2016-08-11 (first published: 2015-06-12)
3,664 reads
2016-08-11 (first published: 2015-06-12)
3,664 reads
2016-08-10 (first published: 2015-05-05)
5,496 reads
This iTVF will produce a calendar table that can be used for complex date manipulation; quickly and effeciently
2016-08-08 (first published: 2014-12-29)
8,561 reads
Generates # of dates in before/after current year.
No more hardcoded date ranges.
2016-08-08 (first published: 2016-07-28)
613 reads
Identify the current statement and its line number within a running batch. Includes a link to the execution plan, if available.
2016-08-08 (first published: 2015-03-03)
5,447 reads
This script will detect and display a quick summary of your SQL Server installation.
2016-08-03 (first published: 2015-01-29)
9,523 reads
This script reports jobs which are running when another job is also running. This could be a reason for performance degradations.
2016-07-29 (first published: 2014-06-21)
2,664 reads
A script we use in our company, that rebuilds online=on by default, but takes care of special exceptions.
2016-07-28 (first published: 2014-06-06)
2,368 reads
Based upon todays date as suffix and chosen string as prefix, select data from predefined table into dynamically named new table. Script is properly error handled with most common errors
2016-07-27 (first published: 2015-05-04)
2,585 reads
2016-07-26 (first published: 2014-11-20)
2,255 reads
By Steve Jones
“Don’t aim to have others like you; aim to have them respect you.” –...
Many years ago, before I joined Oracle, I was working on a major modernisation...
If you work with data pipelines, SQL, notebooks, or machine learning models, a Mac...
Comments posted to this topic are about the item SQL Art, Part 4: Happy...
hi , a new user wants to be able to add sql agent jobs...
Comments posted to this topic are about the item SQL Server Enum Implementation: A...
In SQL Server 2025, I have a table (dbo.UserPermission) that contains this data:
UserID UserPermissions 15 23 37What is returned when I run this code:
select bit_count(UserPermissions) as PermissionCount from dbo.UserPermission where UserID = 3;See possible answers