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Home/ Questions/Q 5950037
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Editorial Team
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Editorial Team
Asked: May 22, 20262026-05-22T17:20:12+00:00 2026-05-22T17:20:12+00:00

If there is one thing I just cant get my head around, it’s regex.

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If there is one thing I just cant get my head around, it’s regex.

So after a lot of searching I finally found this one that suits my needs:

function get_domain_name()
    { 
    aaaa="http://www.somesite.se/blah/sdgsdgsdgs";
    //aaaa="http://somesite.se/blah/sese";
        domain_name_parts = aaaa.match(/:\/\/(.[^/]+)/)[1].split('.');
        if(domain_name_parts.length >= 3){
            domain_name_parts[0] = '';
        }
        var domain = domain_name_parts.join('.');
        if(domain.indexOf('.') == 0)
            alert("1"+ domain.substr(1));
        else
            alert("2"+ domain);
    }

It basically gives me back the domain name, is there anyway I can also get all the stuff after the domain name? in this case it would be /blah/sdgsdgsdgs from the aaaa variable.

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-05-22T17:20:13+00:00Added an answer on May 22, 2026 at 5:20 pm

    Please note that this solution is not the best. I made this just to match the requirements of the OP. I personally would suggest looking into the other answers.

    THe following regexp will give you back the domain and the rest. :\/\/(.[^\/]+)(.*):

    1. http://www.google.com
    2. /goosomething

    I suggest you studying the RegExp documentation here: http://www.regular-expressions.info/reference.html

    Using your function:

    function get_domain_name()
        { 
        aaaa="http://www.somesite.se/blah/sdgsdgsdgs";
        //aaaa="http://somesite.se/blah/sese";
            var matches = aaaa.match(/:\/\/(?:www\.)?(.[^/]+)(.*)/);
            alert(matches[1]);
            alert(matches[2]);
        }
    
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