Red-Gate Jumps the Moon
After this morning’s keynote, I managed to make it down to the expo hall JUST in time to catch Red-Gate’s...
2011-10-13
688 reads
After this morning’s keynote, I managed to make it down to the expo hall JUST in time to catch Red-Gate’s...
2011-10-13
688 reads
Welcome to PASS Summit 2011 Day 1!! Be sure to get your very own copy of the hot-off-the-presses MVP Deep Dives...
2011-10-12
1,413 reads
Today was a good day. I posted about the first part of today earlier, and we left off there with...
2011-10-11
510 reads
Today was also a good day. Sean and I both popped awake at 4:30am local time (6:30am personal time), for...
2011-10-06
510 reads
Summer camp has begun. Sure, people are still en route, but a whole bunch of us are already here and...
2011-10-06
567 reads
I made you something wonderful. It’s a cheatsheet for the PASS Summit, which has info about us (of course), about...
2011-10-06
618 reads
Hidy-ho there, neighbors! This is an update of last year’s Attend the PASS Summit from Home blog. I’ll be updating this...
2011-10-03
1,535 reads
Hello and welcome to my live blog/recap of SQL Saturday #97 in Austin! Wes Brown and the crew from CACTUSS...
2011-10-01
492 reads
When I get in to work in the morning, I like a good cup of coffee waiting for me, a...
2011-09-27
1,200 reads
On Tuesday, September 20 at 5p CST (6pm Charleston time), I will present “Unraveling Tangled Code” via LiveMeeting at the...
2011-09-19
597 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