I have a question about regex. I don't know why I cannot do the following.
Sample sentence:
"This is a test string with five t's"
The regex I use:
^(.*?(?=t)){3}
I want the regex to match the following.
"This is a test s"
But it doesn't work, does anyone know why?
The point here is that the whole .*?(?=t) group pattern can match an empty string. It stops before the first t and cannot "hop thru" because it remains where it is when the lookahead pattern (a non-consuming pattern) matches.
You cannot do it like this, you must consume (and move the regex index) at least one char.
An alternative solution for this concrete case is
^(?:[^t]*t){2}[^t]*
See the regex demo, the ^(?:[^t]*t){2}[^t]* matches the start of string (^), then consumes two occurrences ({2}) of any chars other than t ([^t]*) followed with t, and then again consumes two occurrences ({2}) of any chars other than t.
Or, a general case solution (if t is a multicharacter string):
^(?:.*?t){2}(?:(?!t).)*
See another regex demo. The (?:.*?t){2} pattern matches two occurrences of any 0+ chars, as few as possible, up to the first t, and then (?:(?!t).)* matches any char, 0+ occurrences, that does not start a t char sequence.
As said by @CertainPerformance, .* will match zero or more characters in the pattern, but you use its lazy version .*?.
The lazy version of a quantifier will have it match as few characters as possible.
With a quantifier that matches the empty string, this will always lead to a zero-length match.
You need to use the + quantifier instead`, in order to prevent an empty string match.
Demonstration with Python:
>>> import re
>>> s = "This is a test string with five t's"
>>> r = r'^(.+?(?=t)){3}'
>>> re.match(r, s)
<_sre.SRE_Match object; span=(0, 16), match='This is a test s'>