Skip to main content
Solved

StringReplacer to find last space


dos_gis
Contributor
Forum|alt.badge.img+3

Hi, Regex help please. I'm trying to replace the last space character in a list of names with a comma. Some of the names have multiple spaces in them:

 

Example:

John Doe

John Charles Doe

John Jimmy Charles Doe

 

The result I'm looking for:

John,Doe

John Charles,Doe

John Jimmy Charles,Doe

Thanks!

 

Best answer by davtorgh

Hi,

 

I think you can set parameter 'Text To Replace' to 

\s+(\w+)$

and 'Replacement Text' to

,\1

If you suspect there may be trailing spaces, add `\s+` before `$`.

 

Hope that helps!

View original
Did this help you find an answer to your question?

3 replies

davtorgh
Contributor
Forum|alt.badge.img+10
  • Contributor
  • Best Answer
  • August 26, 2020

Hi,

 

I think you can set parameter 'Text To Replace' to 

\s+(\w+)$

and 'Replacement Text' to

,\1

If you suspect there may be trailing spaces, add `\s+` before `$`.

 

Hope that helps!


dos_gis
Contributor
Forum|alt.badge.img+3
  • Author
  • Contributor
  • August 26, 2020

Works perfectly! Thank you.

 

I didn't realize that you could add the \\1 qualifier to the replacement text.


mark2atsafe
Safer
Forum|alt.badge.img+44
  • Safer
  • August 27, 2020
davtorgh wrote:

Hi,

 

I think you can set parameter 'Text To Replace' to 

\s+(\w+)$

and 'Replacement Text' to

,\1

If you suspect there may be trailing spaces, add `\s+` before `$`.

 

Hope that helps!

Interesting. There's a LastSectionOfStringRemover on the Hub that does this, but removes everything after that character. I should update it to also do a replacement, using this piece of regex. Thanks for posting a great answer.


Cookie policy

We use cookies to enhance and personalize your experience. If you accept you agree to our full cookie policy. Learn more about our cookies.

 
Cookie settings