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Wednesday, January 21, 2009

How to get domain name from URL using javascript.

Use following javascript function to get domain name from URL.
function fnGetDomain(url) {
   return url.match(/:\/\/(.[^/]+)/)[1];
}

For ex: <script language="javascript">fnGetDomain("http://mail.google.com/mail/?source=navclient-ff");</script> will return "mail.google.com"
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20 comments:

  1. TibiNovember 4, 2009 at 3:28 PM

    very good function, thank you

    i added a little tweak to remove the 'www.' part

    function fnGetDomain(url) {
    return (url.match(/:\/\/(.[^/]+)/)[1]).replace('www.','');
    }

    ReplyDelete
  2. AnonymousApril 27, 2010 at 12:28 PM

    function getDomain(url)
    {
    return url.match(/:\/\/(www\.)?(.[^/:]+)/)[2];
    }

    ReplyDelete
    Replies
    1. AnonymousSeptember 27, 2012 at 1:36 PM

      Thank you. Great Code for getting domain.

      Delete
  3. Yogesh PitaleMay 27, 2010 at 12:20 PM

    Excellent function utility! Thanks for the help :)

    ReplyDelete
  4. liltosMay 3, 2011 at 8:41 AM

    Hey, the result isn't domain, it's host of a url.

    for url (http://mail.google.com/mail/?xxxxx)
    its host is: mail.google.com
    but the domain is: google.com

    and others:
    http://www.google.co.ck/#hl=en&source=hp&biw=1440&bih=703&q=hello&aq=f&aqi=g10&aql=&oq=&fp=3ab7f9268f5901b
    its host is: www.google.co.ck
    but the domain is: google.co.ck

    Black.Lee

    ReplyDelete
  5. AnonymousMay 12, 2011 at 4:31 PM

    i have to use a website as a input.so i have to cluster the webpages for pattern analysis, so have to convert a web url into a number.... can u tell me how to extract the webpage info from url and convert into integer in JAVASCRIPT.. mail me at aloksalian@yahoo.com

    ReplyDelete
  6. AnonymousSeptember 2, 2011 at 1:17 PM

    Thanx a lot.

    ReplyDelete
  7. TuloideNovember 27, 2011 at 1:13 AM

    This solution take into account also the optional digit after the "www" (e.g. http://www2.ohchr.org/english):

    function getDomain(url) {
    return url.match(/:\/\/(www[0-9]?\.)?(.[^/:]+)/)[2];
    }

    ReplyDelete
  8. AnonymousDecember 12, 2011 at 4:31 PM

    Thanks Nilesh,

    I am new to Javascript and I was looking for this. And found on your blog. Thanks again.

    ReplyDelete
  9. AnonymousJanuary 3, 2012 at 4:10 AM

    You should really update the original post. It's much better than having to read the comments.

    ReplyDelete
  10. AnonymousFebruary 22, 2012 at 11:37 AM

    But how to consider Black Lee's comment ?? Any help ?

    ReplyDelete
  11. AnonymousAugust 7, 2012 at 9:02 PM

    Thanks was very useful.

    ReplyDelete
  12. sailabSeptember 14, 2012 at 8:17 AM

    Guys i am trying to change this url listtop10.com to www.listtop10.com with replace method but there is no effect nor any error, can please someone help me.

    Thanks in advance

    ReplyDelete
  13. feralJanuary 16, 2013 at 9:55 PM

    just what I was after thanks

    ReplyDelete
  14. tv4designFebruary 5, 2013 at 3:05 AM

    Thanks for this post, as it was very helpful. I have a question in regards to parsing out everything after the "/" in the domain. Example if I have this domain http://www.google.com/patents/US20120211565. Is there a way to parse out the US20120211565 using a similar script like the one above?

    Thanks for you assistance in advanced

    ReplyDelete
  15. Tayou Fabrice LylianFebruary 8, 2013 at 3:51 AM

    I have added some errors control
    function getDomain(url) {
    var s = url.substr(0, 8), d;
    try{
    if(s.search('http://')<0 && s.search('https://')<0 && s.search('ftp://')<0){s="http://"+url;}else{s=url;}
    d = s.match(/:\/\/(www[0-9]?\.)?(.[^/:]+)/)[2];}
    catch(e){
    d = "" ;}
    return d ;
    }

    ReplyDelete
  16. AshMarch 10, 2013 at 6:24 PM

    thanks for you assistance in advanced

    ReplyDelete
  17. adamJanuary 15, 2014 at 10:52 PM

    thanks for your post and thank you tayou fabrice it was really work full for me

    ReplyDelete
  18. ashlyJanuary 16, 2014 at 11:58 PM

    very nice faction thanks

    ReplyDelete
  19. nandanyDecember 5, 2014 at 1:06 AM

    if(s.search('http://')<0 && s.search('https://')<0 && s.search('ftp://')<0){s="http://"+url;}else{s=url;}
    d = s.match(/:\/\/(www[0-9]?\.)?(.[^/:]+)/)[2];}
    catch(e){
    d = "" ;}
    return d ;
    }
    thanks for your post and thank you tayou fabrice it was really work full for me

    ReplyDelete

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