2018-10-03
557 reads
2018-10-03
557 reads
Is an index change equivalent to a code change? Steve Jones asks the question and has a few thoughts.
2018-10-02
141 reads
Learn how you can determine which connection is using the Dedicated Administrator Connection.
2018-10-02
2,769 reads
2018-10-02
734 reads
With the need to remove older data, Steve thinks natural keys may be obsolete.
2018-10-01
199 reads
At Redgate, we build tools to help you build software in the entire DevOps cycle. In fact, I love this...
2018-10-01
291 reads
With lots of news coming out of Ignite, Steve talks about a few of the changes.
2018-10-01
42 reads
2018-10-01
733 reads
Another post for me that is simple and hopefully serves as an example for people trying to get blogging as...
2018-09-28 (first published: 2018-09-20)
5,550 reads
I’ve never been to Pittsburgh. Actually, in all of my speaking and traveling, I don’t think I’ve even flown through...
2018-09-28
960 reads
By Steve Jones
The episode on data masking and subetting is out. You can see it here:...
By Brian Kelley
I'm listening to Effortless by Greg McKeon (link to author's page) through Audible.com. He...
This book was making its rounds on social media, and the concept seems interesting...
I declare @Where based on the input parameter in the stored procedure. Set @SQL...
Hi, hoping someone can help. We're in the process of migrating to a new...
I am building an ETL process between these tables in SQL Server 2022 set to 160 compatibility level:
CREATE TABLE Image_Staging ( imageid INT NOT NULL CONSTRAINT Image_StagingPK PRIMARY KEY , imagestatus TINYINT , imagebinary IMAGE); GO CREATE TABLE Images ( imageid INT NOT NULL CONSTRAINT ImagesPK PRIMARY KEY , imagestatus TINYINT , imagemodified DATETIME , imagebinary IMAGE); GOI want to run this query to check if the images already loaded exist. This will help me decide if I need to insert or update an image. What happens with this query?
SELECT i.imageid FROM dbo.Image_Staging AS ist INNER JOIN dbo.Images AS i ON ist.imagebinary = i.imagebinary;See possible answers