在可能不包含协议的字符串上运行parse_url() [英] Running parse_url() on a string that may not contain the protocol

问题描述

我正在尝试从用户输入的URL字符串中获取域名和TLD(无子域)，该字符串可能具有也可能没有协议，目录，子域，文件名等.

I'm trying to get the domain name and TLD (no subdomain) from a user-input URL string which may or may not have protocol, directories, subdomains, filenames, etc.

换句话说，给出以下任意一项:

In other words, given any of the following:

example.com
www.example.com
sub.example.com
example.com/whatever/hey.html
http://example.com
https://subdomain.example.com
ftp://example.com/whatever/hey.html

我应该总是以example.com结尾.

现在这就是我在做什么:

Right now this is what I am doing:

$hostParts = explode('.', parse_url($URL, PHP_URL_HOST));
$tld = array_pop($hostParts);
$domain = array_pop($hostParts);
$domain = $domain . "." . $tld;

但是，如果提供的URL没有协议，则会中断.为什么parse_url在这种情况下无法获取主机?

However, if a URL is provided without the protocol, it breaks. Why is parse_url failing to get the host in this situation?

推荐答案

根据定义，URL包含协议或方案.检查////，如果不存在，则//放在字符串的前面.这在PHP<中可能有所不同. 5.4.7，如果没有协议，则可以添加http://.

By definition a URL contains a protocol or scheme. Check for // and if not present prepend // to the string. This may be different in PHP < 5.4.7 so maybe add http:// if no protocol.

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