2015-03-03
5 reads
2015-03-03
5 reads
2015-03-03
12 reads
In this session Prakash will walk through the process implemented to alleviates many of the pain points for deploying, managing,...
2015-03-03
553 reads
I am pleased to be presenting a highly interactive session tomorrow evening that is a full end-to-end setup and management...
2015-03-03
427 reads
In part 3 of this series of blogs on "getting more agile" I am going to look at unit testing...
2015-03-03
410 reads
In part 3 of this series of blogs on “getting more agile” I am going to look at unit testing and why we need to do it.
My code is...
2015-03-03
19 reads
In part 3 of this series of blogs on “getting more agile” I am going to look at unit testing...
2015-03-03
67 reads
In part 3 of this series of blogs on “getting more agile” I am going to look at unit testing...
2015-03-03
58 reads
Traditionally, we want our Clustered Index to have the following attributes:
Narrow: So that our clustered index and the non-clustered indexes that point...
2015-03-03
680 reads
That’s right I’m cashing in on the “50 Shades” franchise and twisting it to meet my own sordid needs to...
2015-03-03
669 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