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