How to match text between words using regex?

Question

I would like to match text text between pre-processor defines in Python.

In this example I'd like to match to remove text so lines 2..4 would be removed, eg:

#if 1
#  if 0
Remove me
#  endif
Keep me
#endif

Using this regex, it removes text, but the .* doesn't stop at the first #endif:

def remove_if0(string):
    pattern = r"(^\s*#\s*if\s+0\b.*^\s*#\s*endif\b)"
    regex = re.compile(pattern, re.MULTILINE | re.DOTALL)
    return regex.sub("", string)

Is there a way to match against pairs without the DOTALL reading past a term? eg ^\s*#\s*endif\b.

I tried (?!word), eg: (?!^\s*#\s*endif\b)* - but it didn't work.

Possible duplicate of [RegEx: Smallest possible match or nongreedy match](https://stackoverflow.com/questions/1919982/regex-smallest-possible-match-or-nongreedy-match) — revo, Mar 03 '18 at 12:33
The duplicate has the same answer but the question is much more spesific (also `.net`, although that doesn't always matter). — ideasman42, Mar 03 '18 at 12:53

ideasman42 · Answer 1 · 2018-03-03T12:54:33.463

2

The solution is to use ungreedy .*? (thanks to @bobble-bubble)

Here is a working Python function:

def remove_if0(string):
    pattern = r"(^\s*#\s*if\s+0\b.*?^\s*#\s*endif)"
    regex = re.compile(pattern, re.MULTILINE | re.DOTALL)
    return regex.sub("", string)

edited Mar 03 '18 at 12:54

answered Mar 03 '18 at 12:30

ideasman42

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How to match text between words using regex?

1 Answers1