JavaScript REGEX: Is there a way to match slash after slash char in URL without negative lookbehind?

Question

I'm trying to match slash that is followed by slash in url that is not part part of the protocol or query string.

Is there any other way to do this using REGEX but without lookbehind as it is not supported in all browsers?

My examples:

const urls = `
https://asdf.com//asdf//asdf
http://asdf.com//asdf//asdf
ftp://asdf.com//asdf//asdf
//asdf.com//asdf//asdf
//asdf.com//asdf//asdf?test=//
z39.50s://asdf//
`.replace(/(?<!(^[\w\d-.]{2,}\:|^|\?.*))\/(?=\/)/gim, '');
console.log(urls);

Surprising that [Firefox is behind the curve here](https://stackoverflow.com/questions/50011366/javascript-regex-negative-lookbehind-not-working-in-firefox). — tadman, Nov 02 '18 at 20:09
What are you trying to match here? There's no double colon in your examples as far as I can tell. — tadman, Nov 02 '18 at 20:09
Try `.replace(/^(\S*?\/\/)|(\/)+/gm, "$1$2")`, see [demo](https://regex101.com/r/r72Sed/1). Usually, `https?:` is used instead of `\S*?`, and it might be safer if you can define the protocol pattern. — Wiktor Stribiżew, Nov 02 '18 at 20:11
@WiktorStribiżew Thanks! Nice solution, the only problem remains with the query string — rjanjic, Nov 02 '18 at 20:15
Then it should be `.replace(/^(\S*?\/\/)|(\?.*)$|(\/)+/g, '$1$2$3')`, see [this demo](https://regex101.com/r/r72Sed/2). — Wiktor Stribiżew, Nov 02 '18 at 20:16

score 3 · Accepted Answer · answered Nov 02 '18 at 20:20

You may use

.replace(/^(\S*?\/\/)|(\?.*)$|(\/)+/g, '$1$2$3')

See this regex demo

Details

^(\S*?\/\/) - Group 1 (later referred to with $1 from the replacement pattern): 0 or more non-whitespace chars, as few as possible, from the start of the string, up to the first //
| - or
(\?.*)$ - Group 2 ($2): a ? char and the rest of the string
| - or
(\/)+ - Group 3 ($3) capturing a single / char, 1 or more times (each captured / will overwrite the previous one in the group memory buffer since it is a "repeated capturing group")

score 1 · Answer 2 · edited Apr 09 '19 at 21:51

The normal workaround for the lookbehind deficit is to use a callback function
in the replace part.
The reason is that you have to match the errant part just to move the match position
past it. This requires logic in a callback function.

In %99.99 of the cases, you will have to do it this way if you have different replacements.

For the case you have it doesn't matter because you have a single replacement that is blank.
It gets masked by a combined group replacement where the stripping is controlled by
not having it in a group.

If you were to replace it with anything other than the empty string,
this is the only way to do it.

To that end, here is your (mostly) unaltered regex used with a callback.

     (                             # (1 start)
          (?: ^ [\w\d\-.]{2,} : | ^ | \? .* )
          //
     )                             # (1 end)
  |  /
     (?= / )

var urls = [
'https://asdf.com//asdf//asdf',
'http://asdf.com//asdf//asdf',
'ftp://asdf.com//asdf//asdf',
'//asdf.com//asdf//asdf',
'//asdf.com//asdf//asdf?test=//',
'z39.50s://asdf//' ];

for (var i = 0; i < urls.length; i++)
{
   urls[i] = urls[i].replace(
    /((?:^[\w\d\-.]{2,}:|^|\?.*)\/\/)|\/(?=\/)/gm,
    function(match, Grp1)
    {
      if ( Grp1 )
        return Grp1;
      return '';
    }
  );
}
console.log( urls );

JavaScript REGEX: Is there a way to match slash after slash char in URL without negative lookbehind?

2 Answers2