I was trying to get all quoted (" or ') substrings from a string excluding the quotation marks. I came up with this:
"((?:').*[^'](?:'))|((?:\").*[^\"](?:\"))"
For some reason the matching string still contains the quotation marks in it. Any reason why ?
Sincerely, nikita.utiu.
You could do it with lookahead and lookbehind assertions:
>>> match = re.search(r"(?<=').*?(?=')", "a 'quoted' string. 'second' quote")
>>> print match.group(0)
quoted
Using non-capturing groups doesn’t mean that they are not captured at all. They just don’t create separate capturing groups like normal groups do.
But the structure of the regular expression requires that the quotation marks are part of the match:
"('[^']*'|\"[^\"]*\")"
Then just remove the surrounding quotation marks when processing the matched parts with matched_string[1:-1].
"(?:')(.*[^'](?:'))|((?:\").*)[^\"](?:\")"
foobar and the pattern foo(bar)?, there will be two groups: 0, that holds the match of the whole pattern (foobar); 1, that holds the match of the first subpattern (bar). With foo(?:bar)? you will only get foobar and no separate match of the subpattern.You could try:
import shlex
...
lexer = shlex.shlex(your_input_string)
quoted = [piece.strip("'\"") for piece in lexer if piece.startswith("'") or piece.startswith('"')]
shlex (lexical analysis) takes care of escaped quotes for you. Though note that it does not work with unicode strings.
