My first regex works (extracting "abc", " bed" and " ced"). I wanted to refine by excluding spaces, but adding \s meant only "abc" matched:
const test = "test(abc, bed, ced)";
const regex = /(?<=[\(,])[^,\)] /g
console.log(test.match(regex)); // ["abc", " bed", " ced"]
const regex2 = /(?<=[\(,])[^,\)\s] /g
console.log(test.match(regex2)); // ["abc"]https://regex101.com/r/IJavxn/2
CodePudding user response：
In the pattern (?<=[(,])[^,\)\s] the lookbehind assertion (?<=[(,]) checks that from the current position there is no ( or , directly to the left.
If the following character class [^,\)\s] also does not allow to match a space, then there will only be a match for the first occurrence as there is no space between the parenthesis and abc in (abc
As you are already using a lookbehind, you can exclude the whitespaces as well and use a quantifier. Then start the match also excluding a whitespace char.
The \s* in the lookbehind here (?<=[(,]\s*) allows zero of more spaces to be present.
(?<=[(,]\s*)[^\s,)]
(?<=Positive lookbehind, assert that from the current position what is to the left is[(,]\s*Match a single char other than(or,followed by optional whitespace chars
)Close the lookbehind[^\s,)]Match 1 characters other than a whitespace char or , or)