbered AS (
    SELECT 0 RN, CAST('1900-01-01' AS date) DT
    UNION ALL
    SELECT ROW_NUMBER() OVER (ORDER BY DT) RN, DT
    FROM #tmp
)

SELECT * INTO #tmpTable FROM #numbered;
SELECT @maxRN = MAX(RN) FROM #tmpTable;

INSERT INTO #tmpTable
SELECT @maxRN + 1, CAST('2100-01-01' AS date);

WITH #paired AS (
    SELECT 
    ROW_NUMBER() OVER(ORDER BY TStart.DT) RN, TStart.DT DTS, TEnd.DT DTE
    FROM #tmpTable TStart
    INNER JOIN #tmpTable TEnd 
    ON TStart.RN = TEnd.RN - 1
    AND DATEDIFF(dd,TStart.DT,TEnd.DT) > 1  
)

SELECT TS.DTE, TE.DTs 
FROM #paired TS
INNER JOIN #paired TE ON TS.RN = TE.RN -1
AND TS.DTE <> TE.DTs -- you could remove this filter if you want to have start and end on the same date

DROP TABLE #tmpTable

Replace #tmp data with your actual table.

Answer 6

You can do like this and here is the sqlfiddle

select
  min(ndate) as start_date,
  max(ndate) as end_date
from
(select
  ndate,
  dateadd(day, -row_number() over (order by ndate), ndate) as rnk
 from dates
 ) t
 group by
   rnk

Answer 7

Another simple solution that could work here is -

with tmp as 
(
select
datefield
, dateadd('day',-row_number() over(order by date asc),datefield) as date_group 
from table
)
select
min(datefield) as start_date
, max(datefield) as end_date 
from tmp
group by date_group

Answer 8

with cte as( select start_date, end_date, start_date - row_number() over (order by start_date) as GRN from projects)
select min(start_date), max(end_date) from cte
group by grn order by count(grn),min(start_date)

Answer 9

SELECT InfoDate ,
    CASE
      WHEN TRUNC(InfoDate - 1) = TRUNC(lag(InfoDate,1,InfoDate) over (order by InfoDate))
      THEN NULL
      ELSE InfoDate
    END STARTDATE,
    CASE
      WHEN TRUNC(InfoDate + 1) = TRUNC(lead(InfoDate,1,InfoDate) over (order by InfoDate))
      THEN NULL
      ELSE InfoDate
    END ENDDATE
  FROM TABLE;