用于检索domain.tld的正则表达式

Question

我需要一个Java中的正则表达式,我可以用来从任何url中检索domain.tld部分.所以https://foo.com/bar,HTTP://www.foo.com#bar,http://bar.foo.com将全部返回foo.com.

我写了这个正则表达式,但它匹配整个网址

Pattern.compile("[.]?.*[.x][a-z]{2,3}");

我不确定我是否匹配"." 性格正确.我试过了 "." 但我从netbeans得到一个错误.

更新:

tld不限于2或3个字符,http://www.foo.co.uk/bar应返回foo.co.uk.

Answer 1

这比你想象的要难.您的示例https://foo.com/bar,其中有一个逗号,这是一个有效的URL字符.这是一篇关于一些麻烦的好文章:

https://blog.codinghorror.com/the-problem-with-urls/

https?://([-A-Za-z0-9+&@#/%?=~_()|!:,.;]*[-A-Za-z0-9+&@#/%=~_()|])

是一个很好的起点

关于此主题的"掌握正则表达式"中的一些列表:

http://regex.info/listing.cgi?ed=3&p=207

@sjobe

>>> import re
>>> pattern = r'https?://([-A-Za-z0-9+&@#/%?=~_()|!:,.;]*[-A-Za-z0-9+&@#/%=~_()|])'
>>> url = re.compile(pattern)
>>> url.match('http://news.google.com/').groups()
('news.google.com/',)
>>> url.match('not a url').groups()
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
AttributeError: 'NoneType' object has no attribute 'groups'
>>> url.match('http://google.com/').groups()
('google.com/',)
>>> url.match('http://google.com').groups()
('google.com',)

对不起,这个例子是在python而不是java中,它更简短.Java需要一些无关的正则表达式的逃避.

Answer 2

我会使用java.net.URI类来提取主机名,然后使用正则表达式来提取主机uri的最后两部分.

import java.net.URI;
import java.net.URISyntaxException;
import java.util.regex.Matcher;
import java.util.regex.Pattern;

public class RunIt {

    public static void main(String[] args) throws URISyntaxException {
        Pattern p = Pattern.compile(".*?([^.]+\\.[^.]+)");

        String[] urls = new String[] {
                "https://foo.com/bar",
                "http://www.foo.com#bar",
                "http://bar.foo.com"
        };

        for (String url:urls) {
            URI uri = new URI(url);
            //eg: uri.getHost() will return "www.foo.com"
            Matcher m = p.matcher(uri.getHost());
            if (m.matches()) {
                System.out.println(m.group(1));
            }
        }
    }
}

打印:

foo.com
foo.com
foo.com

Answer 3

如果字符串包含有效的URL,那么您可以使用像(Perl引用)这样的正则表达式:

/^
(?:\w+:\/\/)?
[^:?#\/\s]*?

(
[^.\s]+
\.(?:[a-z]{2,}|co\.uk|org\.uk|ac\.uk|org\.au|com\.au|___etc___)
)

(?:[:?#\/]|$)
/xi;

结果:

url: https://foo.com/bar
matched: foo.com
url: http://www.foo.com#bar
matched: foo.com
url: http://bar.foo.com
matched: foo.com
url: ftp://foo.com
matched: foo.com
url: ftp://www.foo.co.uk?bar
matched: foo.co.uk
url: ftp://www.foo.co.uk:8080/bar
matched: foo.co.uk

对于Java,它将被引用如下:

"^(?:\\w+://)?[^:?#/\\s]*?([^.\\s]+\\.(?:[a-z]{2,}|co\\.uk|org\\.uk|ac\\.uk|org\\.au|com\\.au|___etc___))(?:[:?#/]|$)"

当然你需要更换etc部分.

示例Perl脚本:

use strict;

my @test = qw(
    https://foo.com/bar
    http://www.foo.com#bar
    http://bar.foo.com
    ftp://foo.com
    ftp://www.foo.co.uk?bar
    ftp://www.foo.co.uk:8080/bar
);

for(@test){
    print "url: $_\n";

    /^
    (?:\w+:\/\/)?
    [^:?#\/\s]*?

    (
    [^.\s]+
    \.(?:[a-z]{2,}|co\.uk|org\.uk|ac\.uk|org\.au|com\.au|___etc___)
    )

    (?:[:?#\/]|$)
    /xi;

    print "matched: $1\n";
}