The Financial Crisis
With the economy slowing the the world facing a financial crisis, Steve Jones polls the man on the street this Friday.
2008-10-17
142 reads
With the economy slowing the the world facing a financial crisis, Steve Jones polls the man on the street this Friday.
2008-10-17
142 reads
I've had a bunch of writings based on finances over the last couple weeks, and I've tried to spread them...
2008-10-17
653 reads
The loss of data is getting ridiculous. Steve Jones wants companies and government to do something about it.
2008-10-17
782 reads
The loss of data is getting ridiculous. Steve Jones wants companies and government to do something about it.
2008-10-17
516 reads
2008-10-17
2,112 reads
With the economy slowing the the world facing a financial crisis, Steve Jones polls the man on the street this Friday.
2008-10-16
85 reads
With the economy slowing the the world facing a financial crisis, Steve Jones polls the man on the street this Friday.
2008-10-16
63 reads
With the economy slowing the the world facing a financial crisis, Steve Jones polls the man on the street this Friday.
2008-10-16
72 reads
A great company will be big enough and small enough. Steve Jones talks about finding that balance.
2008-10-15
92 reads
I've always been very concerned about where I work. Well, maybe not always, since early in my career I took...
2008-10-15
627 reads
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...
By Kevin3NF
Flexibility and Scale at the Database Level When SQL Server 2012 introduced Availability Groups...
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