How do I split a string by whitespace and ignoring leading and trailing whitespace into an array of words using a regular expression?

Question

I typically use the following code in JavaScript to split a string by whitespace.

"The quick brown fox jumps over the lazy dog.".split(/\s+/);
// ["The", "quick", "brown", "fox", "jumps", "over", "the", "lazy", "dog."]

This of course works even when there are multiple whitespace characters between words.

"The  quick brown fox     jumps over the lazy   dog.".split(/\s+/);
// ["The", "quick", "brown", "fox", "jumps", "over", "the", "lazy", "dog."]

The problem is when I have a string that has leading or trailing whitespace in which case the resulting array of strings will include an empty character at the beginning and/or end of the array.

"  The quick brown fox jumps over the lazy dog. ".split(/\s+/);
// ["", "The", "quick", "brown", "fox", "jumps", "over", "the", "lazy", "dog.", ""]

It's a trivial task to eliminate such empty characters, but I'd rather take care of this within the regular expression if that's at all possible. Does anybody know what regular expression I could use to accomplish this goal?

Horses for courses. `split` is used to *split* a string, not *mutate* it. See [how to trim a string in JavaScript?](http://stackoverflow.com/questions/498970/how-do-i-trim-a-string-in-javascript). — DCoder, Feb 16 '13 at 16:37
unfortunately javascript doesnt support **lookbehind** and even if you had used **lookbehind**,there would be space in the first split — Anirudha, Feb 16 '13 at 16:40
I never thought of it from that perspective. Thanks for pointing that out! — natlee75, Feb 17 '13 at 00:05

score 124 · Accepted Answer · edited Jun 14 '16 at 23:41

124

If you are more interested in the bits that are not whitespace, you can match the non-whitespace instead of splitting on whitespace.

"  The quick brown fox jumps over the lazy dog. ".match(/\S+/g);

Note that the following returns null:

"   ".match(/\S+/g)

So the best pattern to learn is:

str.match(/\S+/g) || []

edited Jun 14 '16 at 23:41

Andy

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answered Feb 16 '13 at 16:37

kennebec

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Watch out, `' '.match(/\S+/g)` returns `null` instead of `[]`! – Andy Jun 14 '16 at 21:38
I wonder whose solution is faster yours or @Josh 's (if we handle the [""] case). – ibodi Jun 09 '19 at 17:21

score 51 · Answer 2 · answered Feb 16 '13 at 16:36

51

" The quick brown fox jumps over the lazy dog. ".trim().split(/\s+/);

answered Feb 16 '13 at 16:36

Josh

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Thanks for the suggestion. I actually was going to go this route until I remembered that it requires a browser that supports JavaScript 1.8. That's fine for a majority of our users, but we still support older browsers such as Internet Explorer 7 and 8 whose JavaScript engines don't include this functionality. – natlee75 Feb 17 '13 at 00:09
11

Watch out, `' '.trim().split(/\s+/)` returns `[""]`! – Andy Jun 14 '16 at 21:39
I wonder whose solution is faster yours (if we handle the [""] case) or @kennebec 's – ibodi Jun 09 '19 at 17:22

score 16 · Answer 3 · answered Feb 16 '13 at 16:37

16

Instead of splitting at whitespace sequences, you could match any non-whitespace sequences:

"  The quick brown fox jumps over the lazy dog. ".match(/\S+/g)

answered Feb 16 '13 at 16:37

Gumbo

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aris · Answer 4 · 2020-04-18T11:41:04.523

Not elegant as others code but very easy to understand:

    countWords(valOf)
    {
        newArr[];
        let str = valOf;
        let arr = str.split(" ");

        for (let index = 0; index < arr.length; index++) 
       {
           const element = arr[index];
           if(element)
           {
              newArr.push(element);
           }
       }
       const NumberOfWords = newArr.length;

       return NumberOfWords;
   }

How do I split a string by whitespace and ignoring leading and trailing whitespace into an array of words using a regular expression?

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