[SOLVED] Using Parse_URL To Cut All of HTTP_REFERER Out Except "domain.com"

[phpQuestioner]

Ok I learn a couple of days ago how to use parse_url to remove a certin part of HTTP_REFERER.

 

Like So:

 

<?php

  $fowarded = $_SERVER['HTTP_REFERER'];
  $url = "$fowarded";
  $parts = parse_url($url);

// use as needed
// print "http://";

  print $parts["host"];

?> 

 

 

But now I want to be able to remove any sub-domain or sub-directory before actual domain.

 

Example

 

HTTP_REFERER Address: sub1.domain.com

 

What I Want To Display: domain.com (I want to strip off the "sub1" part).

 

 

 

Any one know how I can do this? Do I still use parse_url and if so, how?

[neel_basu]

$url_r = explode('.'.$url);
print_r($url_r);

I hope it will work

[phpQuestioner]

I tried this; but I did not get anything. So what am I doing wrong?

 

<?php

  $fowarded = $_SERVER['HTTP_REFERER'];

  $url = "$fowarded";

  $url_r = explode('.'.$url);

  print_r($url_r);

?>

 

And This.........

 

<?php

  $fowarded = $_SERVER['HTTP_REFERER'];
  $url = "$fowarded";
  $parts = parse_url($url);

  $url_r = $parts["host"];

  $url_r = explode('.'.$url);
  print_r($url_r);

?>

[neel_basu]

$url = 'www.sub.domain.com';
$url_r = explode('.'.$url);
print_r($url_r);

Try this.

[neel_basu]

OOPS there was a mistake

<?php
$url = 'www.sub.domain.com';
$url_r = explode('.',$url);//
print_r($url_r);
?>

Use this it would work

[jitesh]

OOPS there was a mistake

<?php
$url = 'www.sub.domain.com';
$url_r = explode('.',$url);//
print_r($url_r);
?>

Use this it would work

 

Yes

echo $url_r[1].".".$url_r[2];

[phpQuestioner]

That Will Work - Thanks For The Help