PHP how to get the base domain/url?

Question

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.

Your snippet looks very similar to stackoverflow.com/a/14500505/2943403 — mickmackusa
– mickmackusa ♦, Commented Jan 11, 2021 at 0:28

Bad Wolf · Accepted Answer · 2013-06-19 21:01:06Z

122

Use SERVER_NAME.

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

answered Jun 19, 2013 at 21:01

Bad Wolf

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5 Comments

Robin van Baalen Over a year ago

@barbushin's answer below is better nowadays.

k0pernikus Over a year ago

Here's the link to barbushin's answer: stackoverflow.com/a/43911112/457268 It presupposes knowledge of the domain though, and is not useful if one wants to dynamically determine the host name.

JNAK Over a year ago

@barbushin's answer is not better. Its flat out wrong. Hardcoding the domain is not the correct answer.

hamburger Over a year ago

it outputs for me: example.com not www.example.com

jave.web Over a year ago

Although this can be spoofed if not configured properly, it's still better than gethostname() which does not work when on virtual host, I've posted the comparison and solution/config as a new answer: stackoverflow.com/a/77815115/1835470

user1864610 · Accepted Answer · 2013-06-19 21:00:24Z

59

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

user1864610

3 Comments

programaths Over a year ago

$_SERVER['SERVER_NAME'] wasn't an option since we do interactive domain hosting with custom ports. (Why should we make things easy ?)

Deepak Rajpal Over a year ago

I believe parse_url return nothing, if no http. For example: $url = "example.com";

Nico Haase Over a year ago

Please add some explanation to your answer by editing it, such that others can learn from it - where does $url come from?

barbushin · Accepted Answer · 2017-05-11 09:08:30Z

43

Shortest solution:

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

answered May 11, 2017 at 9:08

barbushin

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6 Comments

Naveen DA Over a year ago

This should be accepted as answer, because its cross domain capability

Andrew Magill Over a year ago

This gives you the string "google.com". It's a flat out wrong answer.

undo Over a year ago

@AyexeM You're supposed to change "google.com" to your website's domain.

Andrew Magill Over a year ago

@rahuldotech That makes no sense, copy paste your domain in there and it will return your domain? How is that helpful ? OP wants the domain of the server, not how to parse a domain from a string.

TARKUS Over a year ago

How I understand the question is how to get the domain segment without knowing what the domain is. That is, relative to the page you are loading. What if you are testing locally, then uploading to production? If you already know the domain, you don't need to retrieve it dynamically. I think the assumption is we don't know the domain the resource is loading from.

|

Jasmeen · Accepted Answer · 2017-05-11 06:34:16Z

31

/**
 * 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']."/";
}

edited May 11, 2017 at 6:34

answered May 6, 2016 at 11:17

Jasmeen

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3 Comments

Ultimater Over a year ago

Umm... see php.net/manual/en/reserved.variables.server.php $_SERVER["SERVER_PROTOCOL"] returns a string like HTTP/1.0 or HTTP/1.1 and has nothing to do with https. You're probably looking for something like $_SERVER['HTTPS'] == 'on' or the undocumented REQUEST_SCHEME. See this related question and here's the answer I used: stackoverflow.com/a/14270161/466314

Jasmeen Over a year ago

See, Previously I have wrote 5, It was correct but minor changes should be there like, $protocol = strtolower(substr($_SERVER["SERVER_PROTOCOL"],0,5))=='https'?'https':'http';

Ultimater Over a year ago

No, strtolower(substr($_SERVER["SERVER_PROTOCOL"],0,5))=='https'‌ is always false because $_SERVER["SERVER_PROTOCOL"] doesn't work the way you think it does. It will never contain HTTPS anywhere in it. The example given in the apache documentation <If "%{SERVER_PROTOCOL} != 'HTTPS'"> is wrong as it's always true: httpd.apache.org/docs/2.4/rewrite/remapping.html#canonicalhost

Get Off My Lawn · Accepted Answer · 2013-06-19 21:01:32Z

11

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 Lawn

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1 Comment

Asfandyar Khan Over a year ago

parse_url returns nothing, if there is no http or https, i.e example.com

Dibyendu Mitra Roy · Accepted Answer · 2017-11-28 09:22:51Z

7

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

Dibyendu Mitra Roy

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1 Comment

bksi Over a year ago

what if other subdomain is used instead www?

Aasif khan · Accepted Answer · 2019-05-22 06:35:26Z

3

/* 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;
}

edited May 22, 2019 at 6:35

answered May 22, 2019 at 5:22

Aasif khan

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Comments

Fadi · Accepted Answer · 2019-04-20 20:35:00Z

2

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

Fadi

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1 Comment

mickmackusa Over a year ago

This answer is missing its educational explanation.

Rudolph · Accepted Answer · 2020-04-18 07:52:26Z

2

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

Rudolph

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Comments

lookly Dev · Accepted Answer · 2021-01-11 00:02:26Z

2

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 Dev

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1 Comment

mickmackusa Over a year ago

This answer is missing its educational explanation. Also, I would prepare the url with a single preg call instead of two. ...and please use correct spelling.

Ismail RBOUH · Accepted Answer · 2016-07-14 18:38:32Z

0

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.

edited Jul 14, 2016 at 18:38

Ismail RBOUH

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answered Jul 14, 2016 at 14:49

Charles

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1 Comment

Udara Herath Over a year ago

In here $domain_name = 'sub' not 'example.com'. If need to get 'example.com' , then change the last code to $domain_name = $exploded_uri[0];

jave.web · Accepted Answer · 2024-01-14 12:55:32Z

TL;DR

Specific http server (apache/nginx) config or whitelist of hosts & $_SERVER['SERVER_NAME'].

`gethostname()` ? only for non-virtual host...

https://www.php.net/manual/en/function.gethostname.php

gets the standard host name for the local machine.

... but if you're on a virtual host, this will still return the system host, not the virtual one :-( ...
(typically e.g. shared web hosting)

So what about `$_SERVER['SERVER_NAME']`?

https://www.php.net/manual/en/reserved.variables.server.php

Well, this on the other hand can be spoofed if not configured properly,
but it does return the virtual host if on virtual host.

Note: Under Apache 2, UseCanonicalName = On and ServerName must be set. Otherwise, this value reflects the hostname supplied by the client, which can be spoofed. It is not safe to rely on this value in security-dependent contexts.

Nginx => the same thing, docs just don't mention it.

Solution

Basically you have 2 options:

If you can configure your http server, you can force it to be your value, not value from request, this differs per http server you use:
- Apache: UseCanonicalName (to On) & ServerName directives have to be set - most likely in your <VirtualHost>
- Nginx: you have to set the server_name directive in your server{} block.
If you can't configure your server or can't rely on its setting, you can at least whitelist the allowed hostnames

<?php
    $allowedHosts = [
        'localhost',
        'example.com',
        'www.example.com',
    ];

    if (!\in_array($_SERVER['HTTP_HOST'], $allowedHosts)) {
        exit('You trynna spoof me?');
    }

Oluwaseye · Accepted Answer · 2018-10-02 17:09:01Z

-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

Oluwaseye

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1 Comment

M. Eriksson Over a year ago

^ What they said. You probably also want to check the value of $_SERVER

Elron · Accepted Answer · 2020-05-13 01:44:40Z

-3

If you're using wordpress, use get_site_url:

get_site_url()

answered May 13, 2020 at 1:44

Elron

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So what about `$_SERVER['SERVER_NAME']`?

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