How can i get domain link in php?

function url(){
    if(isset($_SERVER['HTTPS'])){
        $protocol = ($_SERVER['HTTPS'] && $_SERVER['HTTPS'] != "off") ? "https" : "http";
    }
    else{
        $protocol = 'http';
    }
    return $protocol . "://" . $_SERVER['HTTP_HOST'];
}

For example with the function above, it works fine if I work with the same directory, but if I make a sub directory, and work in it, it will give me the location of the sub directory also for example. I just want example.com but it gives me example.com/sub if I'm working in the folder sub. If I'm using the main directory,the function works fine. Is there an alternative to $_SERVER['HTTP_HOST']?

Or how could I fix my function/code to get the main url only? Thanks.

asked Jun 19, 2013 at 20:56

1

Use SERVER_NAME.

echo $_SERVER['SERVER_NAME']; //Outputs www.example.com

answered Jun 19, 2013 at 21:01

Bad WolfBad Wolf

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4

You could use PHP's parse_url() function

function url($url) {
  $result = parse_url($url);
  return $result['scheme']."://".$result['host'];
}

answered Jun 19, 2013 at 21:00

3

Shortest solution:

$domain = parse_url('http://google.com', PHP_URL_HOST);

answered May 11, 2017 at 9:08

barbushinbarbushin

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6

/**
 * Suppose, you are browsing in your localhost 
 * http://localhost/myproject/index.php?id=8
 */
function getBaseUrl() 
{
    // output: /myproject/index.php
    $currentPath = $_SERVER['PHP_SELF']; 

    // output: Array ( [dirname] => /myproject [basename] => index.php [extension] => php [filename] => index ) 
    $pathInfo = pathinfo($currentPath); 

    // output: localhost
    $hostName = $_SERVER['HTTP_HOST']; 

    // output: http://
    $protocol = strtolower(substr($_SERVER["SERVER_PROTOCOL"],0,5))=='https'?'https':'http';

    // return: http://localhost/myproject/
    return $protocol.'://'.$hostName.$pathInfo['dirname']."/";
}

answered May 6, 2016 at 11:17

JasmeenJasmeen

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3

Use parse_url() like this:

function url(){
    if(isset($_SERVER['HTTPS'])){
        $protocol = ($_SERVER['HTTPS'] && $_SERVER['HTTPS'] != "off") ? "https" : "http";
    }
    else{
        $protocol = 'http';
    }
    return $protocol . "://" . parse_url($_SERVER['REQUEST_URI'], PHP_URL_HOST);
}

Here is another shorter option:

function url(){
    $pu = parse_url($_SERVER['REQUEST_URI']);
    return $pu["scheme"] . "://" . $pu["host"];
}

answered Jun 19, 2013 at 21:01

Get Off My LawnGet Off My Lawn

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1

Step-1

First trim the trailing backslash (/) from the URL. For example, If the URL is http://www.google.com/ then the resultant URL will be http://www.google.com

$url= trim($url, '/');

Step-2

If scheme not included in the URL, then prepend it. So for example if the URL is www.google.com then the resultant URL will be http://www.google.com

if (!preg_match('#^http(s)?://#', $url)) {
    $url = 'http://' . $url;
}

Step-3

Get the parts of the URL.

$urlParts = parse_url($url);

Step-4

Now remove www. from the URL

$domain = preg_replace('/^www\./', '', $urlParts['host']);

Your final domain without http and www is now stored in $domain variable.

Examples:

http://www.google.com => google.com

https://www.google.com => google.com

www.google.com => google.com

http://google.com => google.com

answered Nov 28, 2017 at 9:22

1

2 lines to solve it

$actual_link = (isset($_SERVER['HTTPS']) && $_SERVER['HTTPS'] === 'on' ? "https" : "http") . "://$_SERVER[HTTP_HOST]$_SERVER[REQUEST_URI]";
$myDomain = preg_replace('/^www\./', '', parse_url($actual_link, PHP_URL_HOST));

answered Apr 20, 2019 at 20:35

FadiFadi

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1

/* Get sub domain or main domain url
 * $url is $_SERVER['SERVER_NAME']
 * $index int remove subdomain if acceess from sub domain my current url is https://support.abcd.com ("support" = 7 (char))
 * $subDomain string 
 * $issecure string https or http
 * return url
 * call like echo getUrl($_SERVER['SERVER_NAME'],7,"payment",true,false);
 * out put https://payment.abcd.com
 * second call echo getUrl($_SERVER['SERVER_NAME'],7,null,true,true);
*/
function getUrl($url,$index,$subDomain=null,$issecure=false,$www=true) {
  //$url=$_SERVER['SERVER_NAME']
  $protocol=($issecure==true) ?  "https://" : "http://";
  $url= substr($url,$index);
  $www =($www==true) ? "www": "";
  $url= empty($subDomain) ? $protocol.$url : 
  $protocol.$www.$subDomain.$url;
  return $url;
}

answered May 22, 2019 at 5:22

Aasif khanAasif khan

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Use this code is whork :

if (!preg_match('#^http(s)?://#', $url)) {
         $url = 'http://' . $url;
}
$urlParts = parse_url($url);
$url = preg_replace('/^www\./', '', $urlParts['host']);

answered Jan 11, 2021 at 0:02

lookly Devlookly Dev

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1

This works fine if you want the http protocol also since it could be http or https. $domainURL = $_SERVER['REQUEST_SCHEME']."://".$_SERVER['SERVER_NAME'];

answered Apr 18, 2020 at 7:52

RudolphRudolph

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Please try this:

$uri = $_SERVER['REQUEST_URI']; // $uri == example.com/sub
$exploded_uri = explode('/', $uri); //$exploded_uri == array('example.com','sub')
$domain_name = $exploded_uri[1]; //$domain_name = 'example.com'

I hope this will help you.

Ismail RBOUH

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answered Jul 14, 2016 at 14:49

CharlesCharles

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1

Tenary Operator helps keep it short and simple.

echo (isset($_SERVER['HTTPS']) ? 'http' : 'https' ). "://" . $_SERVER['SERVER_NAME']  ;

answered Oct 2, 2018 at 17:09

OluwaseyeOluwaseye

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1

If you're using wordpress, use get_site_url:

get_site_url()

answered May 13, 2020 at 1:44

ElronElron

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