Soon will start 24 Hrs of PASS – free SQL knowledge stuff
Yes, 24 Hours of PASS will start soon two days in row, 7-8 September 2011! Just for info for those...
2011-09-06
757 reads
Yes, 24 Hours of PASS will start soon two days in row, 7-8 September 2011! Just for info for those...
2011-09-06
757 reads
2011-06-13
It's late but I need to say some words about our local tech event, it is the only one that...
2011-06-10
1,990 reads
Today Marlon Ribunal twittes come with new info about MS Certifications exams are rising…
This is not a good news, so...
2011-04-28
745 reads
Today Marlon Ribunal twittes come with new info about MS Certifications exams are rising...
This is not a good news, so...
2011-04-26
912 reads
Hello everybody, some tech events are announced to be held in Kosova soon …
Soon other details will be posted and...
2011-04-19
524 reads
Free tools are very welcome to the community, while some of the creators of these kind of the tools are...
2010-12-28
1,525 reads
SAC (Surface Are Configuration) was the great possibility to configure the SQL Server 2005, so there were some options to...
2010-12-17
1,478 reads
Update from Microsoft for SQL Server 2008 & 2008 R2. There is the graphical map of the system views included SQL...
2010-12-09
1,030 reads
Yesterday, I have received the official email from the PASS HQ, and the confirmation that now I'm the leader of...
2010-11-25
816 reads
By gbargsley
We’ve all been there. Someone walks up and asks, “Is SQL Server having issues?”...
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...
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