T-SQL Tuesday #61 – Giving Back
The Season of Giving
The annual PASS Summit (otherwise known as the #SQLFamily reunion) is over. Here in the United States,...
2014-12-08 (first published: 2014-12-02)
6,427 reads
The Season of Giving
The annual PASS Summit (otherwise known as the #SQLFamily reunion) is over. Here in the United States,...
2014-12-08 (first published: 2014-12-02)
6,427 reads
You know, these 1 hour sessions that are at most SQL Saturdays are just too short sometimes – you just get...
2014-10-16
567 reads
You know, these 1 hour sessions that are at most SQL Saturdays are just too short sometimes – you just get...
2014-08-13
669 reads
Well, here it is again. The second Tuesday of the month, which means that it’s T-SQL Tuesday. T-SQL Tuesday… that...
2014-08-13
611 reads
You know, these 1 hour sessions that are at most SQL Saturdays are just too short sometimes – you just get...
2014-08-11
687 reads
Well, here it is again. The second Tuesday of the month, which means that it’s T-SQL Tuesday. This month, Dev...
2014-07-08
793 reads
You’re utilizing the database mirroring high-availability feature, configured to be in the High Safety mode with automatic failover, which means...
2014-05-28 (first published: 2014-05-22)
6,146 reads
After installing SSAS, using the tabular model, you receive the error “The service cannot be started: The following system error...
2014-05-19 (first published: 2014-05-13)
4,117 reads
Wow, what a month. On March 22, 2014, the SQL Saturday train came into Richmond VA (RVA). I was one...
2014-03-28
895 reads
There’s a new SQL Server user group starting up over in Lynchburg, VA. It’s first meeting is February 27, 2014...
2014-02-17
723 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