Javascript/RegEx: Split a string by commas but ignore commas within double-quotes

Question

I know similar questions are available but I could not find this case.

CASE 1: 'a,b,c,d,e'

OUTPUT: ["a", "b", "c", "d", "e"]

CASE 2: 'a,b,"c,d", e'

OUTPUT: ["a", "b", "c,d", "e"]

CASE 3: 'a,,"c,d", e'

OUTPUT: ["a", "", "c,d", "e"]

RegEx that I tried: (".*?"|[^",]+)(?=\s*,|\s*$)

RegEx Link: https://regex101.com/r/xImG4i/1

This regex works well with CASE1 and CASE2 But is failing for CASE3. Insead it works for

'a,,"c,d", e' , giving output as ["a", " ", "c,d", "e"]

which is also fine but need to work for CASE3 also.

Thanks in advance

Answer 1

You might take optional whitespace chars between 2 comma's if a lookbehind is supported.

"[^"]*"|[^\s,'"]+(?:\s+[^\s,'"]+)*|(?<=,)\s*(?=,)

Regex demo

 const regex = /"[^"]*"|[^\s,'"]+(?:\s+[^\s,'"]+)*|(?<=,)\s*(?=,)/g; [ `'a,b,c,d,e'`, `'a,b,"c,d", e'`, `'a,,"c,d", e'`, ` xz a,, b, c, "d, e, f", g, h`, `'a,,"c,d", e'`, ].forEach(s => console.log(s.match(regex)) )

If you don't want the double quotes you can use a capture group with matchAll and check for the group in the callback.

 const regex = /"([^"]*)"|[^\s,'"]+(?:\s+[^\s,'"]+)*|(?<=,)\s*(?=,)/g; [ `'a,b,c,d,e'`, `'a,b,"c,d", e'`, `'a,,"c,d", e'`, ` xz a,, b, c, "d, e, f", g, h`, `'a,,"c,d", e'`, ].forEach(s => console.log(Array.from(s.matchAll(regex), m => m[1]? m[1]: m[0])) )

Answer 2

An alternate solution that uses a regex for splitting instead of matching:

/,\s*(?=(?:(?:[^"]*"){2})*[^"]*$)/

This regex will split on comma followed by optional spaces if those are outside double quotes by using a lookahead to make sure there are even number of quotes after comma+space.

RegEx Demo

Code Sample:

 const re = /,\s*(?=(?:(?:[^"]*"){2})*[^"]*$)/; [ `a,b,"c,d", e`, `a,,"c,d", e`, ` xz a,, b, c, "d, e, f", g, h`, `a,,"c,d", e`, ].forEach(s => { tok = s.split(re); tok.forEach((e, i) => tok[i] = e.replace(/^"|"$/g, '')) console.log(s, '::', tok); })