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

问题

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

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

How to I achieve this, without splitting on the space within --format='(%h) %s'?

Answers in any language is welcome :)

回答1:

As often in life, you have choices.

Use an expression that matches and captures different parts. This can be combined with a replacement function as in

import re
string = "git log --format='(%h) %s' --abbrev=7 HEAD"

rx = re.compile(r"'[^']*'|(\s+)")

def replacer(match):
    if match.group(1):
        return "#@#"
    else:
        return match.group(0)

string = rx.sub(replacer, string)
parts = re.split('#@#', string)
#                 ^^^ same as in the function replacer

You could use the better regex module with (*SKIP)(*FAIL):

import regex as re
string = "git log --format='(%h) %s' --abbrev=7 HEAD"

rx = re.compile(r"'[^']*'(*SKIP)(*FAIL)|\s+")
parts = rx.split(string)

Write yourself a little parser:

def little_parser(string):
    quote = False
    stack = ''

    for char in string:
        if char == "'":
            stack += char
            quote = not quote
        elif (char == ' ' and not quote):
            yield stack
            stack = ''
        else:
            stack += char

    if stack:
        yield stack

for part in little_parser(your_string):
    print(part)

All three will yield

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

回答2:

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

回答3:

I found one possible (albeit ugly) solution in python (which also works with "):

>>> import re
>>> foo = '''git log --format='(%h) %s' --foo="a b" --bar='c d' HEAD'''
>>> re.findall(r'''(\S*'[^']+'\S*|\S*"[^"]+"\S*|\S+)''', foo)
['git', 'log', "--format='(%h) %s'", '--foo="a b"', "--bar='c d'", 'HEAD']

来源：https://stackoverflow.com/questions/60502562/how-do-i-regex-split-by-space-avoiding-spaces-within-apostrophes

标签

regex

split