Viewing 15 posts - 1 through 15 (of 413 total)
Thanks everyone, I learned a lot
The idea was to place the file groups on different disks as the new table B will become...
June 18, 2007 at 6:37 am
Sounds reasonable, although I suspected it was the other way round. Thanks.
June 14, 2007 at 12:44 am
I have tested my code against Ryan's code on SQL Server 2000 and it seems to be equally fast (with the same output).
If the Numbers table is preconstructed, I guess...
January 31, 2007 at 1:44 am
My suggestion, on our SQL Server 2000 installation it finds all primes less than 5000000 in 54 seconds. But of course it depends on the hardware...
January 30, 2007 at 2:59 am
I modified your last query a bit and obtained something that seems to be faster on SQL Server 2000:
SELECT distinct O.Name, O.Type
FROM
(
select
s1.id,
cast(coalesce(s1.text, '') as varchar(8000)) +
cast(coalesce(s2.text, '') as varchar(8000))
as [text]
from...
January 25, 2007 at 5:40 am
Thanks for a great article, Andy. I am in the process of designing a client/server system (.NET and SQL Server 2000/2005) in which GUIDs are used as clustered primary keys throughout the...
July 27, 2006 at 6:15 am
I believe the following function is faster than all the above suggested:
CREATE FUNCTION dbo.fnSeqDates
(
@LowDate DATETIME,
@HighDate DATETIME
)
RETURNS @Dates TABLE
(
SeqDate DATETIME
)
AS
BEGIN
DECLARE @Temp DATETIME
DECLARE @NumberOfDays int
IF @LowDate...
July 20, 2006 at 2:35 am
Do you really need dates that old?
You could use
select datediff(d, '18000106', @dateTo)/7 - datediff(d, '18000107', @dateFrom)/7
instead. Note that
select datediff(d, '18000106', '1900-01-03')/7 - datediff(d,...
July 19, 2006 at 7:08 am
No, no, use the formula above:
select datediff(d, '19000101', '20060724')/7 - datediff(d, '19000102', '20060717')/7
returns 2, i.e 2 Mondays between the 17th and the 24th.
July 19, 2006 at 6:07 am
Calculating the number of Mondays between two dates was the subject of the following thread:
http://www.sqlservercentral.com/forums/shwmessage.aspx?forumid=8&messageid=258968
My suggestion was the following:
select datediff(d, '19000101', @dateTo)/7 - datediff(d, '19000102', @dateFrom)/7
I still believe this...
July 19, 2006 at 4:47 am
It might also be a possibility to replace procedure B with a function that returns a table. It depends on what you are doing in procedure B. If you e.g...
March 9, 2006 at 8:03 am
I agree with paramind, but would like to elaborate a little further (to help myself understand it better ). Consider the following queries...
March 9, 2006 at 7:53 am
That's probably because
select vendorref from intranet.dbo.mprod
returns (among other things) a null value
Try this instead:
select * from appmis.cambio_icg.dbo.nuevos_productos_ingram_temp
where not partno in(select vendorref from intranet.dbo.mprod where vendorref is not null)
March 9, 2006 at 4:01 am
ang, I still don't know what you want...
Do you have a complete route from origin to destination (including all intermediate stops), and want...
March 9, 2006 at 2:26 am
Why not simply create one temp table with the same structure, then add an InstanceId column to it? This column could be 1 for data in the first temp table,...
March 8, 2006 at 7:51 am
Viewing 15 posts - 1 through 15 (of 413 total)