Nginx location: Nginx location block section have a search order, a modifier, an implicit match type and an implicit switch to whether stop the search on match or not. the following array describe it for regex. # ——————————————————————————————————————————————– # Search-Order Modifier Description Match-Type Stops-search-on-match # ——————————————————————————————————————————————– # 1st = The URI must match the specified … Read more
Depending on the regular expression implementation you can use so called non-capturing groups with the syntax (?:…): ((?:a|b)c) Here (?:a|b) is a group but you cannot reference its match. So you can only reference the match of ((?:a|b)c) that is either ac or bc.
Match the double quoted substrings with a mere “[^”]+” regex (if there are no escape sequences to account for) and replace the tabs inside the matches only inside a match evaluator: var str = “A tab\there \”inside\ta\tdouble-quoted\tsubstring\” some\there”; var pattern = “\”[^\”]+\””; // A pattern to match a double quoted substring with no escape sequences … Read more
The answer is: Python re does not support identically named groups. Python PyPi regex module supports identically named named capturing groups: The same name can be used by more than one group, with later captures ‘overwriting’ earlier captures. All of the captures of the group will be available from the captures method of the match … Read more
Making a non-capturing, zero to more matching group, you must append ?. (?: ….. )? ^ ^____ optional |____ group
Use $n (where n is a digit) to refer to captured subsequences in replaceFirst(…). I’m assuming you wanted to replace the first group with the literal string “number” and the second group with the value of the first group. Pattern p = Pattern.compile(“(\\d)(.*)(\\d)”); String input = “6 example input 4”; Matcher m = p.matcher(input); if … Read more
Since we’re all guessing, I might as well give mine: I’ve always thought it stood for Python. That may sound pretty stupid — what, P for Python?! — but in my defense, I vaguely remembered this thread [emphasis mine]: Subject: Claiming (?P…) regex syntax extensions From: Guido van Rossum (gui…@CNRI.Reston.Va.US) Date: Dec 10, 1997 3:36:19 … Read more
Instead of a negated character class, you have to use a negative lookahead: \bvalue\s*=\s*([“‘])(?:(?!\1).)*\1 (?:(?!\1).)* consumes one character at a time, after the lookahead has confirmed that the character is not whatever was matched by the capturing group, ([“”]). A character class, negated or not, can only match one character at a time. As far … Read more
Have a look at re.sub: result = re.sub(r”(\d.*?)\s(\d.*?)”, r”\1 \2″, string1) This is Python’s regex substitution (replace) function. The replacement string can be filled with so-called backreferences (backslash, group number) which are replaced with what was matched by the groups. Groups are counted the same as by the group(…) function, i.e. starting from 1, from … Read more
The answer is: re.sub(r'(foo)’, r’\g<1>123′, ‘foobar’) Relevant excerpt from the docs: In addition to character escapes and backreferences as described above, \g will use the substring matched by the group named name, as defined by the (?P…) syntax. \g uses the corresponding group number; \g<2> is therefore equivalent to \2, but isn’t ambiguous in a … Read more