URL extraction from string

Question

I found a regular expression that is suppose to capture URLs but it doesn't capture some URLs.

$("#links").change(function() {

    //var matches = new array();
    var linksStr = $("#links").val();
    var pattern = new RegExp("^(https?:\/\/)?([\da-z\.-]+)\.([a-z\.]{2,6})([\/\w \.-]*)*\/?$","g");
    var matches = linksStr.match(pattern);

    for(var i = 0; i < matches.length; i++) {
      alert(matches[i]);
    }

})

It doesn't capture this url (I need it to):

http://www.wupload.com/file/63075291/LlMlTL355-EN6-SU8S.rar

But it captures this

http://www.wupload.com

it does capture that one :P alert("http://www.wupload.com/file/63075291/LlMlTL355-EN6-SU8S.rar".match(/^(https?:\/\/)?([\da-z\.-]+)\.([a-z\.]{2,6})([\/\w \.-]*)*\/?$/g)) — Joseph Marikle
– Joseph Marikle, Commented Aug 8, 2011 at 16:45
that's odd... idk. i'm using Chrome and it worked in Chrome's js console. — Joseph Marikle
– Joseph Marikle, Commented Aug 8, 2011 at 17:41

Brock Adams · Accepted Answer · 2011-08-09 08:28:16Z

1

Several things:

The main reason it didn't work, is when passing strings to RegExp(), you need to slashify the slashes. So this:

"^(https?:\/\/)?([\da-z\.-]+)\.([a-z\.]{2,6})([\/\w \.-]*)*\/?$"

Should be:

"^(https?:\/\/)?([\\da-z\\.-]+)\\.([a-z\\.]{2,6})([\/\\w \\.-]*)*\/?$"

Next, you said that FF reported, "Regular expression too complex". This suggests that linksStr is several lines of URL candidates.
Therefore, you also need to pass the m flag to RegExp().
The existing regex is blocking legitimate values, eg: "HTTP://STACKOVERFLOW.COM". So, also use the i flag with RegExp().
Whitespace always creeps in, especially in multiline values. Use a leading \s* and $.trim() to deal with it.
Relative links, eg /file/63075291/LlMlTL355-EN6-SU8S.rar are not allowed?

Putting it all together (except for item 5), it becomes:

var linksStr    = "http://www.wupload.com/file/63075291/LlMlTL355-EN6-SU8S.rar  \n"
                + "  http://XXXupload.co.uk/fun.exe \n "
                + " WWW.Yupload.mil ";
var pattern     = new RegExp (
                    "^\\s*(https?:\/\/)?([\\da-z\\.-]+)\\.([a-z\\.]{2,6})([\/\\w \\.-]*)*\/?$"
                    , "img"
                );

var matches     = linksStr.match(pattern);
for (var J = 0, L = matches.length;  J < L;  J++) {
    console.log ( $.trim (matches[J]) );
}

Which yields:

http://www.wupload.com/file/63075291/LlMlTL355-EN6-SU8S.rar
http://XXXupload.co.uk/fun.exe
WWW.Yupload.mil

answered Aug 9, 2011 at 8:28

Brock Adams

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1 Comment

el_pup_le Over a year ago

Thanks, I've used some of uber leet Jeff Atwood's regex - \bhttp://[^\s]+ for now

The Mask · Accepted Answer · 2011-08-08 16:47:59Z

0

Why not do make: URLS = str.match(/https?:[^\s]+/ig);

answered Aug 8, 2011 at 16:47

The Mask

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1 Comment

bcoughlan Over a year ago

If the URL is in a user-supplied text-box and is followed by a comma or full stop (as users often do!), those punctuation marks will be treated as part of the URL

Senad Meškin · Accepted Answer · 2011-08-08 16:48:31Z

0

(https?\:\/\/)([a-z\/\.0-9A-Z_-\%\&\=]*)

this will locate any url in text

answered Aug 8, 2011 at 16:48

Senad Meškin

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