Viewing 15 posts - 13,321 through 13,335 (of 13,849 total)
I agree - DTS is the way to go. Why make things harder for yourself??
September 15, 2005 at 4:09 am
Difficult to tell from the way that the stuff's been formatted for display. What is the field delimiter (ie which character appears between the fields and nowhere else)?
September 15, 2005 at 4:07 am
I've had a quick look, but can't find where this alias is appearing - please direct me to it!
Phil
September 15, 2005 at 4:04 am
Wow, comprehensive answer! So there is a God and he understands SQL Server ... all is well ![]()
September 15, 2005 at 3:28 am
What message are you getting?
September 14, 2005 at 5:22 am
Not in SQL Server - but perhaps it's something to do with Magic Software ???, check out:
September 14, 2005 at 5:20 am
This appears to be a logon error - possibly because the external app is trying to log in as sa and the sa password has been changed on the server. ...
September 14, 2005 at 2:38 am
This gets you some of the way and may give you an idea about how to finish it off:
select b1.*, sum(b2.units) CumUnits
from basetable b1
join basetable b2 on b1.tdate >= b2.tdate
group...
September 13, 2005 at 8:12 am
Why not recast this as a conditional insert?
if not exists (SELECT destination FROM filespath WHERE Destination = @a)
begin
INSERT INTO FilesPath (Path,Destination)
VALUES (@a,@b)
end
September 13, 2005 at 3:57 am
Hi Bob
Can you break the script into several stored procedures and then bring these all together in a master stored procedure? Just use
exec [procname]
within a stored proc to execute another...
September 13, 2005 at 2:36 am
I reckon people would be more keen to help if you posted a copy of pornstar.mdb ![]()
Under what circumstances does the error message appear?...
September 6, 2005 at 3:19 pm
Watch out for the possibility that INSERT statements can add a row that has its clustered index value 'between' existing entries - possibly forcing a major rewrite of data on...
September 6, 2005 at 3:13 pm
Hi Ray, just a quick comment on your response - your final query will, in fact, only return those orders that are placed at the same millisecond as the query...
September 5, 2005 at 4:13 pm
I would have thought that the GROUP BY clause would ensure that city5 would not be displayed 3 times - ** please test and post the results along with the...
September 5, 2005 at 6:12 am
I'm having a bit of trouble working out what you want. Is it something like this?
select t1.city, t1.name, sum(no_of_games)
from test1 t1
join test2 t2 on t1.city = t2.city
join test3...
September 5, 2005 at 4:21 am
Viewing 15 posts - 13,321 through 13,335 (of 13,849 total)