2014-03-31
1,563 reads
2014-03-31
1,563 reads
One of the biggest mistakes we can make when troubleshooting is to confuse the necessary responsibility of observing activity that...
2014-02-11
1,497 reads
We are awaiting Polar Vortex 2 in New England and are set to have more snow dumped on us starting...
2014-01-21
1,072 reads
Back when I was getting ready to take the plunge and begin consulting I sought advice from a friend who...
2014-01-20
782 reads
Just a quick note today on being effective.
For better or for worse, I was raised by parents who had me...
2014-01-16
664 reads
2013-12-24
575 reads
Microsoft publishes the following caution about enabling the CredSSP Windows Group Policy:
Caution: Credential Security Support Provider (CredSSP) authentication, in which...
2013-12-04
902 reads
“There is little value in ensuring the survival of our nation if our traditions do not survive with it. And...
2013-11-21
624 reads
ugggh.
My Drupal site in AWS EC2 was hijacked and is being used by the hijackers to do port scans.
Who said...
2013-11-06
597 reads
This sort of ties in to my last post on denying reality.
Managing expectations is the most important thing to success...
2013-10-11
690 reads
By Steve Jones
ecstatic shock – n. a surge of energy upon catching a glimpse from someone...
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...
Comments posted to this topic are about the item Lessons from the Postmark-MCP Backdoor
Just saw the "Azure Extension for SQL Server" Does anyone has experience with it?...
I've noticed several instances of what looks like a recursive insert with the format:...
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