Question

我有一系列像这样的域名：

domain.com
second.com
www.third.com
www.fourth.fifth.com
sixth.com
seventh.eigth.com

我想要的是一个只返回主机的功能。没有子域名。

此代码是我到目前为止获取主机名的代码：

$parse = parse_url($url);
$domain = $parse['host'];

但这仅返回：

domain.com
second.com
third.com
fourth.fifth.com
sixth.com
seventh.eigth.com

我需要这个输出：

domain.com
second.com
third.com
fifth.com
sixth.com
eigth.com

Answer 1

function giveHost($host_with_subdomain) {
    $array = explode(".", $host_with_subdomain);

    return (array_key_exists(count($array) - 2, $array) ? $array[count($array) - 2] : "").".".$array[count($array) - 1];
}

Answer 2

试试此代码

 <?php
    /**
    * @param string $domain Pass $_SERVER['SERVER_NAME'] here
    * @param bool $debug
    *
    * @debug bool $debug
    * @return string
    */
    function get_domain($domain, $debug = false) {
        $original = $domain = strtolower($domain);     
        if (filter_var($domain, FILTER_VALIDATE_IP)) { return $domain; }    

        $debug ? print('<strong style="color:green">&raquo;</strong> Parsing: '.$original) : false; //DEBUG 

        $arr = array_slice(array_filter(explode('.', $domain, 4), function($value){
                            return $value !== 'www'; }), 0); //rebuild array indexes

        if (count($arr) > 2)    {
            $count = count($arr);
            $_sub = explode('.', $count === 4 ? $arr[3] : $arr[2]);

            $debug ? print(" (parts count: {$count})") : false; //DEBUG

            if (count($_sub) === 2)  { // two level TLD
                $removed = array_shift($arr);
                if ($count === 4) // got a subdomain acting as a domain
                    $removed = array_shift($arr);            
                $debug ? print("<br>\n" . '[*] Two level TLD: <strong>' . join('.', $_sub) . '</strong> ') : false; //DEBUG
            }
            elseif (count($_sub) === 1){ // one level TLD
                $removed = array_shift($arr); //remove the subdomain             
                if (strlen($_sub[0]) === 2 && $count === 3) // TLD domain must be 2 letters
                    array_unshift($arr, $removed);                
                else{
                    // non country TLD according to IANA
                    $tlds = array(    'aero',    'arpa',    'asia',    'biz',    'cat',    'com',    'coop',    'edu',    'gov',    'info',    'jobs',    'mil',    'mobi',    'museum',    'name',    'net',    'org',    'post',    'pro',    'tel',    'travel',    'xxx',    );             
                    if (count($arr) > 2 && in_array($_sub[0], $tlds) !== false) {//special TLD don't have a country
                        array_shift($arr);
                    }
                }
                $debug ? print("<br>\n" .'[*] One level TLD: <strong>'.join('.', $_sub).'</strong> ') : false; //DEBUG
            }
            else { // more than 3 levels, something is wrong
                for ($i = count($_sub); $i > 1; $i--) 
                    $removed = array_shift($arr);

                $debug ? print("<br>\n" . '[*] Three level TLD: <strong>' . join('.', $_sub) . '</strong> ') : false; //DEBUG
            }
        }
        elseif (count($arr) === 2) {
            $arr0 = array_shift($arr);     
            if (strpos(join('.', $arr), '.') === false
                        && in_array($arr[0], array('localhost','test','invalid')) === false) // not a reserved domain
            {
                $debug ? print("<br>\n" .'Seems invalid domain: <strong>'.join('.', $arr).'</strong> re-adding: <strong>'.$arr0.'</strong> ') : false; //DEBUG
                // seems invalid domain, restore it
                array_unshift($arr, $arr0);
            }
        }     

        $debug ? print("<br>\n".'<strong style="color:gray">&laquo;</strong> Done parsing: <span style="color:red">' . $original . '</span> as <span style="color:blue">'. join('.', $arr) ."</span><br>\n") : false; //DEBUG     
        return join('.', $arr);
    }


     //TEST
    $urls = array(
    'www.example.com' => 'example.com',
    'example.com' => 'example.com',
    'example.com.br' => 'example.com.br',
    'www.example.com.br' => 'example.com.br',
    'www.example.gov.br' => 'example.gov.br',
    'localhost' => 'localhost',
    'www.localhost' => 'localhost',
    'subdomain.localhost' => 'localhost',
    'www.subdomain.example.com' => 'example.com',
    'subdomain.example.com' => 'example.com',
    'subdomain.example.com.br' => 'example.com.br',
    'www.subdomain.example.com.br' => 'example.com.br',
    'www.subdomain.example.biz.br' => 'example.biz.br',
    'subdomain.example.biz.br' => 'example.biz.br',
    'subdomain.example.net' => 'example.net',
    'www.subdomain.example.net' => 'example.net',
    'www.subdomain.example.co.kr' => 'example.co.kr',
    'subdomain.example.co.kr' => 'example.co.kr',
    'example.co.kr' => 'example.co.kr',
    'example.jobs' => 'example.jobs',
    'www.example.jobs' => 'example.jobs',
    'subdomain.example.jobs' => 'example.jobs',
    'insane.subdomain.example.jobs' => 'example.jobs',
    'insane.subdomain.example.com.br' => 'example.com.br',
    'www.doubleinsane.subdomain.example.com.br' => 'example.com.br',
    'www.subdomain.example.jobs' => 'example.jobs',
    'test' => 'test',
    'www.test' => 'test',
    'subdomain.test' => 'test',
    'www.detran.sp.gov.br' => 'sp.gov.br',
    'www.mp.sp.gov.br' => 'sp.gov.br',
    'ny.library.museum' => 'library.museum',
    'www.ny.library.museum' => 'library.museum',
    'ny.ny.library.museum' => 'library.museum',
    'www.library.museum' => 'library.museum',
    'info.abril.com.br' => 'abril.com.br',
    '127.0.0.1' => '127.0.0.1',
    '::1' => '::1',
    );

