Viewing 15 posts - 1 through 15 (of 37 total)
Thanks Gaz
I found a solutions. I tried as administrator option and administrator user name and password.
April 25, 2012 at 2:17 pm
I opened IE as an administrator and tried but I got the same issue. 🙁
April 25, 2012 at 10:02 am
which version of SQL Server 2000 you are running ? SP4? later ?
It's RTM version.
September 7, 2009 at 12:51 am
which version of SQL Server 2000 you are running ? SP4? later ?
It's RTM version.
September 7, 2009 at 12:48 am
which version of SQL Server 2000 you are running ? SP4? later ?
It's RTM version.
September 7, 2009 at 12:45 am
This code lists databases available in the local server, not remote server 10.0.172.53. ODS database is available in the remote server. So, ODS database does not listed in the above...
December 10, 2008 at 5:45 am
I get list of databases available in the local server. But I need to retrieve data from Remote server.
December 10, 2008 at 5:03 am
ODS database is not available in the local server. It is available in the remote server. I need get data from remote server.
December 10, 2008 at 4:50 am
Hi Chris Morris,
Thanks for your reply. When I execute your code I get database available in the local server. Not the remote server(10.0.172.53).
December 10, 2008 at 4:29 am
TRy and look up the Use of CTE's, this should point you in the right direction.
Can I use CTE in SQL Server 2000?
August 28, 2008 at 6:48 am
DECLARE @counter as int
SET @counter = 0
WHILE @counter < 10
BEGIN
--Your code here
PRINT @counter
SET @counter = @counter + 1
END
August 13, 2008 at 7:15 am
DECLARE @date AS DATETIME
SET @date = '2007/01/01'
SELECT CONVERT(VARCHAR,@date,112)
August 6, 2008 at 10:10 am
You may do this on triggers.
CREATE TRIGGER insert_balance ON [dbo].[TEST]
FOR INSERT
AS
UPDATE Test
SET Balance = (inserted.Debit - inserted.Credit)
+ COALESCE((SELECT TOP 1 Balance FROM Test
WHERE ID<(SELECT MAX([ID])
FROM...
August 3, 2008 at 5:25 am
Try this,
SELECT COUNT(ACCOUNT) FROM
(SELECT COUNT(DISTINCT ACCOUNT) AS ACCOUNT
FROM dxhistory
WHERE RANK=1
GROUP BY ACCOUNT, RANK
HAVING COUNT(RANK)>1)AS T
July 29, 2008 at 10:43 pm
Viewing 15 posts - 1 through 15 (of 37 total)