Day four of Summit – Keynote 2 #sqlpass
The keynote started out with Bill Graziano taking to the stage in a kilt and declaring that the second day...
2011-10-13
1,107 reads
The keynote started out with Bill Graziano taking to the stage in a kilt and declaring that the second day...
2011-10-13
1,107 reads
Today was our first day to attend regular sessions. With over 189 sessions being offered over 3 days there are...
2011-10-13
692 reads
The second day of the Summit was full of even more pre-con’s. Word on the street last night is that...
2011-10-12
940 reads
Last year we set out some very high goals. 1 million technical training hours, 250k members in the community, and...
2011-10-12
969 reads
My first day in Seattle has ended. My flight arrived around 12:30 and I was in good company with John...
2011-10-11
736 reads
Watching the twitter sphere it is apparent that many of the SQL Nation are in route to Seattle WA today....
2011-10-10
464 reads
Last year at the Summit I missed out on this really cool concept. Donate your unused items from your hotel....
2011-10-06
691 reads
I decided to start monitoring the amount of free space in my database files so that I can make sure...
2011-10-05
1,567 reads
Well it isn’t quite over for me just yet, I still have a flight home tomorrow and can’t wait to...
2011-10-02
470 reads
I flew into Austin Texas today and got to spend time with a lot of great people. The journey started...
2011-10-01
523 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...
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
I am trying to check out elastic query between two test instances we have...
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