Daily SQL Server 2008 New Feature – Day 12
Every day for the next couple of weeks, I aim to highlight one of SQL Server 2008’s new features, simply...
2010-05-21
390 reads
Every day for the next couple of weeks, I aim to highlight one of SQL Server 2008’s new features, simply...
2010-05-21
390 reads
My oldest son graduates high school today. After a long journey, he’s the first of my wife’s and my three...
2010-05-21
691 reads
If you download the May 2010 minutes of the Board meeting (login required) you’ll see that we devoted an entire...
2010-05-21
585 reads
Recently I received a new opportunity via email. Steve Jones at SQLServerCentral sent me an email to see if I...
2010-05-21
647 reads
Introduction
Today’s script is one that I wrote based on the logic outlined in this post by Paul Randal (Blog|Twitter). This...
2010-05-21
552 reads
Today's script is one that I wrote based on the logic outlined in a post by Paul Randal. This script is written for SQL 2000 but, as Paul notes,...
2010-05-21
6 reads
Today's script is one that I wrote based on the logic outlined in a post by Paul Randal. This script is written for SQL 2000 but, as Paul notes,...
2010-05-21
13 reads
Every day for the next couple of weeks, I aim to highlight one of SQL Server 2008’s new features, simply...
2010-05-20
425 reads
2010-05-20
768 reads
A new article for the SQL Standard is now available. You can find the article here. The article is by...
2010-05-20
604 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