Another post for me that is simple and hopefully serves as an example for people trying to get blogging as #SQLNewBloggers. Recently I saw Jason Horner do a presentation...
2020-12-11 (first published: 2020-11-30)
554 reads
I know that this isn’t the correct name, though the by-laws still list this as the corporation. Perhaps this is one more sign of the failure to evolve and...
2020-12-11
45 reads
I started to add a daily coping tip to the SQLServerCentral newsletter and to the Community Circle, which is helping me deal with the issues in the world. I’m...
2020-12-11
28 reads
Today is a day for reflections in Steve's life, looking back at the incredible year of 2020.
2020-12-11
101 reads
2020-12-11
413 reads
I started to add a daily coping tip to the SQLServerCentral newsletter and to the Community Circle, which is helping me deal with the issues in the world. I’m...
2020-12-10
21 reads
As legal requirements grow and change, cloud computing might require a new consideration: where we physically store data.
2020-12-10
153 reads
2020-12-10
412 reads
I started to add a daily coping tip to the SQLServerCentral newsletter and to the Community Circle, which is helping me deal with the issues in the world. I’m...
2020-12-09
16 reads
Steve wonders if any of you see much of a difference between any of the virtual events taking place today.
2020-12-09
83 reads
By Chris Yates
In the beginning, there was OLTP – Online Transaction Processing. Fast, reliable, and ruthlessly...
One thing I’ve always loved about the Scooby-Doo cartoon is that he never solved...
By Kevin3NF
Flexibility and Scale at the Database Level When SQL Server 2012 introduced Availability Groups...
Hello SQL Server 2022 16.0.4212.1 running on a Windows Server 2025 Std,V 24H2, SO...
i have subscription of github copilot which i can access in vs 2022 comunity...
Comments posted to this topic are about the item Password Guidance
What happens if you run the following code in SQL Server 2022+?
declare @t1 table (id int); insert into @t1 (id) values (NULL), (1), (2), (3); select count(*) from @t1 where @t1.id is distinct from NULL;See possible answers