URL validation javascript jquery

Question

I want to validate url and its need to be allow to with and with out http://

that means if user type as http://www.google.com or www.google.com should allow to the user ..

i tried jquery validations class to do this.. no luck

i tried regx as follow its also not working as i want . any help very appreciated

<script>

var myVariable = "http://www.google.com/";
if(/^([a-z]([a-z]|\d|\+|-|\.)*):(\/\/(((([a-z]|\d|-|\.|_|~|[\u00A0-\uD7FF\uF900-\uFDCF\uFDF0-\uFFEF])|(%[\da-f]{2})|[!\$&'\(\)\*\+,;=]|:)*@)?((\[(|(v[\da-f]{1,}\.(([a-z]|\d|-|\.|_|~)|[!\$&'\(\)\*\+,;=]|:)+))\])|((\d|[1-9]\d|1\d\d|2[0-4]\d|25[0-5])\.(\d|[1-9]\d|1\d\d|2[0-4]\d|25[0-5])\.(\d|[1-9]\d|1\d\d|2[0-4]\d|25[0-5])\.(\d|[1-9]\d|1\d\d|2[0-4]\d|25[0-5]))|(([a-z]|\d|-|\.|_|~|[\u00A0-\uD7FF\uF900-\uFDCF\uFDF0-\uFFEF])|(%[\da-f]{2})|[!\$&'\(\)\*\+,;=])*)(:\d*)?)(\/(([a-z]|\d|-|\.|_|~|[\u00A0-\uD7FF\uF900-\uFDCF\uFDF0-\uFFEF])|(%[\da-f]{2})|[!\$&'\(\)\*\+,;=]|:|@)*)*|(\/((([a-z]|\d|-|\.|_|~|[\u00A0-\uD7FF\uF900-\uFDCF\uFDF0-\uFFEF])|(%[\da-f]{2})|[!\$&'\(\)\*\+,;=]|:|@)+(\/(([a-z]|\d|-|\.|_|~|[\u00A0-\uD7FF\uF900-\uFDCF\uFDF0-\uFFEF])|(%[\da-f]{2})|[!\$&'\(\)\*\+,;=]|:|@)*)*)?)|((([a-z]|\d|-|\.|_|~|[\u00A0-\uD7FF\uF900-\uFDCF\uFDF0-\uFFEF])|(%[\da-f]{2})|[!\$&'\(\)\*\+,;=]|:|@)+(\/(([a-z]|\d|-|\.|_|~|[\u00A0-\uD7FF\uF900-\uFDCF\uFDF0-\uFFEF])|(%[\da-f]{2})|[!\$&'\(\)\*\+,;=]|:|@)*)*)|((([a-z]|\d|-|\.|_|~|[\u00A0-\uD7FF\uF900-\uFDCF\uFDF0-\uFFEF])|(%[\da-f]{2})|[!\$&'\(\)\*\+,;=]|:|@)){0})(\?((([a-z]|\d|-|\.|_|~|[\u00A0-\uD7FF\uF900-\uFDCF\uFDF0-\uFFEF])|(%[\da-f]{2})|[!\$&'\(\)\*\+,;=]|:|@)|[\uE000-\uF8FF]|\/|\?)*)?(\#((([a-z]|\d|-|\.|_|~|[\u00A0-\uD7FF\uF900-\uFDCF\uFDF0-\uFFEF])|(%[\da-f]{2})|[!\$&'\(\)\*\+,;=]|:|@)|\/|\?)*)?$/i.test('www.google.com/')) {
  alert("valid url");
} else {
  alert("invalid url");
}
</script>

