Reputation: 4756
I have the following query
SELECT Id, Request, BookingDate, BookingId FROM Table ORDER BY Request DESC, Date
If a row has a similar ForeignKeyId, I would like that to go in before the next ordered row like:
Request Date ForeignKeyId
Request3 01-Jun-11 56
Request2 03-Jun-11 89
NULL 03-Jun-11 89
Request1 05-Jun-11 11
NULL 20-Jul-11 57
I have been looking at RANK and OVER but haven't found a simple fix.
EDIT
I've edited above to show the actual fields and pasted data using the following query from Andomar's answer
select *
from (
select row_number() over (partition by BookingId order by Request DESC) rn
, Request, BookingDate, BookingID
from Table
WHERE Date = '28 aug 11'
) G
order by
rn
, Request DESC, BookingDate
1 ffffff 23/01/2011 15:57 350821
1 ddddddd 10/01/2011 16:28 348856
1 ccccccc 13/09/2010 14:44 338120
1 aaaaaaaaaa 21/05/2011 20:21 364422
1 123 17/09/2010 16:32 339202
1 NULL NULL
2 gggggg 08/12/2010 14:39 346634
2 NULL NULL
2 17/09/2010 16:32 339202
2 NULL 10/04/2011 15:08 361066
2 NULL 02/05/2011 14:12 362619
2 NULL 11/06/2011 13:55 366082
3 NULL NULL
3 16/10/2010 13:06 343023
3 22/10/2010 10:35 343479
3 30/04/2011 10:49 362435
The booking ID's 339202 should appear next to each other but don't
Upvotes: 4
Views: 139
Reputation: 238048
You could partition
by ForeignKeyId, then sort each second or lower row below their "head". With the "head" defined as the first row for that ForeignKeyId
. Example, sorting on Request
:
; with numbered as
(
select row_number() over (partition by ForeignKeyID order by Request) rn
, *
from @t
)
select *
from numbered n1
order by
(
select Request
from numbered n2
where n2.ForeignKeyID = n1.ForeignKeyID
and n2.rn = 1
)
, n1.Request
The subquery is required because SQL Server doesn't allow row_number
in an order by
clause.
Full example at SE Data.
Upvotes: 2