php 从完整 URL 获取域名
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Get domain name from full URL
提问by ramo
say someone enters a URL like this:
假设有人输入这样的 URL：
http://i.imgur.com/a/b/c?query=value&query2=value
And I want to return: imgur.com
我想回来： imgur.com
not i.imgur.com
不是 i.imgur.com
This is code I have right now
这是我现在拥有的代码
$sourceUrl = parse_url($url);
$sourceUrl = $sourceUrl['host'];
But this returns i.imgur.com
但这返回 i.imgur.com
回答by stanev01
Check the code below, it should do the job fine.
检查下面的代码，它应该可以正常工作。
<?php
function get_domain($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'
?>
回答by Oleksandr Fediashov
You need package that using Public Suffix List. Yes, you can use string functions arround parse_url() or regex, but they will produce incorrect result in complex URLs.
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。
I recomend TLDExtractfor domain parsing, here is sample code:
我推荐使用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'
回答by Muhammad Sanaullah
I have found a very useful library using publicsuffix.org, PHP Domain Parseris a Public Suffix List based domain parser implemented in PHP.
我使用 publicsuffix.org 找到了一个非常有用的库， PHP Domain Parser是一个在 PHP 中实现的基于公共后缀列表的域解析器。
https://github.com/jeremykendall/php-domain-parser
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($parser->getRegistrableDomain('www.scottwills.co.uk'));
?>
string(16) "scottwills.co.uk"
字符串（16）“scottwills.co.uk”
回答by DMVerfurth
The code below should be perfect for the job.
下面的代码应该非常适合这项工作。
function get_domain($url){
$charge = explode('/', $url);
$charge = $charge[2]; //assuming that the url starts with http:// or https://
return $charge;
}
echo get_domain('http://www.example.com/example.php');
回答by mahipal purohit
if(substr_count($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];
}
回答by Marcelo Viana
use this:
用这个：
$uri = "$_SERVER[REQUEST_URI]";<br>
print($uri);
Example:
例子：
http://exemple.com/?directory<br>
Result:
/?diretory
The command get the directory and not domain.
该命令获取目录而不是域。
回答by Andreas Veith
If you only want the domain name, try the following:
如果您只想要域名，请尝试以下操作：
$domain = $_SERVER['SERVER_NAME'];
echo $domain;
