Does rtrim remove whitespace other than spaces?

This select yields the results below.  rtrim doesn't seem to trim the last character.  I'm wondering if rtrim removes spaces but not tabs or something.

select len(loginwindows) as loginlen, len(rtrim(loginwindows)) as loginlenrtrim, left(loginwindows, 8) as login8, loginwindows from general.employees where lastname = 'langley'

loginlen      loginlentrim    login8           loginwindows
9                9                      LL004838      LL004838 
Barry SweezeySoftware EngineerAsked:
Who is Participating?

[Product update] Infrastructure Analysis Tool is now available with Business Accounts.Learn More

x
I wear a lot of hats...

"The solutions and answers provided on Experts Exchange have been extremely helpful to me over the last few years. I wear a lot of hats - Developer, Database Administrator, Help Desk, etc., so I know a lot of things but not a lot about one thing. Experts Exchange gives me answers from people who do know a lot about one thing, in a easy to use platform." -Todd S.

Scott PletcherSenior DBACommented:
RTRIM/LTRIM remove only spaces.
0
Barry SweezeySoftware EngineerAuthor Commented:
How do I remove all whitespace?
0
Scott PletcherSenior DBACommented:
Easiest way to remove others is to use REPLACE ... but if you only want to replace trailing values it's quite a bit trickier.

To remove *all* tabs, carriage returns and line feeds, you can do this:

UPDATE
SET column = REPLACE(REPLACE(REPLACE(column, CHAR(9), ''), CHAR(13), ''), CHAR(10), '')
WHERE
    column LIKE '%' + CHAR(9) + '%' OR
    column LIKE '%' + CHAR(13) + '%' OR
    column LIKE '%' + CHAR(10) + '%'
0
Acronis True Image 2019 just released!

Create a reliable backup. Make sure you always have dependable copies of your data so you can restore your entire system or individual files.

Scott PletcherSenior DBACommented:
If you want to be really thorough, you could also check for CHAR(11) and CHAR(15), although the ones above are the most common in Windows.
0
Barry SweezeySoftware EngineerAuthor Commented:
Thanks for the quick response.

I ran the query below and received the message (0 row(s) affected).  The original query yields the sames results.  How can I identify the last character in the string?


UPDATE general.employees
SET LoginWindows = REPLACE(REPLACE(REPLACE(REPLACE(REPLACE(LoginWindows, CHAR(9), ''), CHAR(10), ''), CHAR(11), ''), CHAR(13), ''), CHAR(15), '')
WHERE
    LoginWindows LIKE '%' + char(9) + '%' OR
    LoginWindows LIKE '%' + char(10) + '%' OR
    LoginWindows LIKE '%' + char(10) + '%' OR
    LoginWindows LIKE '%' + char(13) + '%' OR
    LoginWindows LIKE '%' + char(15) + '%'
0
Scott PletcherSenior DBACommented:
SELECT
    DISTINCT
    ASCII(RIGHT(LoginWindows, 1))
FROM general.employees
WHERE
    ASCII(RIGHT(LoginWindows, 1)) < 32 OR
    ASCII(RIGHT(LoginWindows, 1)) > 127
0
Barry SweezeySoftware EngineerAuthor Commented:
I corrected a line; no difference.

    LoginWindows LIKE '%' + char(10) + '%' OR
    LoginWindows LIKE '%' + char(11) + '%' OR
0
Scott PletcherSenior DBACommented:
Re-post of query:

Try this query to identify the trailing char(s) :

SELECT
    DISTINCT
    ASCII(RIGHT(LoginWindows, 1))
FROM general.employees
WHERE
    ASCII(RIGHT(LoginWindows, 1)) < 32 OR
    ASCII(RIGHT(LoginWindows, 1)) > 127
0

Experts Exchange Solution brought to you by

Your issues matter to us.

Facing a tech roadblock? Get the help and guidance you need from experienced professionals who care. Ask your question anytime, anywhere, with no hassle.

Start your 7-day free trial
Barry SweezeySoftware EngineerAuthor Commented:
Results:

(No column name)
160

So it's a Non-breaking space.

I updated the update query (below) and it worked.

UPDATE general.employees
SET LoginWindows = REPLACE(REPLACE(REPLACE(REPLACE(REPLACE(REPLACE(LoginWindows, CHAR(9), ''), CHAR(10), ''), CHAR(11), ''), CHAR(13), ''), CHAR(15), ''), CHAR(160), '')
WHERE
    LoginWindows LIKE '%' + char(9) + '%' OR
    LoginWindows LIKE '%' + char(10) + '%' OR
    LoginWindows LIKE '%' + char(11) + '%' OR
    LoginWindows LIKE '%' + char(13) + '%' OR
    LoginWindows LIKE '%' + char(15) + '%' OR
    LoginWindows LIKE '%' + char(160) + '%'
0
Barry SweezeySoftware EngineerAuthor Commented:
I appreciate ScottPletcher's hanging with me until the problem was solved.
0
It's more than this solution.Get answers and train to solve all your tech problems - anytime, anywhere.Try it for free Edge Out The Competitionfor your dream job with proven skills and certifications.Get started today Stand Outas the employee with proven skills.Start learning today for free Move Your Career Forwardwith certification training in the latest technologies.Start your trial today
Microsoft SQL Server 2005

From novice to tech pro — start learning today.