Query to get tables with no clustered indexes
Query to get tables with no clustered indexes(2K5,2K8,2K8 R2)
2011-02-14 (first published: 2011-01-28)
1,860 reads
Query to get tables with no clustered indexes(2K5,2K8,2K8 R2)
2011-02-14 (first published: 2011-01-28)
1,860 reads
2011-02-11 (first published: 2011-01-31)
2,960 reads
Find quickly all stored procedures, tables and views that are somehow related to the search term.
2011-02-09 (first published: 2011-01-28)
2,777 reads
Generates a list of Server, Database, Table , Column Names and properties
2011-02-08 (first published: 2011-01-31)
1,401 reads
This Stored procedure re-sends the failed reports subscriptions on Demand.
2011-01-31 (first published: 2011-01-12)
1,126 reads
2011-01-28 (first published: 2011-01-11)
1,819 reads
This stored Procedure grabs the reports scheduled to run on today's date and reruns the subscriptions on demand.
2011-01-26 (first published: 2011-01-12)
891 reads
2011-01-21 (first published: 2011-01-11)
2,922 reads
This func returns the max value,min value and count of values from collection of values
2011-01-20 (first published: 2011-01-08)
1,789 reads
2011-01-17 (first published: 2011-01-05)
7,225 reads
By Steve Jones
I caught this interesting item over on Pinal Dave’s blog: Eleven Interview Questions that...
By Hemantgiri
I use color-coded connections in SSMS to distinguish Production, Pre-Production, UAT, and Development. But...
Comments posted to this topic are about the item Split Large Queries in Athena...
Comments posted to this topic are about the item Finding Trailing Spaces
I have some data in a SQL Server 2025 database. It looks like this for the dbo.Customer table:
CustomerID CustomerName PreferredName 1 Steve Steve 2 Andy Andy 3 Brian Brian 4 Allan Allan 5 Devin Devin 6 Steve Steve 7 Sally SallyI want to detect which names have a single trailing space. The CustomerName is a varchar() and the PreferredName is a CHAR(). Does this query detect the problem rows?
SELECT
CustomerID,
CustomerName,
PreferredName
FROM Customer
WHERE CustomerName <> RTRIM(CustomerName)
OR PreferredName <> RTRIM(PreferredName); See possible answers