Connect by level for SQL Server 2008 like Oracle - sql

Question

nDays := Round( dEndTime - dStartTime ) + 1;
For i in 1..7 Loop
nDay := i + 1;
if i = 7 Then
nDay := 1;
End If;
SELECT To_Date(To_Char((dStartTime+Level-1),'DD.MM.YYYY')||' 00:00','DD.MM.YYYY HH24:MI'),
To_Date(To_Char((dStartTime+Level-1),'DD.MM.YYYY')||' 23:59','DD.MM.YYYY HH24:MI')
FROM DUAL
WHERE To_Char( dStartTime + Level -1 , 'd' ) = To_Char(nDay)
CONNECT BY Level <= nDays;
End Loop;
output:
22-JUL-12
23-JUL-12
18-JUL-12
19-JUL-12
20-JUL-12
21-JUL-12
I need to convert this query to SQL Server 2008, please help to find workaround with the same......
I have tried above output with single query with nDay from 1 to 7.....

Answer 1

You can do this with a recursive CTE. However, that syntax can be hard to remember, so when I need a handful of items, I do something like:
select *
from (select row_number() over (order by (select NULL)) as seqnum
from information_schema.columns
) t
where seqnum < <value>
Any table can be used. I just put in the INFORMATION_SCHEMA.columns table, because it is easy and generally has dozens or hundreds of rows.
Your output doesn't match the query. The following generates all dates between two values that are reasonably close together, Here is an example:
declare #dstarttime date = '2012-07-11', #dendtime date= '2012-07-13';
with const as (select #dstarttime as dStartTime, #dendtime as dendTime)
SELECT DATEADD(d, seqnum - 1, dStartTime)
FROM (select *
from (select row_number() over (order by (select NULL)) as seqnum, const.*
from information_schema.columns cross join const
) t
where seqnum <= DATEDIFF(d, dStartTime, dendTime) + 1
) t
As I said, you can also do this with recursive CTEs or, if you have one, with a calendar table.