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
         


        

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