I would need help with my code:
I have a function which only replaces thee www. with a blank space.
For example:
If I add the url: www.testek.com
The user will see testek.com
But if I add the url: s.dada.testek.com
The user will see s.dada.testek.com
So if we use the domain s.dada.testek.com I would like that the end user sees only testek.com.
But I would like to get only the main domain without any subdomains.
Code:
function getdomain($url){
$parsed = parse_url($url);
return str_replace('www.','', strtolower($parsed['host']));
}
I saw a post but it won't work for me.
Thanks for the help!
Now I've changed the code to:
function getdomain($url){
$parsed = parse_url($url);
$bits = explode(".",$parsed["host"]);
$mainDomain = array_filter($bits, function ($i) use ($bits) {
return $i >= count($bits)-2;
}, array(
'www.rover.ebay.com' => 'ebay.com',
's.click.aliexpress.com' => 'aliexpress.com', );
return implode(".", $mainDomain);
}
Am I thinking the right way?
Because now the end user sees like this:
http://i.stack.imgur.com/JddKB.jpg
If you simply want to get the last 2 segments of a URL main domain name then you can do the following:
function getdomain($url){
$parsed = parse_url($url);
$bits = explode(".",$parsed["host"]);
$mainDomain = array_filter($bits, function ($i) use ($bits) {
return $i >= count($bits)-2;
}, ARRAY_FILTER_USE_KEY );
return implode(".", $mainDomain);
}
See how it works in https://eval.in/636860
Unfortunately most of the times there's no "catch all" solution and you have to do a lot of hard-coded things. e.g. the UK has .co.uk but France just .fr so depending on that you may need the last 3 or even 4 segments.
I've fixed it like this:
function getdomain($url){
$parsed = parse_url($url);
$replace = array ("rover.", "www.", "s.click.");
return str_replace($replace,'', strtolower($parsed['host']));
}
I've created an array with the "subdomains" which I don't want to be shown.
And now it works ok.
apokryfos thanks for your support and for opening my mind :)
Related
How can I take out the part of the url after the view name
Example:
The URL:
http://localhost/winner/container.php?fun=page&view=eims
Extraction
eims
This is called a GET parameter. You can get it by using
<?php
$view = $_GET['view'];
If this is for a URL which is not part of your website (e.g. Not your domain), but you wish to parse it. Something like this will work
$url = "http://example.com/index.php?foo=bar&acme=baz&view=asdf";
$params = explode('?', $url)[1]; // This gets the text AFTER the ? Note: If using PHP 5.3 or less, this may not work. You would then need to split it into two lines with the [1] happening on $params.
$pairs = explode('&', $params);
foreach($pairs as $p => $pair) {
list($keys[$p], $values[$p]) = explode('=', $pair);
$splits[$keys[$p]] = $values[$p];
}
echo $splits['view'];
<?php echo $_GET['view']; ?> //eims
If that's current URL, the simple and rock solid approach is to use the filter functions:
filter_input(INPUT_GET, 'view')
Otherwise, you can use parse_url() with PHP_URL_QUERY as second argument. The resulting string can be split with e.g. parse_str().
Are sure you are writing this "echo $_GET['view'];" in the container.php file?
Maybe write why do you need that "view".
I have difficulties in remove some slug from the url.
I have some urls like these below
http://domain1.com/so-section/upload/image1.jpg
http://domain2example.com/so-section/upload/image2.jpg
http://domain3place.com/so-section/upload/image3.jpg
http://domain4code.com/so-section/upload/image4.jpg
http://domain5action.com/so-section/upload/image5.jpg
http://domain6rack.com/so-section/upload/image5.jpg
Obviously, you will see domain are unsame, So I only would like to get "/so-section/upload/imagename.jpg" from the url. Would that be possible to remove whatever domain name come! Please help!
You can use parse_url and then get the path:
$url = parse_url("http://domain1.com/so-section/upload/image1.jpg");
echo $url["path"]; ///so-section/upload/image1.jpg
$pieces = explode('/', $url);
echo '/'. $pieces[4].'/'.$pieces[5].'/'.$pieces[6];
A simple solution to get everything after the domain name.
function GetJunk($url)
{
$start = explode("/",$url); //get everything between "/"
array_shift($start);array_shift($start);array_shift($start); //shift sections
$junk = implode("/",$start); //get everything after the new shifted sections
return $junk; // so-section/upload/image2.jpg
};
Then to use you simply:
echo GetJunk("http://domain2example.com/so-section/upload/image2.jpg");
Here is a working example: http://phpfiddle.org/main/code/4sr-zrm
Hope this helps!
Is there any predefined method in PHP to get sub-domain from url if any?
url pattern may be:
http://www.sd.domain.com
http://domain.com
http://sd.domain.com
http://domain.com
where sd stands for sub-doamin.
Now method must return different values for every case:
case 1 -> return sd
case 2 -> return false or empty
case 3 -> return sd
case 4 -> return false or empty
I found some good links
PHP function to get the subdomain of a URL
Get subdomain from url?
but not specifically apply on my cases.
Any help will be most appreciable.
Thanks
Okay, here I create a script :)
$url = $_SERVER['HTTP_HOST'];
$host = explode('.', $url);
if( !empty($host[0]) && $host[0] != 'www' && $host[0] != 'localhost' ){
$domain = $host[0];
}else{
$domain = 'home';
}
So, there are several possibilities...
