Forum Discussion

Qrew Captain

2 months ago

Solved

Using RegexReplace()

Does this function support capturing groups, and if yes, how does one access the captured data?

Intent is to use a this function looking for punctuation from a text field, and then add a new line to this.

regex: ([.!?])\s*

substitution: $1\n

Documentation doesn't state one way or the other. I've tried escaping the new line value without success. Perhaps I'm just missing something simple.

Mez
2 months ago
Found the solution: capturing is referred to with backslash 1, which needs to be escaped.
RegexReplace( [Additional Notes], "([.!?])\\s*", "\\1\n" )

4 Replies

MarkShnier
Qrew Cadet
2 months ago
Can you give an example of the before and after results you are looking for?
- Mez
  Qrew Captain
  2 months ago
  Sure. If capturing is supported for this formula, I would like, rather expect that we could invoke the variable nomenclature of $1 to refer to it. It does not appear to be supported.
  Since the substitution or replace input is a string and you have to escape certain characters I tried "\\n" to create a newline - no luck; tried to refer to the capture variable - $1 - highlights as yellow - not supported
  RegexReplace( [Notes text], "([.!?])\\s*", $1&"\\n")
  Looks like this might be an enhancement request.
  *edit: for some reason the solution isn't showing in the thread; posting here. Access captured data with \1, which has to be escaped.
  RegexReplace( [Notes text], "([.!?])\\s*", "\\1\n")
  - Mez
    Qrew Captain
    2 months ago
    Found the solution: capturing is referred to with backslash 1, which needs to be escaped.
    RegexReplace( [Additional Notes], "([.!?])\\s*", "\\1\n" )
Mez
Qrew Captain
2 months ago
Posting this again.
Found the solution: capturing is referred to with backslash 1, which needs to be escaped.
RegexReplace( [Additional Notes], "([.!?])\\s*", "\\1\n" )