Viewing 15 posts - 26,146 through 26,160 (of 26,487 total)
Here is a rough outline:
1. Create an empty database with all requisit tables, call it reportdb_new.
2. Load reportdb_new.
3. If load successful:
Drop all active connections to reportdb
Rename reportdb to...
April 4, 2007 at 3:07 pm
As this is posted under SQL Server 2005, you could also use SSIS and the filesystem task to move files around based on information in your database.
April 4, 2007 at 9:53 am
What version of SQL Server 2005 are you trying to install?
April 4, 2007 at 9:41 am
Or try this:
select dateadd(dd,datediff(dd,getdate()),0)
April 4, 2007 at 9:40 am
We are building a data warehouse where I work, and we are using schemas to identify the source systems in our staging environment. Once the data actually moves to the...
April 4, 2007 at 8:57 am
Be sure you read the license agreement. If I remember correctly, if you install Reporting Services on a machine other than the one where you installed SQL Server 2005, you...
April 4, 2007 at 8:42 am
I had my pattern and expression reversed. Also note that I had to add a single quote to the sixth string due to the single quote in the company name:
create...
April 2, 2007 at 12:32 pm
Worth a shot. That was the only way we could get our DTS packages to run as scheduled jobs at my last company.
If I think of anything else, I add...
March 30, 2007 at 10:15 pm
If you use Windows Server 2003 R2 as the host OS, you can use the same license for up to 4 Virtual Servers on the same box. Not sure if...
March 30, 2007 at 5:49 pm
Just for the heck of it, set the owner of the job to sa and see what happens.
March 30, 2007 at 4:24 pm
Try replacing the variable @string with the column you are trying to split into 2 seperate columns.
March 30, 2007 at 8:13 am
There is a way, unfortunately, I can't remember how or where I saw it. I will, during my free time, try and see if I can find it again and...
March 29, 2007 at 3:04 pm
Oops, meant this:
select substring(@string, 1, patindex(@string, '% %') - 1), substring(@string, patindex(@string, '% %') + 1, len(@string) - patindex(@string, '% %'))
March 29, 2007 at 2:47 pm
Try this:
select substring(@string, 1, patindex(@string, '% %') - 1), substring(@string, patindex(@string, '% %') - 1, len(@string) - patindex(@string, '% %'))
March 29, 2007 at 2:46 pm
I would say because you were trying to make it more complicated than it really was. You probably could have gotten your way to work, but you should should have...
March 29, 2007 at 2:41 pm
Viewing 15 posts - 26,146 through 26,160 (of 26,487 total)