regex to match until a colon is found with or without quotes present

Question

I'm looking for a regex to achieve the following:

When my input is root:Folder1:fname I want to get fname

and when the input string is root:Folder1:'fname' to get again fname (no quotes!) .

For the former [^:?]*$ seems to get the job done but I cannot find a single regex to match both cases.

Is it possible to do it with a single regex? Tried to achieve it with conditional regex but I couldn't make it.

Any other tip/solution is welcome.

Thank you.

edit: The format of the paths is not fixed, quotes might appear elsewhere (eg root:'Folder with spaces1':fname or root:'Folder with spaces1':'fname'). The solution I guess is to search backwards as the filename we want to capture is always after the last colon.

Answer 1

As you tagged pcre, you might also use a branch reset group to get group 1 between single quotes or not between single quotes.

^.*:(?|'(\w+)'|(\w+))$

• ^ Start of string
• .*: Match until the last :
• (?| Branch reset group
  • '(\w+)' Capture 1+ word chars between single quotes | Or (\w+) Capture 1+ word chars
• ) Close branch reset group
• $ End of string

Regex demo

Answer 2

This will work if your "grammar" is strict:

(\w+:)*'?(\w+)'?

• First we repeat anyword: from 0 to infinite times
• then we consume a single quote if it exists
• then we record the last word
• then we consume the last quote if it exists

Watch the regex debugger here: https://regex101.com/r/SjoWYP/1

Answer 3

If you want to enforce paired quotes, you need to capture the first quote.

(\w+:)*('?)(\w+)\2$

The expression inside the third pair of parentheses will be available as the third captured group ( $3 or \3 or match.group(3) or what have you).

If your regex dialect permits non-grouping parentheses, you can rearticulate this as

(?:\w+:)*('?)(\w+)\1$

and/or also perhaps extend the quote group to permit either single or double quote ( ['"]? pro '? ). Then of course the final group will be $2 instead of $3 .