regex expression for string.split() that splits a string on urls

Question

I have a regex expression for finding url's in a string, but when I use it with String.prototype.split() it returns undefined's.

const regex = /(http(s)?:\/\/.)?(www\.)?[-a-zA-Z0-9@:%._\+~#=]{2,256}\.[a-z]{2,6}\b([-a-zA-Z0-9@:%_\+.~#?&//=]*)/g;
const str = "yay http://www.google.com woo www.google.com haha google.com";

console.log(str.match(regex));
// [ 'http://www.google.com', 'www.google.com', 'google.com' ]

console.log(str.split(regex));
// [ 'yay ','http://w',undefined,undefined,'',' woo ',undefined,undefined,'www.','',' haha ',undefined,undefined,undefined,'','' ]

After some research it appears that this has to do with capturing groups. I attempted adding :? to all the capturing groups (parts wrapped in parenthesis) and it removed the undefined's.

const reg2 = /(?:http(?:s)?:\/\/.)?(?:www\.)?[-a-zA-Z0-9@:%._\+~#=]{2,256}\.[a-z]{2,6}\b(?:[-a-zA-Z0-9@:%_\+.~#?&//=]*)/g

const str = "yay http://www.google.com woo www.google.com haha google.com";

console.log(str.split(reg2));
// [ 'yay ', ' woo ', ' haha ', '' ]

But it is omitting the urls. I am looking to return:

[ 'yay ', 'http://www.google.com', ' woo ', 'www.google.com', ' haha ', 'google.com' ]

Tim Biegeleisen · Accepted Answer · 2020-02-20 01:42:46Z

2

You might be able to just split on whitespace here:

var str = "yay http://www.google.com woo www.google.com haha google.com";
var parts = str.split(/\s+/);
console.log(parts);

If the leading/trailing whitespace are really required here, then try matching on the pattern:

<URL>|\s*\S+\s*

This would match either a URL, or a series of non word characters, with surrounding whitespace. Consider:

var str = "yay http://www.google.com woo www.google.com haha google.com";
console.log(str.match(/(?:http(s)?:\/\/.)?(?:www\.)?[-a-zA-Z0-9@:%._\+~#=]{2,256}\.[a-z]{2,6}\b(?:[-a-zA-Z0-9@:%_\+.~#?&//=]*)|\s*\S+\s*/g));

This uses an alternation trick to first try to selectively consume a URL, with no surrounding whitespace. That failing, the fallback pattern is \s*\S\s*, i.e. any other word with the leading/trailing whitespace.

edited Feb 20, 2020 at 1:42

answered Feb 20, 2020 at 1:37

Tim Biegeleisen

526k32 gold badges323 silver badges399 bronze badges

Sign up to request clarification or add additional context in comments.

Collectives™ on Stack Overflow

regex expression for string.split() that splits a string on urls

1 Answer 1

Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

1 Answer 1

Comments

Your Answer

Sign up or log in

Post as a guest

Related