41

假设有人输入这样的 URL：

http://i.imgur.com/a/b/c?query=value&query2=value

我想回来：imgur.com

不是i.imgur.com

这是我现在拥有的代码

$sourceUrl = parse_url($url);
$sourceUrl = $sourceUrl['host'];

但这会返回i.imgur.com

4

8 回答 8

82

检查下面的代码，它应该可以正常工作。

<?php

function get_domain($url)
{
  $pieces = parse_url($url);
  $domain = isset($pieces['host']) ? $pieces['host'] : $pieces['path'];
  if (preg_match('/(?P<domain>[a-z0-9][a-z0-9\-]{1,63}\.[a-z\.]{2,6})$/i', $domain, $regs)) {
    return $regs['domain'];
  }
  return false;
}

print get_domain("http://mail.somedomain.co.uk"); // outputs 'somedomain.co.uk'

?>
于 2013-04-16T01:10:28.883 回答
8

您需要使用Public Suffix List的包。是的，您可以在 parse_url() 或 regex 周围使用字符串函数，但它们会在复杂的 URL 中产生不正确的结果。

我推荐TLDExtract进行域解析，这里是示例代码：

$url = 'http://i.imgur.com/a/b/c?query=value&query2=value';

parse_url($url, PHP_URL_HOST); // will return 'i.imgur.com'

$extract = new LayerShifter\TLDExtract\Extract();
$result = $extract->parse($url);
$result->getFullHost(); // will return 'i.imgur.com'
$result->getSubdomain(); // will return 'i'
$result->getRegistrableDomain(); // will return 'imgur.com'
$result->getSuffix(); // will return 'com'
于 2016-06-23T09:20:24.173 回答
2

我使用 publicsuffix.org 找到了一个非常有用的库， PHP 域解析器是一个用 PHP 实现的基于公共后缀列表的域解析器。

https://github.com/jeremykendall/php-domain-parser

 <?php 
 // this will do the job

 require_once '../vendor/autoload.php';

 $pslManager = new Pdp\PublicSuffixListManager();
 $parser = new Pdp\Parser($pslManager->getList());
 var_dump($parser->getRegistrableDomain('www.scottwills.co.uk'));
 ?>

string(16) "scottwills.co.uk"

于 2017-09-30T22:42:53.120 回答
2

下面的代码应该非常适合这项工作。

function get_domain($url){
  $charge = explode('/', $url);
  $charge = $charge[2]; //assuming that the url starts with http:// or https://
  return $charge;
}

echo get_domain('http://www.example.com/example.php');
于 2018-04-09T00:39:24.830 回答
0

检查简单代码您可以获得主机、子域、域、扩展名

$urls = array("https://www.face.com","www.asdasd.asd","sasdas.com/asdas","sdfsdf.sdf","https://app.abcdlink.com/user/test/");
function getDomainname($a)
{
   $r = "(?P<host>(?:(?P<subdomain>[\w\.]+)\.)?" . "(?P<domain>\w+\.(?P<extension>\w+)))";
   $r = "!$r!";// Delimiters
   preg_match($r, $a, $out);

// if you need only domain then return $out['domain'];
// if you need only host then return $out['host'];
// if you need only subdomain then return $out['subdomain'];
// if you need only extension then return $out['extension'];

// Full Data array
    return $out;

}

$urls = array_map('getDomainname', $urls);
于 2022-02-10T08:04:24.980 回答
-1 
     if(substr_count($original_url, 'http://')) {
    if(substr_count($original_url, 'www.')) {
        // url style would be 'http://www.abc.xxx/page?param' or http://www.abc.xxx.xx/page?param
        // extract 'abc'
        $temp = explode('.', $original_url);

        $store_url = $temp[1];
        // now 
        // $temp[2] = xxx or xxx/page?param 
        // $temp[3] = null or xx/page?param 

        //if ($temp[3] == null) { // then we are sure that $temp[2]== "xxx/page?param"
                    if(sizeof($temp) > 3) {
            // extract "xxx" from "xxx/page?param" and append to store url so it will be "abc.xxx"  
            $temp = explode('/',$temp[2]);
            $store_url .= '.'.$temp[0];
        }
        else { 
            // then we are sure that $temp[2]== "xxx" and then $temp[3] == "xx/page?param"
            //                  or   $temp[2]== xxx/page?stripped-link from second dot(.)
            if(substr_count($temp[2], '/')) { // in case  $temp[2]== xxx/page?stripped-link from second dot(.)
                // extract "xxx" from "xxx/page?stripped-link" and appent to store url so it will be "abc.xxx"
                $temp = explode('/',$temp[2]);
                $store_url .= '.'.$temp[0]; // "abc".="xxx" ==> abc.xxx
            }
            else { // in case $temp[2]== "xxx" and then $temp[3] == "xx/page?param"
                $store_url .= '.'.$temp[2]; // "abc".="xxx" ==> abc.xxx
                // extract "xx" from "xx/page?param" and appent to store url so it will be "abc.xxx.xx"
                $temp = explode('/',$temp[3]);
                if(strlen($temp[0])==2) {
                    $store_url .= '.'.$temp[0];
                }
            }
        }
    }
    else {
        // url style would be 'http://abc.xxx/page?param' or 'http://abc.xxx.xx/page?param'
        // remove 'http://'
        $temp = substr($original_url, 7);
        // now temp would be either 'abc.xxx/page?param' or 'abc.xxx.xx/page?param'
        // explode with '/'
        $temp = explode('/', $temp);
        $store_url = $temp[0];
    }
}
else if(substr_count($original_url, 'www.')) {
    // url style would be 'www.abc.xxx/page?param' or 'www.abc.xxx.xx/page?param'
    // remove 'www.'
    $temp = substr($original_url, 4);
    // now, $temp would be either "abc.xxx/page?param" or "abc.xxx.xx/page?param"
    // explode with '/'
    $temp = explode('/', $temp);
    $store_url = $temp[0];
}
else {
    // url style would be 'abc.xxx/page?param' or 'abc.xxx.xx/page?param'
    //explode with '/'
    $temp = explode('/', $original_url);
    $store_url = $temp[0];
}
于 2013-07-29T05:27:39.930 回答
-5

用这个：

$uri = "$_SERVER[REQUEST_URI]";<br>
print($uri);

例子：

http://exemple.com/?directory<br>
Result:
/?diretory

该命令获取目录而不是域。

于 2014-07-27T02:53:37.203 回答
-5

如果您只想要域名，请尝试以下操作：

$domain = $_SERVER['SERVER_NAME'];

echo $domain;
于 2014-12-01T11:49:41.157 回答