php 从 URL 解析域
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Parsing domain from a URL
提问by zuk1
I need to build a function which parses the domain from a URL.
我需要构建一个从 URL 解析域的函数。
So, with
所以，与
http://google.com/dhasjkdas/sadsdds/sdda/sdads.html
http://google.com/dhasjkdas/sadsdds/sdda/sdads.html
or
或者
http://www.google.com/dhasjkdas/sadsdds/sdda/sdads.html
http://www.google.com/dhasjkdas/sadsdds/sdda/sdads.html
it should return google.com
它应该返回 google.com
with
和
http://google.co.uk/dhasjkdas/sadsdds/sdda/sdads.html
http://google.co.uk/dhasjkdas/sadsdds/sdda/sdads.html
it should return google.co.uk.
它应该返回google.co.uk。
回答by Owen
Check out parse_url():
退房parse_url()：
$url = 'http://google.com/dhasjkdas/sadsdds/sdda/sdads.html';
$parse = parse_url($url);
echo $parse['host']; // prints 'google.com'
parse_urldoesn't handle really badly mangled urls very well, but is fine if you generally expect decent urls.
parse_url不能很好地处理严重损坏的 url，但如果您通常期望不错的 url，那就没问题了。
回答by Alix Axel
$domain = str_ireplace('www.', '', parse_url($url, PHP_URL_HOST));
This would return the google.comfor both http://google.com/... and http://www.google.com/...
这将返回google.com两个http://google.com/...和http://www.google.com/...
回答by philfreo
From http://us3.php.net/manual/en/function.parse-url.php#93983
来自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:
出于某种奇怪的原因，当输入 url 中没有提供方案时，parse_url 返回主机（例如 example.com）作为路径。所以我写了一个快速函数来获取真正的主机：
function getHost($Address) {
$parseUrl = parse_url(trim($Address));
return trim($parseUrl['host'] ? $parseUrl['host'] : array_shift(explode('/', $parseUrl['path'], 2)));
}
getHost("example.com"); // Gives example.com
getHost("http://example.com"); // Gives example.com
getHost("www.example.com"); // Gives www.example.com
getHost("http://example.com/xyz"); // Gives example.com
回答by Shaun
The code that was meant to work 100% didn't seem to cut it for me, I did patch the example a little but found code that wasn't helping and problems with it. so I changed it out to a couple of functions (to save asking for the list from Mozilla all the time, and removing the cache system). This has been tested against a set of 1000 URLs and seemed to work.
本来可以 100% 工作的代码似乎对我来说并没有削减它，我确实对示例进行了一些修补，但发现代码没有帮助并且存在问题。所以我把它改成了几个功能（为了避免一直从 Mozilla 请求列表，并删除缓存系统）。这已经针对一组 1000 个 URL 进行了测试，并且似乎有效。
function domain($url)
{
global $subtlds;
$slds = "";
$url = strtolower($url);
$host = parse_url('http://'.$url,PHP_URL_HOST);
preg_match("/[^\.\/]+\.[^\.\/]+$/", $host, $matches);
foreach($subtlds as $sub){
if (preg_match('/\.'.preg_quote($sub).'$/', $host, $xyz)){
preg_match("/[^\.\/]+\.[^\.\/]+\.[^\.\/]+$/", $host, $matches);
}
}
return @$matches[0];
}
function get_tlds() {
$address = 'http://mxr.mozilla.org/mozilla-central/source/netwerk/dns/effective_tld_names.dat?raw=1';
$content = file($address);
foreach ($content as $num => $line) {
$line = trim($line);
if($line == '') continue;
if(@substr($line[0], 0, 2) == '/') continue;
$line = @preg_replace("/[^a-zA-Z0-9\.]/", '', $line);
if($line == '') continue; //$line = '.'.$line;
if(@$line[0] == '.') $line = substr($line, 1);
if(!strstr($line, '.')) continue;
$subtlds[] = $line;
//echo "{$num}: '{$line}'"; echo "<br>";
}
$subtlds = array_merge(array(
'co.uk', 'me.uk', 'net.uk', 'org.uk', 'sch.uk', 'ac.uk',
'gov.uk', 'nhs.uk', 'police.uk', 'mod.uk', 'asn.au', 'com.au',
'net.au', 'id.au', 'org.au', 'edu.au', 'gov.au', 'csiro.au'
), $subtlds);
$subtlds = array_unique($subtlds);
return $subtlds;
}
Then use it like
然后像这样使用它
$subtlds = get_tlds();
echo domain('www.example.com') //outputs: example.com
echo domain('www.example.uk.com') //outputs: example.uk.com
echo domain('www.example.fr') //outputs: example.fr
I know I should have turned this into a class, but didn't have time.
