Parsing domain from a URL

Check out parse_url():

$url = 'http://google.com/dhasjkdas/sadsdds/sdda/sdads.html';
$parse = parse_url($url);
echo $parse['host']; // prints 'google.com'

parse_url doesn't handle really badly mangled urls very well, but is fine if you generally expect decent urls.

$domain = str_ireplace('www.', '', parse_url($url, PHP_URL_HOST));

This would return the google.com for both http://google.com/... and http://www.google.com/...

From http://us3.php.net/manual/en/function.parse-url.php#93983

for some odd reason, parse_url returns the host (ex. example.com) as the path when no scheme is provided in the input url. So I've written a quick function to get the real host:

function getHost($Address) { 
   $parseUrl = parse_url(trim($Address)); 
   return trim($parseUrl['host'] ? $parseUrl['host'] : array_shift(explode('/', $parseUrl['path'], 2))); 
} 

getHost("example.com"); // Gives example.com 
getHost("http://example.com"); // Gives example.com 
getHost("www.example.com"); // Gives www.example.com 
getHost("http://example.com/xyz"); // Gives example.com 

Please consider replacring the accepted solution with the following:

parse_url() will always include any sub-domain(s), so this function doesn't parse domain names very well. Here are some examples:

$url = 'http://www.google.com/dhasjkdas/sadsdds/sdda/sdads.html';
$parse = parse_url($url);
echo $parse['host']; // prints 'www.google.com'

echo parse_url('https://subdomain.example.com/foo/bar', PHP_URL_HOST);
// Output: subdomain.example.com

echo parse_url('https://subdomain.example.co.uk/foo/bar', PHP_URL_HOST);
// Output: subdomain.example.co.uk

Instead, you may consider this pragmatic solution. It will cover many, but not all domain names -- for instance, lower-level domains such as 'sos.state.oh.us' are not covered.

function getDomain($url) {
    $host = parse_url($url, PHP_URL_HOST);

    if(filter_var($host,FILTER_VALIDATE_IP)) {
        // IP address returned as domain
        return $host; //* or replace with null if you don't want an IP back
    }

    $domain_array = explode(".", str_replace('www.', '', $host));
    $count = count($domain_array);
    if( $count>=3 && strlen($domain_array[$count-2])==2 ) {
        // SLD (example.co.uk)
        return implode('.', array_splice($domain_array, $count-3,3));
    } else if( $count>=2 ) {
        // TLD (example.com)
        return implode('.', array_splice($domain_array, $count-2,2));
    }
}

// Your domains
    echo getDomain('http://google.com/dhasjkdas/sadsdds/sdda/sdads.html'); // google.com
    echo getDomain('http://www.google.com/dhasjkdas/sadsdds/sdda/sdads.html'); // google.com
    echo getDomain('http://google.co.uk/dhasjkdas/sadsdds/sdda/sdads.html'); // google.co.uk

// TLD
    echo getDomain('https://shop.example.com'); // example.com
    echo getDomain('https://foo.bar.example.com'); // example.com
    echo getDomain('https://www.example.com'); // example.com
    echo getDomain('https://example.com'); // example.com

// SLD
    echo getDomain('https://more.news.bbc.co.uk'); // bbc.co.uk
    echo getDomain('https://www.bbc.co.uk'); // bbc.co.uk
    echo getDomain('https://bbc.co.uk'); // bbc.co.uk

// IP
    echo getDomain('https://1.2.3.45');  // 1.2.3.45

Finally, Jeremy Kendall's PHP Domain Parser allows you to parse the domain name from a url. League URI Hostname Parser will also do the job.

function getTrimmedUrl($link)
{
    $str = str_replace(["www.","https://","http://"],[''],$link);
    $link = explode("/",$str);
    return strtolower($link[0]);                
}

None of this solutions worked for me when I use this test cases:

public function getTestCases(): array
{
    return [
        //input                              expected
        ['http://google.com/dhasjkdas',      'google.com'],
        ['https://google.com/dhasjkdas',     'google.com'],
        ['https://www.google.com/dhasjkdas', 'google.com'],
        ['http://www.google.com/dhasjkdas',  'google.com'],
        ['www.google.com/dhasjkdas',         'google.com'],
        ['google.com/dhasjkdas',             'google.com'],
    ];
}

but wrapping this answer into function worked in all cases: https://stackoverflow.com/a/65659814/5884988