Is it possible to replace characters only within selected a group?

Question

Consider: word1.word2.worda-wordb-wordc.ext

Is there a regular expression capture and replace string that can be used to obtain as result: worda wordb wordc using perl compatible regular expressions?

I know you can capture the group of words a,b,c with: /.+?\..+?\.(.+?)\.ext$/$1/, but I don't know how to additionally replace dash (-) characters by space ( ) characters only within that group.

---

you can assume that:

• word groups are separated by period .
• words within a group are separated by dash -
• words are made up of alphanumeric characters [A-Za-z0-9]

looking for a one line /regex/replace/, not a script.

Answer 1

You can use this regex search-replace:

(?:^(?:[^.\n]+\.)*?(?=[^.\n]+\.ext$)|(?!^)\G)([^\n-]+)-(?:([^\n.-]+)\..+)?

And replace it with:

$1 $2

Output:

worda wordb wordc

RegEx Demo

RegEx Details:

• (?:: Start non-capture group
  • ^: Start
  • (?:[^.\n]+\.)*?: Match 1+ non-dot strings followed by a dot. Repeat this group 0 or more times (lazy match)
  • (?=[^.\n]+\.ext$): Must be followed by a <anything-but-dot.ext> and end position
  • |: OR
  • (?!^)\G: Start the next match from end of the previous match
• ): End non-capture group
• (: Start capture group #1
  • [^\n-]+: Match and capture a non-hyphen substring in capture group #1
• ): End capture group #1
• -: Match a -
• (?:: Start non-capture group
  • ([^\n.-]+): Match and capture non-dot, non-hyphen substring in capture group #2
  • \.: Match a dot
  • .+: Followed by extension
• )?: End non-capture group, ? makes this an optional match

PS: You can remove \n everywhere from above regex when matching a single line input.