Regex - replace all odd numbered occurrences of a comma

Question

If I have a string formatted like this:

"name", "bob", "number", 16, "place", "somewhere" 

And I want, instead, to have a string like this:

"name": "bob", "number": 16, "place": "somewhere" 

Also, the test cases do have some examples of strings like this:

"name", "bob", "hello, world", true

That would need to be formatted like this:

"name" : "bob", "hello, world" : true

...with every odd comma being replaced by a colon (so long as that comma falls outside of quotes), how on Earth would I do that via regex?

I've found the following regex via Google: /(,)(?=(?:[^"]|"[^"]*")*$)/,':' , which matches the first comma instance. How do I alternate every other one from there on out?

Edit for more info:

What I'm trying to do is take this string where each value is delineated by a comma and format it like a JS object using .replace(). So, in this case, "name", "number" and "place" represent key values. They're not fixed, this is simply an example.

Answer 1

Regex:

,(.*?(?:,|$))

Replacement string:

:$1

DEMO

Example:

> '"name", "bob", "number", 16, "place", "somewhere" '.replace(/,(.*?(?:,|$))/g, ':$1');
'"name": "bob", "number": 16, "place": "somewhere" '

Update:

If the field names are comma seperated then you could try the below regex,

> '"name", "bob", "hello, world", true'.replace(/("(?:\S+?|\S+ \S+)"), ("[^"]*"|\S+)/g, '$1: $2');
'"name": "bob", "hello, world": true'

Answer 2

You can go with this Regular Expression that covers all matches together.

(?=(?:[^"]*"[^"]*")*[^"]*$)(,)(.*?,|)(?=.*?(?:,|$))

Replacement is: :$2

Live demo