I have the following code that splits a combination of names on either the word and or the & ampersand:

    var name = 'Ron & Peggy Sue'; //or 'Ron and Peggy Sue';

    if ( name.indexOf('&') > -1 ){
        names = name.split(/ & /g);

    if ( name.toLowerCase().indexOf(' and ') > -1 ) {
        names = name.split(/ and /gi);

Its returns an array of names like:

names = ['Ron', 'Peggy Sue']; //regardless of `&` or `and` separator

I was trying to combine these two conditions into one, but the results were not what I wanted. I know there must be a better way. How can the above code be improved?


1 Answer 1


You were almost there. You can use pipe (a|b) to designate the regex to look for both a and b. With that knowledge your split is as easy as:

var name = 'Ron & Peggy Sue and Darin Douglass';
names = name.split(/\s(?:and|&)\s/ig);

EDIT: The syntax (?:...) is whats called non-capturing parens. These match the regex provided as '...' but it does not remember the matches. The problem with the simple /\s(and|&)\s/ig/ was that the words and and & would be matched. This new syntax 'matches' them but does not remember (i.e. report) them. Thanks to @cbojar for the pointer.

Also, there is no need to check for either delimiter in name. If none exist it will simply return a single-element array with name as its element.

  • \$\begingroup\$ Totally nit-picking, but rather than repeat the spaces surrounding the two words, I'd say / (and|&) /ig - or /\b(and|&)\b/ig \$\endgroup\$
    – Flambino
    May 27, 2014 at 13:43
  • \$\begingroup\$ Agreed. Thats a better solution in any case :D \$\endgroup\$ May 27, 2014 at 13:45
  • \$\begingroup\$ The problem with using (and|&) is that it is captured: ["Ron & Peggy Sue ", "and", " Darin Douglass"] \$\endgroup\$
    – chovy
    May 27, 2014 at 17:27
  • 1
    \$\begingroup\$ You could try with non-capturing parens (?:and|&) and the whitespace delimiter \w. \$\endgroup\$
    – cbojar
    May 27, 2014 at 18:30
  • 1
    \$\begingroup\$ \w is a word delimiter, not a whitespace delimiter, afaik. \$\endgroup\$
    – chovy
    May 29, 2014 at 3:51

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