Find word before opening parenthesis
I have the following line of code:
let regexp: RegExp = new RegExp('(?=\\s(.+))');
What I am trying to do is find the word that comes before a
(With what I have, it matches more than what I would like.
this is a test (
Matches :
is a test (testfor(
No matches were found.
So, how can I get the word that comes before the parenthesis?
The (?=\\s(.+)) regex matches an empty location (because the whole pattern presents a positive lookahead, a zero-width assertion consuming no characters) that is followed with 1 whitespace and 1+ any characters other than a newline (that are captured into Group 1). Thus, it may match more than one word.
You can use
\S+(?=\s+\()
or
\w+(?=\s*\()
See the regex demo
Pattern explanation:
\w+- 1 or more word chars (or 1+ non-whitespace chars if\S+is used)....(?=\s*\()- before 0+ whitespaces (\s*) followed with a literal(char (or before 1+ whitespaces followed with a(if(?=\s+\()is used).
var re = /\w+(?=\s*\()/;
var str = 'this is a test (';
var res = (m = str.match(re)) ? m : "";
console.log(res[0]);Alternative: Capturing Groups
You can achieve the same without any lookarounds. I like lookaheads since they are not that costly in most cases and the match structure is cleaner (you just have a single group equal to the whole match), but in this scenario, a mere capturing mechanism can be leveraged with
/(\w+)\s*\(/
where the value we need is captured with (\w+), a parenthesized part of the pattern. See the regex demo
var re = /(\w+)\s*\(/;
var str = 'this is a test (';
var res = (m = str.match(re)) ? m[1] : "";
console.log(res);