A Lack of SQL
Steve Jones talks about a company looking to write all their stored procedures using the CLR in SQL Server. Is this a good idea?
2008-09-15
256 reads
Steve Jones talks about a company looking to write all their stored procedures using the CLR in SQL Server. Is this a good idea?
2008-09-15
256 reads
I'm starting a series of blog posts from the Business of Software conference
that I attended last week in Boston. If...
2008-09-15
743 reads
A bug in the SQL Server 2008 upgrade process has Steve Jones questioning the coding practices at Microsoft.
2008-09-15
72 reads
I haven't done as much reading on the Kindle since I've been home from my last trip, mainly because I've...
2008-09-15
815 reads
2008-09-15
3,274 reads
A new data mining thrilled from Jeffrey Deaver has Steve Jones concerned about the centralization of data mining.
2008-09-15
27 reads
A new data mining thrilled from Jeffrey Deaver has Steve Jones concerned about the centralization of data mining.
2008-09-15
82 reads
A new data mining thrilled from Jeffrey Deaver has Steve Jones concerned about the centralization of data mining.
2008-09-15
68 reads
Steve Jones talks about a company looking to write all their stored procedures using the CLR in SQL Server. Is this a good idea?
2008-09-14
88 reads
Steve Jones talks about a company looking to write all their stored procedures using the CLR in SQL Server. Is this a good idea?
2008-09-14
67 reads
Setting page visibility and the active page are often overlooked last steps when publishing...
By Steve Jones
It’s time for T-SQL Tuesday again and this time Todd Kleinhans has a great...
By Steve Jones
Recently I was working in VS Code and I saw a walkthrough for the...
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