Generate index on foreign key columns
Script find all foreign key without index in database and create new one for them.
2012-03-29 (first published: 2008-01-25)
3,380 reads
Script find all foreign key without index in database and create new one for them.
2012-03-29 (first published: 2008-01-25)
3,380 reads
Query Servers using 1 sql statement. Will make your job easy!!!
2012-03-28 (first published: 2008-01-16)
3,357 reads
Reindex all tables, showing statistics before and after reindexing, table name optional.
2012-03-27 (first published: 2008-01-30)
3,109 reads
2012-03-26 (first published: 2008-01-23)
3,235 reads
You Can Find Some Date Functions and extracting the different Date Formats using the Convert and Cast
2012-03-23 (first published: 2007-10-26)
3,682 reads
Split Function for T-SQL Server to break the Job History Message into Table.
2012-03-22 (first published: 2012-03-05)
1,075 reads
A small retake on the popular percent complete script which tell how much more time an executing command will take.
2012-03-21 (first published: 2012-03-07)
1,609 reads
The job will send an HTML email with the latest error at specific job.
2012-03-15 (first published: 2012-03-05)
890 reads
Stored procedure to update SSRS Subscription owners to avoid email errors.
2012-03-13 (first published: 2012-02-28)
1,445 reads
Loads temporary stored procs that assist in running and monitoring server-side tracing.
2012-03-12 (first published: 2012-02-21)
1,095 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