从URL解析域 [英] Parsing domain from a URL

问题描述

我需要构建一个从URL解析域的函数.

I need to build a function which parses the domain from a URL.

所以，用

http://google.com/dhasjkdas/sadsdds/sdda/sdads.html

或

http://www.google.com/dhasjkdas/sadsdds/sdda/sdads.html

它应该返回google.com

与

http://google.co.uk/dhasjkdas/sadsdds/sdda/sdads.html

它应该返回google.co.uk.

推荐答案

查看 parse_url() :

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

parse_url不能很好地处理错误整齐的网址，但是如果您通常希望有不错的网址，那就很好了.

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

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