regex for ignoring character if inside () parenthesis?

December 30, 2022

I was doing some regex, but I get this bug:

I have this string for example "+1/(1/10)+(1/30)+1/50" and I used this regex /\+.[^\+]*/g

and it working fine since it gives me ['+1/(1/10)', '+(1/30)', '+1/50']

BUT the real problem is when the + is inside the parenthesis ()

like this: "+1/(1+10)+(1/30)+1/50"

because it will give ['+1/(1', '+10)', '+(1/30)', '+1/50']

which isn’t what I want :(… the thing I want is ['+1/(1+10)', '+(1/30)', '+1/50']

so the regex if it see \(.*\) skip it like it wasn’t there…

how to ignore in regex?

my code (js):

const tests = {
      correct: "1/(1/10)+(1/30)+1/50",
      wrong  : "1/(1+10)+(1/30)+1/50"
}

function getAdditionArray(string) {
      const REGEX = /\+.[^\+]*/g; // change this to ignore the () even if they have the + sign
      const firstChar = string[0];

      if (firstChar !== "-") string = "+" + string;

      return string.match(REGEX);
}

console.log(
    getAdditionArray(test.correct),
    getAdditionArray(test.wrong),
)

>Solution :

You can exclude matching parenthesis, and then optionally match (...)

\+[^+()]*(?:\([^()]*\))?

The pattern matches:

\+ Match a +
[^+()]* Match optional chars other than + ( )
(?: Non capture group to match as a whole part
- \([^()]*\) Match from (...)
)? Close the non capture group and make it optional