Reblog: April 18 to April 24
Photo credit – JStove
Old posts are like unicorns. You might see them just once, but it’d sure be nice to have...
2014-04-25
1,320 reads
Photo credit – JStove
Old posts are like unicorns. You might see them just once, but it’d sure be nice to have...
2014-04-25
1,320 reads
It’s Monday time for this week’s weekly link round-up. If you want to catch these links “live” (so exciting), follow...
2014-04-22
989 reads
Photo credit – Travis
Ever go digging through your backyard for dinosaur bones? Maybe not, but it’s worthwhile from time to time...
2014-04-18
489 reads
It’s Monday time for this week’s weekly link round-up. If you want to catch these links “live” (so exciting), follow...
2014-04-14
710 reads
Photo credit – Husso
I’ve always been a fan of the feeling when I find an old blog post that’s got just...
2014-04-11
351 reads
Similar to yesterday’s post, I have a follow-up for another Pragmatic WorksTraining on the T’s session that I delivered last month....
2014-04-10
831 reads
Last month, I presented a session for Pragmatic WorksTraining on the T’s titled Introduction to Clustered Indexes and Heaps. I’d meant to...
2014-04-09
760 reads
It’s Monday time for this week’s weekly link round-up. If you want to catch these links “live” (so exciting), follow...
2014-04-07
684 reads
image source
Make sure you start the month right by taking the time to make certain your SQL Server environment is...
2014-04-15 (first published: 2014-04-07)
3,727 reads
It’s Monday time for this week’s weekly link round-up. If you want to catch these links “live” (so exciting), follow...
2014-03-31
1,944 reads
By Chris Yates
The New Arena of Leadership The role of the Chief Data Officer is no...
Presenting you with an updated version of our sp_snapshot procedure, allowing you to easily...
SELECT * feels convenient, but in SQL Server it bloats I/O, burns network bandwidth,...
I've noticed several instances of what looks like a recursive insert with the format:...
Comments posted to this topic are about the item Cleaning Up the Cloud
Comments posted to this topic are about the item The Maximum Value in the...
I have a table with this data:
TravelLogID CityID StartDate EndDate 1 1 2025-01-01 2025-01-06 2 2 2025-01-01 2025-01-06 3 3 2025-01-01 2025-01-06 4 4 2025-01-01 2025-01-06 5 5 2025-01-01 2025-01-06I run this code:
SELECT IDENT_CURRENT('TravelLog')I get the value 5 back. Now I do this:
SET IDENTITY_INSERT dbo.TravelLog ON INSERT dbo.TravelLog ( TravelLogID, CityID, StartDate, EndDate ) VALUES (25, 5, '2025-09-12', '2025-09-17') SET IDENTITY_INSERT dbo.TravelLog OFFI now run this code.
DBCC CHECKIDENT(TravelLog) GO INSERT dbo.TravelLog ( CityID, StartDate, EndDate ) VALUES (4, '2025-10-14', '2025-10-17') GOWhat is the value for TravelLogID for the row I inserted for CityID 4 and dates starting on 14 Oct 2025? See possible answers