2010-04-01 (first published: 2010-03-31)
4,395 reads
2010-04-01 (first published: 2010-03-31)
4,395 reads
With this view you can obtain last failed job steps without using visual interface.
2010-03-30 (first published: 2010-03-19)
2,386 reads
2010-03-29 (first published: 2010-03-24)
1,849 reads
Here is a one more very quick way to find row count for a table.
2010-03-26 (first published: 2010-03-23)
5,406 reads
Evaluate credit card numbers based on ISO 2894 algorithm.
2010-03-24 (first published: 2010-03-17)
1,937 reads
This script displays Phonetic Pronunciation of a given password
2010-03-22 (first published: 2010-03-12)
1,403 reads
2010-03-17 (first published: 2010-03-04)
1,545 reads
A super-powered EXEC on steroids, the Power Tool every DBA wants for Christmas.
2010-03-16 (first published: 2010-03-04)
2,971 reads
This Script is used to List out the objects lying in the box which were un used from the day of the sql server recycled.
2010-03-09 (first published: 2010-02-18)
1,938 reads
Returns all properties from ServerProperty, also has a case function for EditionID and EngineEdition.
2010-03-08 (first published: 2010-02-18)
1,400 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