Using T-SQL, how do I generate a result that shows a range of dates - sql

Question

Using SQL Server, how do I generate a result set that shows a range of dates, like so:
StartDate EndDate
01/01/2014 01/04/2014
01/08/2014 01/11/2014
01/14/2014 01/15/2014
The original data had the dates in this format:
ColumnA DateColumn
blah 01/01/2014
blah 01/02/2014
blah 01/03/2014
blah 01/04/2014
blah 01/08/2014
blah 01/09/2014
blah 01/10/2014
blah 01/11/2014
blah 01/14/2014
blah 01/15/2014
Currently, I have a bunch of queries that does this, but I'm wondering if I can do something in less code:
SELECT ROW_NUMBER() OVER(ORDER BY DateColumn) AS rownum,
DateColumn
INTO #main
FROM MyTable
SELECT m1.DateColumn AS TBegin,
m2.DateColumn AS TEnd,
COALESCE(DATEDIFF(day, m2.TimePk, m1.TimePk), 0) AS Gap
INTO #Gap
FROM #main m1
LEFT OUTER JOIN #main m2
ON m1.rownum = m2.rownum + 1
ORDER BY m1.DateColumn
SELECT ROW_NUMBER() OVER(ORDER BY i_id, TBegin) AS rownum,
TBegin
INTO #Begin
FROM #Gap
WHERE Gap <> 1
ORDER BY TBegin
SELECT ROW_NUMBER() OVER(ORDER BY i_id, TEnd) AS rownum,
TEnd
INTO #End
FROM (
SELECT TEnd
FROM #Gap
WHERE Gap > 1
UNION
SELECT MAX(TBegin)
FROM #Gap
) as t
ORDER BY TEnd
SELECT b.TBegin,
e.TEnd
FROM #Begin b
INNER JOIN #End e
ON b.i_id = e.i_id
AND b.rownum = e.rownum
ORDER BY b.TBegin
Any ideas on how to simplify or approach this in an entirely different way?

Answer 1

My approach to these is to identify the first date that has no date preceding it. This is the beginning of a group. Then I take the cumulative sum of that as a group identifier, and do the aggregation.
SQL Server 2008 doesn't have lag or cumulative sums, so I use correlated subqueries for this:
with mt as (
select t.*,
(case when (select top 1 t2.dateColumn
from MyTable t2
where t2.ColumnA = t.ColumnA and
t2.dateColumn < t.dateColumn
order by t2.dateColumn desc
) = dateadd(day, -1, t.datecolumn)
then 0
else 1
end) as IsStart
from MyTable t
),
mtcum as (
select mt.*,
(select sum(mt2.IsStart)
from mt mt2
where mt2.ColumnA = mt.ColumnA and
mt2.dateColumn <= mt.DateColumn
) as grpId
from mt
)
select ColumnA, min(dateColumn) as StartDate, max(dateColumn) as EndDate
from mtcum
group by ColumnA, grpId;
EDIT:
An easier way of approach this is with the observation that the difference between a sequence of dates and a sequence of numbers is constant.
select columnA, min(dateColumn) as StartDate, max(dateColumn) as EndDate
from (select mt.*, row_number() over (partition by ColumnA order by datecolumn) as seqnum
from mytable mt
) t
group by columnA, dateadd(day, - seqnum, datecolumn);

Answer 2

This will work for you. It's still fairly complex though. It uses inner queries to find the first date that is after a gap for each date. This way all days belonging to the same group of dates can be grouped together.
select MIN(DateColumn) StartDate, MAX(DateColumn) EndDate from
(select X.DateColumn, MIN(Y.DateColumn) MinOverGap from
(select DateColumn, ROW_NUMBER() OVER (ORDER BY DateColumn) RowNumber
from MyTable) X
left join
(select DateColumn, ROW_NUMBER() OVER (ORDER BY DateColumn) RowNumber
from MyTable) Y
on DATEADD(d, Y.RowNumber - 1, X.DateColumn) <> DATEADD(d, X.RowNumber -1, Y.DateColumn) AND X.DateColumn < Y.DateColumn
group by x.DateColumn) grouped
group by MinOverGap
order by 1