SQL how to convert row with date range to many rows with each date

Question

If I have a table that looks like this

begin date      end date        data
 2013-01-01     2013-01-04       7
 2013-01-05     2013-01-06       9

How can I make it be returned like this...

    date         data
 2013-01-01       7
 2013-01-02       7
 2013-01-03       7
 2013-01-04       7
 2013-01-05       9
 2013-01-06       9

One thing I was thinking of doing is to have another table that just has all the dates and then join the table with just dates to the above table using date>=begin date and date<=end date but that seems a little clunky to have to maintain that extra table with nothing but repetitive dates.

In some instances I don't have a data range but just an as of date which basically looks like my first example but with no end date . The end date is implied by the next row's 'as of' date (ie end date should be the next row's as of -1). I had a "solution" for this that uses the row_number() function to get the next value but I suspect that methodology, which the way I'm doing it has a bunch of nested self joins, contributes to very long query times.

Answer 1

Using some sample data...

create table data (begindate datetime, enddate datetime, data int);
insert data select 
 '20130101', '20130104', 7 union all select
 '20130105', '20130106', 9;

The Query : (Note: if you already have a numbers/tally table - use it)

select dateadd(d,v.number,d.begindate) adate, data
  from data d
  join master..spt_values v on v.type='P'
       and v.number between 0 and datediff(d, begindate, enddate)
order by adate;

Results :

|                       COLUMN_0 | DATA |
-----------------------------------------
| January, 01 2013 00:00:00+0000 |    7 |
| January, 02 2013 00:00:00+0000 |    7 |
| January, 03 2013 00:00:00+0000 |    7 |
| January, 04 2013 00:00:00+0000 |    7 |
| January, 05 2013 00:00:00+0000 |    9 |
| January, 06 2013 00:00:00+0000 |    9 |

---

Alternatively you can generate a number table on the fly (0-99) or as many numbers as you need

;WITH Numbers(number) AS (
  select top(100) row_number() over (order by (select 0))-1
  from sys.columns a
  cross join sys.columns b
  cross join sys.columns c
  cross join sys.columns d
  )
select dateadd(d,v.number,d.begindate) adate, data
  from data d
  join Numbers v on v.number between 0 and datediff(d, begindate, enddate)
order by adate;

SQL Fiddle Demo

Answer 2

You can use recursive CTE to get all the dates between two dates. Another CTE is to get ROW_NUMBERs to help you with those missing EndDates.

DECLARE @startDate DATE
DECLARE @endDate DATE

SELECT @startDate = MIN(begindate) FROM Table1
SELECT @endDate = MAX(enddate) FROM Table1

;WITH CTE_Dates AS 
(
    SELECT @startDate AS DT
    UNION ALL
    SELECT DATEADD(DD, 1, DT)
    FROM CTE_Dates
    WHERE DATEADD(DD, 1, DT) <= @endDate
)
,CTE_Data AS 
(
    SELECT *, ROW_NUMBER() OVER (ORDER BY BeginDate) AS RN FROM Table1

)
SELECT DT, t1.data FROM CTE_Dates d
LEFT JOIN CTE_Data t1 on d.DT 
BETWEEN t1.[BeginDate] AND COALESCE(t1.EndDate, 
        (SELECT DATEADD(DD,-1,t2.BeginDate) FROM CTE_Data t2 WHERE t1.RN + 1 = t2.RN))

SQLFiddle DEMO