How to split string while ignoring portion in parentheses?

Question

I have a string that I want to split into an array by using the commas as delimiters. I do not want the portions of the string that is between parentheses to be split thoug

Answer 1

var regex = /,(?![^(]*\)) /;
var str = "bibendum, morbi, non, quam (nec, dui, luctus), rutrum, nulla"; 

var splitString = str.split(regex);

Here you go. An explanation of the regex:

,     //Match a comma
(?!   //Negative look-ahead. We want to match a comma NOT followed by...
[^(]* //Any number of characters NOT '(', zero or more times
/)    //Followed by the ')' character
)     //Close the lookahead.

Answer 2

I don't like to have large opaque regular expressions in my code so I used a different solution. It still makes use of regex but I think is somewhat more transparent.

I use a simpler regex to replace any commas within parenthesis with a special string. I then split the string by commas, then within each resulting token I replace the special string with a comma.

    splitIgnoreParens(str: string): string[]{
    const SPECIAL_STRING = '$REPLACE_ME';

    // Replaces a comma with the special string
    const replaceComma = s => s.replace(',',SPECIAL_STRING);
    // Vice versa
    const replaceSpecialString = s => s.replace(SPECIAL_STRING,',');

    // Selects any text within parenthesis
    const parenthesisRegex = /\(.*\)/gi;

    // Withing all parenthesis, replace comma with special string.
    const cleanStr = str.replace(parenthesisRegex, replaceComma);
    const tokens = cleanStr.split(',');
    const cleanTokens = tokens.map(replaceSpecialString);

    return cleanTokens;
}

Answer 3

var start = "bibendum, morbi, non, quam (nec, dui, luctus), rutrum, nulla";
start = start.replace(/ /g,'');
console.log(start);

var front = start.substring(0,start.lastIndexOf('(')).split(',');
var middle = '('+start.substring(start.lastIndexOf('(')+1,start.lastIndexOf(')'))+')';
var end = start.substring(start.lastIndexOf(')')+2,start.length).split(',');
console.log(front)
console.log(middle)
console.log(end)
return front.concat(middle,end);

Answer 4

You don't need fancy regular expressions for this.

s="bibendum, morbi, non, quam (nec, dui, luctus), rutrum, nulla" 
var current='';
var parenthesis=0;
for(var i=0, l=s.length; i<l; i++){ 
  if(s[i] == '('){ 
    parenthesis++; 
    current=current+'(';
  }else if(s[i]==')' && parenthesis > 0){ 
    parenthesis--;
    current=current+')';
  }else if(s[i] ===',' && parenthesis == 0){
    console.log(current);current=''
  }else{
    current=current+s[i];
  }   
}
if(current !== ''){
  console.log(current);
}

Change console.log for an array concatenation or what ever you want.

Answer 5

Instead of focusing on what you do not want it's often easier to express as a regular expression what you want, and to match that with a global regex:

var str = "bibendum, morbi, non, quam (nec, dui, luctus), rutrum, nulla";
str.match(/[^,]+(?:\(+*?\))?/g) // the simple one
str.match(/[^,\s]+(?:\s+\([^)]*\))?/g) // not matching whitespaces