我知道我应该把它变成一门课，但没有时间。
回答by nikmauro
function get_domain($url = SITE_URL)
{
preg_match("/[a-z0-9\-]{1,63}\.[a-z\.]{2,6}$/", parse_url($url, PHP_URL_HOST), $_domain_tld);
return $_domain_tld[0];
}
get_domain('http://www.cdl.gr'); //cdl.gr
get_domain('http://cdl.gr'); //cdl.gr
get_domain('http://www2.cdl.gr'); //cdl.gr
回答by Oleksandr Fediashov
If you want extract host from string http://google.com/dhasjkdas/sadsdds/sdda/sdads.html, usage of parse_url() is acceptable solution for you.
如果你想从 string 中提取主机http://google.com/dhasjkdas/sadsdds/sdda/sdads.html，使用 parse_url() 是你可以接受的解决方案。
But if you want extract domain or its parts, you need package that using Public Suffix List. Yes, you can use string functions arround parse_url(), but it will produce incorrect results sometimes.
但是，如果您想提取域或其部分，则需要使用Public Suffix List 进行打包。是的，您可以在 parse_url() 周围使用字符串函数，但有时会产生不正确的结果。
I recomend TLDExtractfor domain parsing, here is sample code that show diff:
我建议使用TLDExtract进行域解析，这里是显示差异的示例代码：
$extract = new LayerShifter\TLDExtract\Extract();
# For 'http://google.com/dhasjkdas/sadsdds/sdda/sdads.html'
$url = 'http://google.com/dhasjkdas/sadsdds/sdda/sdads.html';
parse_url($url, PHP_URL_HOST); // will return google.com
$result = $extract->parse($url);
$result->getFullHost(); // will return 'google.com'
$result->getRegistrableDomain(); // will return 'google.com'
$result->getSuffix(); // will return 'com'
# For 'http://search.google.com/dhasjkdas/sadsdds/sdda/sdads.html'
$url = 'http://search.google.com/dhasjkdas/sadsdds/sdda/sdads.html';
parse_url($url, PHP_URL_HOST); // will return 'search.google.com'
$result = $extract->parse($url);
$result->getFullHost(); // will return 'search.google.com'
$result->getRegistrableDomain(); // will return 'google.com'
回答by fatih
I've found that @philfreo's solution (referenced from php.net) is pretty well to get fine result but in some cases it shows php's "notice" and "Strict Standards" message. Here a fixed version of this code.
我发现@philfreo 的解决方案（从 php.net 引用）可以很好地获得良好的结果，但在某些情况下，它显示了 php 的“通知”和“严格标准”消息。这是此代码的固定版本。
function getHost($url) {
$parseUrl = parse_url(trim($url));
if(isset($parseUrl['host']))
{
$host = $parseUrl['host'];
}
else
{
$path = explode('/', $parseUrl['path']);
$host = $path[0];
}
return trim($host);
}
echo getHost("http://example.com/anything.html"); // example.com
echo getHost("http://www.example.net/directory/post.php"); // www.example.net
echo getHost("https://example.co.uk"); // example.co.uk
echo getHost("www.example.net"); // example.net
echo getHost("subdomain.example.net/anything"); // subdomain.example.net
echo getHost("example.net"); // example.net
回答by Kristoffer Bohmann
Please consider replacring the accepted solution with the following:
请考虑使用以下内容替换已接受的解决方案：
parse_url() will always include any sub-domain(s), so this function doesn't parse domain names very well. Here are some examples:
parse_url() 将始终包含任何子域，因此该函数不能很好地解析域名。这里有些例子：
$url = 'http://www.google.com/dhasjkdas/sadsdds/sdda/sdads.html';
$parse = parse_url($url);
echo $parse['host']; // prints 'www.google.com'
echo parse_url('https://subdomain.example.com/foo/bar', PHP_URL_HOST);
// Output: subdomain.example.com
echo parse_url('https://subdomain.example.co.uk/foo/bar', PHP_URL_HOST);
// Output: subdomain.example.co.uk
Instead, you may consider this pragmatic solution. It will cover many, but not all domain names -- for instance, lower-level domains such as 'sos.state.oh.us' are not covered.
相反，您可以考虑这种务实的解决方案。它将涵盖许多域名，但不是所有域名——例如，不包括诸如“sos.state.oh.us”之类的较低级别的域。
function getDomain($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 Parserallows you to parse the domain name from a url. League URI Hostname Parserwill also do the job.
最后，Jeremy Kendall 的PHP Domain Parser允许您从 url 解析域名。League URI Hostname Parser也将完成这项工作。
回答by Michael
$domain = parse_url($url, PHP_URL_HOST);
echo implode('.', array_slice(explode('.', $domain), -2, 2))
回答by Oleg Matei
You can pass PHP_URL_HOST into parse_url function as second parameter
您可以将 PHP_URL_HOST 作为第二个参数传递给 parse_url 函数
$url = 'http://google.com/dhasjkdas/sadsdds/sdda/sdads.html';
$host = parse_url($url, PHP_URL_HOST);
print $host; // prints 'google.com'
