How do I remove all whitespaces except those between words with Regular Expression in Javascript

Question

  var str=" hello world! , this is Cecy ";
  var r1=/^\s|\s$/g;
  console.log(str.match(r1));
  console.log(str.replace(r1,''))

Here, the output I expect is "hello world!,this is Cecy", which means remove whitespaces in the beginning and the end of the string, and the whitespace before and after the non-word characters. The output I have right now is"hello world! , this is Cecy", I don't know who to remove the whitespace before and after "," while keep whitespace between "o" and "w" (and between other word characters).

p.s. I feel I can use group here, but don't know who

Answer 1

Method 1

See regex in use here

\B\s+|\s+\B

• \B Matches a location where \b doesn't match
• \s+ Matches one or more whitespace characters

const r = /\B\s+|\s+\B/g
const s = ` hello world! , this is Cecy `

console.log(s.replace(r, ''))

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Method 2

See regex in use here.

(?!\b\s+\b)\s+

• (?!\b +\b) Negative lookahead ensuring what follows doesn't match
  • \b Assert position as a word boundary
  • \s+ Match any whitespace character one or more times
  • \b Assert position as a word boundary
• \s+ Match any whitespace character one or more times

const r = /(?!\b\s+\b)\s+/g
const s = ` hello world! , this is Cecy `

console.log(s.replace(r, ''))

Answer 2

You can use the RegEx ^\s|\s$|(?<=\B)\s|\s(?=\B)

• ^\s handles the case of a space at the beginning

• \s$ handles the case of a space at the end

• (?<=\B)\s handles the case of a space after a non-word char

• \s(?=\B) handles the case of a space before a non-word char

Demo.

EDIT : As ctwheels pointed out, \b is a zero-length assertion, therefore you don't need any lookbehind nor lookahead.

Here is a shorter, simpler version : ^\s|\s$|\B\s|\s\B

var str = " hello world! , this is Cecy ";
console.log(str.replace(/^\s|\s$|\B\s|\s\B/g, ''));