Last Book Chapter Submitted
Whew, I just got my last chapter submitted to the editorial team for our newest book. I still have to...
2010-07-09
341 reads
Whew, I just got my last chapter submitted to the editorial team for our newest book. I still have to...
2010-07-09
341 reads
Hey guys, I’m preparing a few new article series for you but in the meantime I wanted to let you...
2010-07-09
317 reads
Hey there SQL Lunchers. Your co-host Adam Jorgensen here. I wanted to let you know, for those of you whose...
2010-07-09
342 reads
Hey folks. I have yet to explore the world of Wolfram Alpha and still don’t claim to be an expert...
2010-07-09
390 reads
Check out our latest book on Amazon. It’s not for sales yet, but just seeing the title up there is...
2010-07-06
334 reads
I have posted several entries lately about events that I’ve been attending, speaking and listening at and one of the...
2010-05-24
696 reads
Well, for those of you who attended, SQL Saturday Jacksonville, was, well I’ll say it, EPIC. Between all the great...
2010-05-24
307 reads
This is long overdue and goes out to my buddies from the SSAS Class I taught late last year. Two...
2010-05-24
400 reads
Hey there gang ! – It’s been a while since I’ve written and I apologize. It’s shameful I know. I just completed...
2010-05-10
348 reads
SQL Lunch has been a great resource for a while now, delivering lunchtime presentations. I was recently asked to be...
2010-03-02
342 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...
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