Shortest match in regex from end

Question

Given an input string fooxxxxxxfooxxxboo I am trying to write a regex that matches fooxxxboo i.e. starting from the second foo till the last boo.

I tried the following

foo.*?boo matches the complete string fooxxxxxxfooxxxboo

foo.*boo also matches the complete string fooxxxxxxfooxxxboo

I read this Greedy vs. Reluctant vs. Possessive Quantifiers and I understand their difference, but I am trying to match the shortest string from the end which matches the regex i.e. something like the regex to be evaluated from back. Is there any way I can match only the last portion?

Answer 1

Use negative lookahead assertion.

foo(?:(?!foo).)*?boo

DEMO

(?:(?!foo).)*? - Non-greedy match of any character but not of foo zero or more times. That is, before matching each character, it would check that the character is not the letter f followed by two o's. If yes, then only the corresponding character will be matched.

Why the regex foo.*?boo matches the complete string fooxxxxxxfooxxxboo?

Because the first foo in your regex matches both the foo strings and the following .*? will do a non-greedy match upto the string boo, so we got two matches fooxxxxxxfooxxxboo and fooxxxboo. Because the second match present within the first match, regex engine displays only the first.

Answer 2

.*(foo.*?boo)

Try this. Grab the capture i.e $1 or \1.

See demo.

https://regex101.com/r/nL5yL3/9