Regex - nested lookahead assertion
regex, regex-lookarounds
Solution
Many tools exist to automatically explain what your regex does, character by character.
The idea behind it, though, is that you want to check `one` is followed by `</out>` while forbidding to enter a new `out` tag: if there's a `...</out>` following and we haven't entered a new `<out>...</out>` structure, we know we are inside one already.
So the regex will match `one` if it is followed by `</out>` and if there's no `<out>` between the two.
The work is done by `(?:(?!<out>).)*`: the `.` matches only if it is not the first `<` in `<out>`. So we can go up to `</out>` only by consuming characters that are not this `<` followed by `out>`.
A speed improvement would be:
one(?=(?:[^<]*+|<(?!out>))*+</out>)
Stepping inside the negative lookahead at each character greatly increases the cost of matching this character. Here `[^<]*+` will match directly up to the next suspicious `<`, and we perform the negative look ahead check only when we have to.
Problem
suppose we want to match all `one`(s) between `<out>...</out>` in this text (option: dot matches all): ``` <out>hello!</out> <nx1>home one</nx1> <nx2>living</nx2> <out>one text text one continues and at last here ends one</out> <m2>dog one</m2> <out>bye!</out> ``` let's say we use this pattern: ``` one(?=(?:(?!<out>).)*</out>) ``` I really appreciate it if someone explains how regex engine process that pattern step-by-step and where it would be(position in the original text) in every phase of processing;(something like accepted @Tim Pietzcker's helpful explanation for this question: Regex - lookahead assertion)