First, regular expressions of course:
(http://)?(www\.)?([^\.]*?)\.?([^\.]+)\.([^\.]+)
The entry in the third parenthesis will be your subdomain. Of course, if your url would be https:// or www2 (seen it all...) the regex would break. So this is just a first draft to start working with.
My second idea is, just as yours, explodeing the url. I thought of something like this:
function getSubdomain($url) {
$parts = explode('.', str_replace('http://', '', $url));
if(count($parts) >= 3) {
return $parts[count($parts) - 3];
}
return null;
}
My idea behind this function was, that if an url is splitted by . the subdomain will almost always be the third last entry in the resulting array. The protocol has to be stripped first (see case 3). Of course, this certainly can be done more elegant.
I hope I could give you some ideas.
Try this.
[update] We have a constant defined _SITE_ADDRESS such as www.mysite.com you could use a literal for this.
It works well in our system for what seems like that exact purpose.
public static function getSubDomain()
{
if($_SERVER["SERVER_NAME"] == str_ireplace('http://','',_SITE_ADDRESS)) return ''; //base domain
$host = str_ireplace(array("www.", _SITE_ADDRESS), "", strtolower(trim($_SERVER["HTTP_HOST"])));
$sub = preg_replace('/\..*/', '', $host);
if($sub == $host) return ''; //this is likely an ip address
return $sub;
}
There is an external note on that function but no link, So sorry to any original developer who's code this is based on.
I am trying to make a user submit link box. I've been trying all day and can't seem to get it working.
The goal is to make all of these into example.com... (ie. remove all stuff before the top level domain)
Input is $url =
Their are 4 types of url:
www.example.com...
example.com...
http://www.example.com...
http://example.com...
Everything I make works on 1 or 2 types, but not all 4.
How one can do this?
You can use parse_url for that. For example:
function parse($url) {
$parts = parse_url($url);
if ($parts === false) {
return false;
}
return isset($parts['scheme'])
? $parts['host']
: substr($parts['path'], 0, strcspn($parts['path'], '/'));
}
This will leave the "www." part if it already exists, but it's trivial to cut that out with e.g. str_replace. If the url you give it is seriously malformed, it will return false.
Update (an improved solution):
I realized that the above would not work correctly if you try to trick it hard enough. So instead of whipping myself trying to compensate if it does not have a scheme, I realized that this would be better:
function parse($url) {
$parts = parse_url($url);
if ($parts === false) {
return false;
}
if (!isset($parts['scheme'])) {
$parts = parse_url('http://'.$url);
}
if ($parts === false) {
return false;
}
return $parts['host'];
}
Your input can be
www.example.com
example.com
http://www.example.com
http://example.com
$url_arr = parse_url($url);
echo $url_arr['host'];
output is example.com
there's a few steps you can take to get a clean url.
Firstly you need to make sure there is a protocol to make parse_url work correctly so you can do:
//Make sure it has a protocol
if(substr($url,0,7) != 'http://' || substr($url,0,8) != 'https://')
{
$url = 'http://' . $url;
}
Now we run it through parse_url()
$segments = parse_url($url);
But this is where it get's complicated because the way domain names are constructed is that you can have 1,2,3,4,5,6 .. .domain levels, meaning that you cannot detect the domain name from all urls, you have to have a pre compiled list of tld's to check the last portion of the domain, so you then can extract that leaving the website's domain.
There is a list available here : http://mxr.mozilla.org/mozilla-central/source/netwerk/dns/effective_tld_names.dat?raw=1
But you would be better of parsing this list into mysql and then select the row where the tld matches the left side of the domain string.
Then you order by length and limit to 1, if this is found then you can do something like:
$db_found_tld = 'co.uk';
$domain = 'a.b.c.domain.co.uk';
$domain_name = substr($domain,0 - strlen($db_found_tld));
This would leave a.b.c.domain, so you have removed the tld, now the domain name would be extracted like so:
$parts = explode($domain_name);
$base_domain = $parts[count($parts) - 1];
now you have domain.
this seems very lengthy but I hope now you know that its not easy to get just the domain name without tld or sub domains.
Given this variable:
$variable = foo.com/bar/foo
What function would trim $variable to foo.com ?
Edit: I would like the function to be able to trim anything on a URL that could possibly come after the domain name.
Thanks in advance,
John
Working for OP:
$host = parse_url($url, PHP_URL_HOST);
The version of PHP I have to work with doesn't accept two parameters (Zend Engine 1.3.0). Whatever. Here's the working code for me - you do have to have the full URL including the scheme (http://). If you can safely assume that the scheme is http:// (and not https:// or something else), you could just prepend that to get what you need.
Working for me:
$url = 'http://foo.com/bar/foo';
$parts = parse_url($url);
$host = $parts['host'];
echo "The host is $host\n";
I'm using http://www.google.com/asdf in my example
If you're fine with getting the subdomain as well, you could split by "//" and take the 1th element to effectively remove the protocol and get www.google.com/asdf
You can then split by "/" and get the 0th element.
That seems ugly. Just brainstorming here =)
Try this:
function getDomain($url)
{
if(filter_var($url, FILTER_VALIDATE_URL, FILTER_FLAG_HOST_REQUIRED) === FALSE)
{
return false;
}
/*** get the url parts ***/
$parts = parse_url($url);
/*** return the host domain ***/
return $parts['scheme'].'://'.$parts['host'];
}
$variable = 'foo.com/bar/foo';
echo getDomain($variable);
You can use php's parse_url function and then access the value of the key "host" to get the hostname