我知道有无数的线程提出这个问题,但我找不到能帮助我解决这个问题的线程.
我基本上试图解析大约10,000,000个URL的列表,确保它们符合以下条件,然后获取根域URL.此列表包含您可以想象的所有内容,包括(和预期的格式化URL)之类的内容:
biy.ly/test [VALID] [return - bit.ly]
example.com/apples?test=1&id=4 [VALID] [return - example.com]
host101.wow404.apples.test.com/cert/blah [VALID] [return - test.com]
101.121.44.xxx [**inVALID**] [return false]
localhost/noway [**inVALID**] [return false]
www.awesome.com [VALID] [return - awesome.com]
i am so awesome [**inVALID**] [return false]
http://404.mynewsite.com/visits/page/view/1/ [VALID] [return - mynewsite.com]
www1.151.com/searchresults [VALID] [return - 151.com]
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有没有人对此有任何建议?
Tom*_*lak 14
^(?:https?://)?(?:[a-z0-9-]+\.)*((?:[a-z0-9-]+\.)[a-z]+)
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说明
^ # start-of-line
(?: # begin non-capturing group
https? # "http" or "https"
:// # "://"
)? # end non-capturing group, make optional
(?: # start non-capturing group
[a-z0-9-]+\. # a name part (numbers, ASCII letters, dashes) & a dot
)* # end non-capturing group, match as often as possible
( # begin group 1 (this will be the domain name)
(?: # start non-capturing group
[a-z0-9-]+\. # a name part, same as above
) # end non-capturing group
[a-z]+ # the TLD
) # end group 1
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http://rubular.com/r/g6s9bQpNnC