What do you call a "valid URL"? Is it a validly formatted string, or a URL that actually links to a resource? — RobG
– RobG, Commented May 31, 2011 at 4:32
hi i want google.com or wwww.google.com but not as www.google — Roshan Wijesena
– Roshan Wijesena, Commented May 31, 2011 at 4:34
The scheme-matching part seems incorrect - the way I read RFC 3986, a scheme must have at least 2 characters. RegExp seems to allow just one. A better RegExp might be: ^[a-zA-Z][a-zA-Z0-9\+\.-]+: — RobG
– RobG, Commented May 31, 2011 at 4:59
I'm curious if you're just focused on google. because if not, then you will have a problem with sites like youtu.be, fb.me, goo.gl which are valid... — Reigel Gallarde
– Reigel Gallarde, Commented May 31, 2011 at 5:14
www.google is a validly formated URI (assuming a scheme like HTTP and :// is prepended), the only way you can know it's not a real URI is to test it against a list of all valid URIs (a very long list) or send a request, presumbaly HTTP. That would most likely need to come from the server. So send the URI to the server and have it tested there. — RobG
– RobG, Commented May 31, 2011 at 5:15

user557846 · Accepted Answer · 2011-05-31 04:23:40Z

4

<?php
$url = "http://www.example.com";

if(!filter_var($url, FILTER_VALIDATE_URL))
  {
  echo "URL is not valid";
  }
else
  {
  echo "URL is valid";
  }
?>

answered May 31, 2011 at 4:23

user557846

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Comments

Brett Zamir · Accepted Answer · 2011-05-31 04:24:45Z

1

How about trying something like http://phpjs.org/functions/parse_url:485 which is based on Steve Levithan's well-tested, standards-aware parseUri function.

answered May 31, 2011 at 4:24

Brett Zamir

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1 Comment

RobG Over a year ago

Or something like: ntfs://[email protected]:/what/the/home.dtb?thing=whatever&species=goose#beak

Sujit Agarwal · Accepted Answer · 2011-05-31 04:39:17Z

0

function checkURL(value) {
  var urlregex = new RegExp(
        "^((http|https|ftp)\://)*([a-zA-Z0-9\.\-]+(\:[a-zA-Z0-9\.&amp;%\$\-]+)*@)*((25[0-5]|2[0-4][0-9]|[0-1]{1}[0-9]{2}|[1-9]{1}[0-9]{1}|[1-9])\.(25[0-5]|2[0-4][0-9]|[0-1]{1}[0-9]{2}|[1-9]{1}[0-9]{1}|[1-9]|0)\.(25[0-5]|2[0-4][0-9]|[0-1]{1}[0-9]{2}|[1-9]{1}[0-9]{1}|[1-9]|0)\.(25[0-5]|2[0-4][0-9]|[0-1]{1}[0-9]{2}|[1-9]{1}[0-9]{1}|[0-9])|([a-zA-Z0-9\-]+\.)*[a-zA-Z0-9\-]+\.(com|edu|gov|int|mil|net|org|biz|arpa|info|name|pro|aero|coop|museum|[a-zA-Z]{2}))(\:[0-9]+)*(/($|[a-zA-Z0-9\.\,\?\'\\\+&amp;%\$#\=~_\-]+))*$");
  if(urlregex.test(value))
  {
    return(true);
  }
  return(false);
}

edited May 31, 2011 at 4:39

answered May 31, 2011 at 4:23

Sujit Agarwal

12.5k11 gold badges50 silver badges79 bronze badges

13 Comments

RobG Over a year ago

There are many schemes other than HTTP, HTTP and FTP (e.g. file). Host names don't have to start with www and URLs can include a port number (e.g. :80).

Roshan Wijesena Over a year ago

thank you but there is a problem when i type the www.google it also getting as valid url how could i avoid this

Sujit Agarwal Over a year ago

I have changed the regex. Now please check.

Sujit Agarwal Over a year ago

jsfiddle.net/Zqg7M/2 , i again updated the answer and fiddle. CHeck them for www.google.com

Sujit Agarwal Over a year ago

whoohooo... i guess this question should be raised as a community wiki... because there can be so many possiblities of a url.

|

Lito · Accepted Answer · 2011-06-23 10:41:43Z

0

<?php
// PHP 5.3.5-1ubuntu7.2
$url = "http://www.example.com:80i/"; // Yes, I have an "i" after port 80, then is not a valid URL

if (filter_var($url, FILTER_VALIDATE_URL)) {
  echo "URL is valid";
} else {
  echo "URL is NOT valid";
}

$url = parse_url($url);

echo $url['port']; // Returns 80 (without the "i")

// BUG...
?>

Using filter_var with FILTER_VALIDATE_URL doesn't validate correctly the URL

answered Jun 23, 2011 at 10:41

Lito

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