用于在斜杠上拆分的正则表达式

Question

I am trying to split URLs to get the domain name.

example.com                => example.com
example.com/dir/index.html => example.com

The regular expression I am trying to us is

(.+?)(/|$)

When I use it in python like this:

import re
m = re.search('(.+?)(/|$)', url)

It works for the first one, but for the second example I always get example.com/ . How do I get rid of the backslash?

edit: I am very sorry, I forgot to include one important information. I need a regular expression, because I need to write this in Oracle SQL. Fortunately, Oracle supports regex, but nothing like urlparse . I am just using python for testing. Sorry about that!

Answer 1

The easy way to do this is to use the urlparse function in the stdlib:

>>> from urllib.parse import urlparse
>>> url = 'http://example.com/dir/index.html'
>>> p = urlparse(url)
>>> p.netloc
'example.com'

Besides being a whole lot simpler, it handles cases that you haven't thought of in a well-defined and clearly-documented way (eg, what if there's a port as well as a host?), whereas with your code, who knows what happens with any cases you didn't anticipate?

---

If you really want to treat the URL as a string instead of a URL, the easy way to split on slashes is to split on slashes:

>>> bits = url.split('/')
>>> bits[2]
example.com

---

If you really want to use regexps to split on slashes, you could use re.split instead of trying to figure out a way to trick re.search into splitting for you:

>>> bits = re.split('/', url)
>>> bits[2]
example.com

---

Finally, if you want to do it with match or search , and you don't want to capture the / , don't put the / in a capturing group, and look at the group you went out of your way to capture instead of at the whole string:

>>> url = 'example.com/dir/index.html'
>>> m = re.search('(.+?)(/|$)', url)
>>> m.groups()
('example.com', '/')
>>> m = re.search('(.+?)(?:/|$)', url)
>>> m.groups()
('example.com',)

Answer 2

尝试匹配非froward斜杠，如([^/]+?)(/|$)