Problems With Database Problems
The GitLab crisis has Steve Jones thinking about two things: competence and care.
2017-02-13
114 reads
The GitLab crisis has Steve Jones thinking about two things: competence and care.
2017-02-13
114 reads
From a well-built cockney who knew how to avoid wheel clamps, to an ex-member of an elite anti-riot unit, some of the best DBAs have a surprising diversity of skills and experiences, according to Phil Factor.
2017-02-13
93 reads
An updated editorial. The five years are almost up for the prediction that IT departments would be eliminated. Read Steve Jones thoughts on where we are today.
2017-02-10
101 reads
DevOps is supposed to help us build better software, faster. Steve Jones looks at one of the other benefits: security.
2017-02-09
90 reads
2017-02-08
153 reads
Today Steve Jones looks at how we can better build software by considering the cloud.
2017-02-06
70 reads
Microsoft's Customer Experience Improvement Program for SQL Server: once easy to avoid, now hard to ignore, but Rodney Landrum sees an upside.
2017-02-06
1,520 reads
This week Steve Jones asks what parts of SQL Server would you improve and why.
2017-02-03
70 reads
A look at the good, and bad, of the IoT world along with the potential future.
2017-02-02
97 reads
We all have different levels of skills, and it's not always because we don't try. Sometimes it's a question of timing.
2017-01-31
123 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