Regex: Remove quotes from within starting and ending quotes - javascript

Question

I want to remove the single quote below after 10 and replace with empty string.

So 100,50,"bla 10" bla",20,"another string"

becomes

100,50,"bla 10 bla",20,"another string"

I was so close but the following is selecting 0"

/[^"]"[^$"]/g

Any thoughts? Thank you!

Answer 1

You may use

 var s = '100,,50,"bla 10" bla",20,"another string","string"'; var splits = s.split(/(?:^|,)(".*?")(?:,|$)|,/); splits = splits.filter(function (x) { return x !== undefined;}); console.log(splits); var res = splits.map(function(x) { return x.replace(/^,*(")|("),*$|"/g, '$1$2'); }).join(','); console.log(res);

Details

• (?:^|,)" - match , or start of string
• ( - start of the capturing group
• " - a "
• .*? - any 0+ chars, as few as possible
• " - a "
• ) - end of the capturing group
• (?:,|$) - a , or end of string
• | - or
• , - a comma

After we get all the matches, the .replace(/^,*(")|("),*$|"/g, '$1$2' removes all " but at the start/end of the matches. Its details:

• ^,*(")| - matches , 1 or more times at the start of the string ( ^ ) and then captures a " into Group 1, or
• ("),*$| - captures a " at the end of the string ( $ ) and then 0+ , s into Group 2 or
• " - matches any other " .

The $1$2 replacement pattern "restores" the " s captured into Group 1 and 2 (ie keeps the " s that are either at the start of the string with an optional , in front or end of the string and removes all others).