PASS Board of Directors Part 3
Yesterday I posted about deciding to throw in my hat as far as running for the board, today I'll add...
2008-11-05
438 reads
Yesterday I posted about deciding to throw in my hat as far as running for the board, today I'll add...
2008-11-05
438 reads
A couple weeks ago I wrote about trying to decide whether to run for the PASS board. I talked about...
2008-11-04
533 reads
I'm not trying to talk politics here, but I don't want to be so politically correct that I avoid important...
2008-11-03
450 reads
This question came up during lunch this past week, and it's a good question. I'll start with my stock 'It...
2008-11-02
550 reads
A month or so ago I posted Blog Review - Yours & Mine to see what I might get for feedback about...
2008-10-30
1,494 reads
This is an unpaid/unsolicited comment about a product now owned by Red Gate Software. Purely my opinions, and I'll try...
2008-10-29
1,594 reads
For the past few years I've been using Frontpage 2003 for basic HTML work, usually to write content for SQLServerCentral....
2008-10-28
1,351 reads
Yesterday I posted an overview of the event, today I want to go over some areas where we excelled, and...
2008-10-27
1,325 reads
By any measure the event was a success. We had a solid 275 attendees this year, up about 50 from last...
2008-10-25
1,495 reads
In Good Ideas Take Time or How to Brainstorm - Part 1 I wrote about how I come up with ideas...
2008-10-23
1,598 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