Displays Identity Column Values
Displays identity column values, number of rows in the table, and the difference between the two. The difference can indicate the amount of deletions that could be going on.
2001-10-21
962 reads
Displays identity column values, number of rows in the table, and the difference between the two. The difference can indicate the amount of deletions that could be going on.
2001-10-21
962 reads
Displays identity values, number of rows in the table, and the difference between the two. The difference can indicate the amount of deletions that could be going on.
2001-09-06
369 reads
Display any stored procedures using input and output parameters. I took the view, PARAMETERS that ships with SQL 2000 and cleaned it up a bit.
2001-09-04
1,278 reads
2001-09-04
557 reads
2001-08-22
424 reads
Displays database properties. Was this the best way of doing it? Probably not, but it works.
2001-08-22
565 reads
Displays Maintance plan history if any. I used distinct to filter out and return only what is relevent to what I need. You may not wish this for your purposes. Feel free to make changes
2001-08-22
411 reads
2001-08-22
757 reads
2001-08-22
603 reads
2001-08-22
767 reads
By Steve Jones
I caught this interesting item over on Pinal Dave’s blog: Eleven Interview Questions that...
By Hemantgiri
I use color-coded connections in SSMS to distinguish Production, Pre-Production, UAT, and Development. But...
Comments posted to this topic are about the item Split Large Queries in Athena...
Comments posted to this topic are about the item Finding Trailing Spaces
I have some data in a SQL Server 2025 database. It looks like this for the dbo.Customer table:
CustomerID CustomerName PreferredName 1 Steve Steve 2 Andy Andy 3 Brian Brian 4 Allan Allan 5 Devin Devin 6 Steve Steve 7 Sally SallyI want to detect which names have a single trailing space. The CustomerName is a varchar() and the PreferredName is a CHAR(). Does this query detect the problem rows?
SELECT
CustomerID,
CustomerName,
PreferredName
FROM Customer
WHERE CustomerName <> RTRIM(CustomerName)
OR PreferredName <> RTRIM(PreferredName); See possible answers