User access management on DB & object level
It will help to find and manage the users access on DB level
2018-07-05 (first published: 2018-06-26)
432 reads
It will help to find and manage the users access on DB level
2018-07-05 (first published: 2018-06-26)
432 reads
2018-06-15 (first published: 2018-01-04)
2,122 reads
Este bloco de código tem o objetivo de demonstrar como podemos inserir um quantidade de linhas de registro de dados aleatórios de uma tabela no Microsoft SQL Server.
2018-06-14 (first published: 2018-06-12)
1,491 reads
2018-06-12 (first published: 2018-01-03)
1,994 reads
2018-06-11 (first published: 2018-01-03)
2,693 reads
2018-06-08 (first published: 2018-01-03)
2,555 reads
2018-06-07 (first published: 2018-01-13)
2,394 reads
2018-06-05 (first published: 2018-01-11)
2,208 reads
Solution for Backup, Integrity Check, Index and Statistics Maintenance in SQL Server 2005, SQL Server 2008, SQL Server 2008 R2, SQL Server 2012, SQL Server 2014, SQL Server 2016, and SQL Server 2017.
2018-06-04 (first published: 2008-02-23)
48,321 reads
The procedure generates an import script of diagrams.
2018-05-31 (first published: 2017-03-22)
6,747 reads
By Steve Jones
Here are the resources for my talk at Day of Data Boston. Slides –...
By Steve Jones
I’ve covered the values in a number of previous posts on the Book of...
By Arun Sirpal
When the evidence names the database, it says so. When it doesn’t, it stops....
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