    $failed = 0;
    $total = count($urls);

    foreach ($urls as $from => $expected){
        $from = get_domain($from, true);
        if ($from !== $expected){
            $failed++;
            print("<div style='color:fuchsia;'>expected {$from} to be {$expected}</div>");
        }
    }    
    if ($failed)    
        print("{$failed} tests failed out of {$total}");    
    else    
        print("Success");

所有功劳都归pocesar

Answer 3

这是适用于所有域名的域名，包括那些具有二级域名的域名，例如“co.uk”

function strip_subdomains($url){

    # credits to gavingmiller for maintaining this list
    $second_level_domains = file_get_contents("https://raw.githubusercontent.com/gavingmiller/second-level-domains/master/SLDs.csv");

    # presume sld first ...
    $possible_sld = implode('.', array_slice(explode('.', $url), -2));

    # and then verify it
    if (strpos($second_level_domains, $possible_sld)){
        return  implode('.', array_slice(explode('.', $url), -3));
    } else {
        return  implode('.', array_slice(explode('.', $url), -2));
    }
}

Answer 4

我更喜欢使用正则表达式来做这件事，对我来说更容易理解：

$url = "http://www.domain.com";
$parse = parse_url($url);
echo preg_replace("/^([a-zA-Z0-9].*\.)?([a-zA-Z0-9][a-zA-Z0-9-]{1,61}[a-zA-Z0-9]\.[a-zA-Z.]{2,})$/", '$2', $parse['host']);

Answer 5

尝试使用str_replace（）;

$parse = parse_url($url);
$domain = str_replace('www.','',$parse['host']);

Answer 6

我正在使用Baptiste Donaux的版本，直到需要检查“localhost＆＃39;”。我认为Shanoop的版本更可靠。

我已经在190个断言的测试套件中测试了这两个版本，并且对性能没有太大影响。如果仍有问题，您可以使用Redis或类似方法将结果缓存到生产中。

这是Shanoop的答案版本，但是没有调试行和一点清理：

function stripSubdomain($domain) 
{
    $domain = strtolower($domain);
    if (isIp($domain)) ? return $domain;
    return stripArray( buildArray($domain) );
}

function isIp($domain)
{
    return (filter_var($domain, FILTER_VALIDATE_IP));
}

function buildArray($domain)
{
    return array_slice(array_filter(explode('.', $domain, 4), function($value){
                                                                  return $value !== 'www';
                                                              }), 0);
}

function stripArray($arr)
{
    // TLD Domains
    if (count($arr) > 2) {
        $count = count($arr);
        $_sub = $this->retrieveSubdomain($arr);

        // two level TLD
        if (count($_sub) === 2)  {
            array_shift($arr);
            if ($count === 4) array_shift($arr);
        }

        // one level TLD
        elseif (count($_sub) === 1){
            $removed = array_shift($arr);
            if (strlen($_sub[0]) === 2 && $count === 3) array_unshift($arr, $removed);

            else {
                // non country TLD according to IANA
                $tlds = ['aero', 'arpa', 'asia', 'biz', 'cat', 'com', 'coop',
                         'edu', 'gov', 'info', 'jobs', 'mil', 'mobi', 'museum',
                         'name', 'net', 'org', 'post', 'pro', 'tel', 'travel', 'xxx'];

                if (count($arr) > 2 &&
                    in_array($_sub[0], $tlds) !== false) array_shift($arr);
            }

        }

        // more than 3 levels, something is wrong
        else
            for ($i = count($_sub); $i > 1; $i--) array_shift($arr);

    }

    // Special Domains
    elseif (count($arr) === 2) {
        $removed = array_shift($arr);
        $reserved = ['localhost','test','invalid'];
        if (strpos(join('.', $arr), '.') === false && in_array($arr[0], $reserved) === false)
            array_unshift($arr, $removed); // seems invalid domain, restore it
    }

    return join('.', $arr);
}

function retrieveSubdomain($arr)
{
    return explode('.', (count($arr) === 4 ? $arr[3] : $arr[2]) );
}

Answer 7

这个适用于大多数域名，如果我必须自己这样说，那就很优雅！

public static function StripSubdomain($Domain) {

    $domain_array = explode('.', $Domain);
    $domain_array = array_reverse($domain_array);

    return $domain_array[1] . '.' . $domain_array[0];
}

Answer 8

尝试一下。我喜欢这种简单性，并且它在大多数用例中都对我有用。

$domains = [
    "domain.com",
    "second.com",
    "www.third.com",
    "www.fourth.fifth.com",
    "openhours.colyn.dev"
];

$domains = array_map(function ($domain) {
    $parts = explode('.', $domain);
    return implode('.', array_slice($parts, count($parts)-2));
}, $domains);

/**
[
    "domain.com",
    "second.com",
    "third.com",
    "fifth.com",
    "colyn.dev",
]
*/

不是超级健壮，但对我来说，它工作正常。

如果找到子域，则从url字符串中删除子域

8 个答案: