2022-08-08
611 reads
2022-08-08
611 reads
One of the sales managers at Redgate Software posted an origin story about how he came to work at Redgate Software. It's an interesting story, and while I don't work in sales or have the same origin story, I don't know that I could have seen the place I'm in now early in my career. […]
2022-08-08
201 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...
2022-08-05
33 reads
I was honored to attend the first SQL Saturday in New Jersey last weekend. I consulted with and helped the organizers get the event going and executed. I made...
2022-08-05 (first published: 2022-07-26)
242 reads
2022-08-05
448 reads
Microsoft reads your feedback, but doesn't always build it. However, your votes can sway them at times. Today Steve asks what you'd like to see added to SQL Server.
2022-08-05
160 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...
2022-08-04
32 reads
2022-08-03
467 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...
2022-08-03
10 reads
One place where Artificial Intelligence systems might shine is the medical field.
2022-08-03
174 reads
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...
Setting page visibility and the active page are often overlooked last steps when publishing...
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
Comments posted to this topic are about the item Using table variables in T-SQL
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