PASSMN May Meeting
If you are going to be visiting Minnesota on May 17th or are there on more permanent basis, you can...
2011-05-09
624 reads
If you are going to be visiting Minnesota on May 17th or are there on more permanent basis, you can...
2011-05-09
624 reads
As I mentioned in the introductory post, I’m summarizing posts from previous years in the the past week. Some posts...
2011-05-06
562 reads
SQLRally is a week away. Are you ready? It’s going to be a good time to learn and freshen up...
2011-05-04
478 reads
Earlier today, Thomas LaRock (Blog | @SQLRockstar) tagged me for his Meme Monday post. The question for today is: how many...
2011-05-02
566 reads
As I mentioned in the introductory post, I’m summarizing posts from previous years in the the past week. Some posts...
2011-04-29
1,450 reads
As I mentioned in the introductory post, I’m summarizing posts from previous years in the the past week. Some posts...
2011-04-22
561 reads
I’ve been counting down the days until I get the score from my second attempt at the SQL Server MCM...
2011-04-19
498 reads
I mentioned the April PASSMN meeting the other day… well now that meeting is tomorrow. Take a chance to invest...
2011-04-18
573 reads
As I mentioned in the introductory post, I’m summarizing posts from previous years in the the past week. Some posts...
2011-04-15
568 reads
Over the last couple months, I’ve been trying to figure out what to do with some of my old posts. ...
2011-04-14
446 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