Scripts

Technical Article

Find and Replace a value across all tables in database

End user has entered ‘goodmorning’ and ‘good morning’ interchangeably and it was causing logic failure at application level. So task was to replace ‘goodmorning’ with ‘good morning’ in database wherever it is there.

★ ★ ★ ★ ★ ★ ★ ★ ★ ★

(9)

You rated this post out of 5. Change rating

2017-04-28 (first published: )

1,121 reads

Blogs

Goodbye, Microsoft

By

A few years ago I took a new job at Microsoft, working as a...

The SUM of Nothing: #SQLNewBlogger

By

I caught this interesting item over on Pinal Dave’s blog: Eleven Interview Questions that...

How to Move the SSMS Status Bar to the Top and Color-Code SQL Server Connections

By

I use color-coded connections in SSMS to distinguish Production, Pre-Production, UAT, and Development. But...

Read the latest Blogs

Forums

Alamat BCA KCP Sentral Cikini Telp:08218520154

By R4nt4u

WhatsApp CS 628218520154 Jl. Pegangsaan Timur No.1, Cikini, Kec. Menteng, Kota Jakarta Pusat, Daerah...

Alamat BCA KCP Pintu Air Telp:08218520154

By m4rt1n4

WhatsApp CS 628218520154 Jl. Pintu Air Raya No.36 Q, RT.6/RW.1, Ps. Baru, Kecamatan Sawah...

Today's AI

By Steve Jones - SSC Editor

Comments posted to this topic are about the item Today's AI

Visit the forum

Question of the Day

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        Sally
I 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