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1

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>
share|improve this question
What do you call a "valid URL"? Is it a validly formatted string, or a URL that actually links to a resource? – RobG May 31 '11 at 4:32
hi i want google.com or wwww.google.com but not as www.google – Roshan Wijesena May 31 '11 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 May 31 '11 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 May 31 '11 at 5:14
1  
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 May 31 '11 at 5:15

4 Answers

up vote 0 down vote accepted 
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);
}
share|improve this answer
2  
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). – RobG May 31 '11 at 4:30
thank you but there is a problem when i type the www.google it also getting as valid url how could i avoid this – Roshan Wijesena May 31 '11 at 4:31
1  
jsfiddle.net/Zqg7M/1 , check this fiddle – PsyCoder May 31 '11 at 4:37
1  
OHK> LEMME CHECK. – PsyCoder May 31 '11 at 4:47
1  
whoohooo... i guess this question should be raised as a community wiki... because there can be so many possiblities of a url. – PsyCoder May 31 '11 at 5:23
show 13 more comments
up vote 4 down vote 
<?php
$url = "http://www.example.com";

if(!filter_var($url, FILTER_VALIDATE_URL))
  {
  echo "URL is not valid";
  }
else
  {
  echo "URL is valid";
  }
?>
share|improve this answer
up vote 1 down vote

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.

share|improve this answer
Or something like: ntfs://[email protected]:/what/the/home.dtb?thing=whatever&species=goose#beak – RobG May 31 '11 at 5:20
up vote 0 down vote 
<?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

share|improve this answer

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