How do I regex split by space, avoiding spaces within apostrophes?

Question

I want \"git log --format=\'(%h) %s\' --abbrev=7 HEAD\" to be split into

[
  \"git\", 
  \"log\",
  \"--format=\'(%h) %s\'\",
  \"--abbrev=7\",
  \         


        

Answer 1

As I understand, the idea is to split the string on contiguous spaces except where the spaces are part of a substring surrounded by single quotes. I believe this will work:

/(?:[^ ']*(?:'[^']+')?[^ ']*)*/

but invite readers to subject it to careful scrutiny.

demo

This regex can be made self-documenting by writing it in free-spacing mode:

/
(?:         # begin a non-capture group
  [^ ']*    # match 0+ chars other than spaces and single quotes
  (?:       # begin non-capture group
    '[^']+' # match 1+ chars other than single quotes, surrounded
            # by single quotes 
  )?        # end non-capture group and make it optional
  [^ ']*    # match 0+ chars other than spaces and single quotes
)*          # end non-capture group and execute it 0+ times
/x          # free-spacing regex definition mode

This obviously will not work if there are nested single quotes.

@n.'pronouns'm. suggested an alternative regex that also works:

/([^ "']|'[^'"]*')*/

demo