I am trying to capture a string that can contain any character but must always be followed by ';' I want to capture it and trim the white space around it. I've tried using positive lookahead but that does not seem to exclude the whitespace. Example:
this is a match ;
this is not a match
regex:
.+(?=\s*;)
result:
"this is a match " gets captured with trailing white space behind.
expected result:
"this is a match" (without whitespace)
You have to make sure the first and the last characters of your match are not spaces. Thus we use the non-whitespace character match (\S) before and after the all character match (.*). As spaces might be optional, the any character match (.) must be optional, thus we use * instead of +.
\S.*\S(?=\s*;)
If the string can start with space use .*\S(?=\s*;).
Thanks to @CarySwoveland for improving the answer.
You can match
.*(?<!\s)(?=\s*;)
provided the regex engine supports negative lookbehinds.
Note that this returns an empty string if the string is " ;".
You can make the dot non greedy and start the match with a non whitespace character:
\S.*?(?=\s*;)
If the non whitespace character itself should also not be a semicolon:
[^\s;].*?(?=\s*;)