Capturing part of a url

I'm having some difficulty writing a regular expression. My input will be a url, looking like this:

http://www.a.com/farms/important-stuff-here#ignorable-stuff

I wanted to capture (some-stuff-here), which is everything between the last forward slash, and the first # sign (or just the ending, if the # sign extra content does not exist. I thought this might do it:

(http://www.a.com/farms/)

([anything but a # character]*)

(.*)

I'm not sure how to express the 2nd group ([anything but a # character]*).

开发者_JS百科

Thanks

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"Anything but" is called a negated character class, and, in your case, is spelled

[^#]

Your regex would be

http://www.a.com/farms/([^#]+)

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For most re engines you probably want [^#] (the ^ negates a character class).

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depending on your language, you might want to use modules/libraries that can parse url nicely for you. eg in PHP, you can use parse_url

$url = "http://www.a.com/farms/important-stuff-here#ignorable-stuff";
$parsed = parse_url($url);
print $parsed['path'];

with Python, urlparse() eg:

>>> import urlparse
>>> s=""http://www.a.com/farms/important-stuff-here#ignorable-stuff"
>>> urlparse.urlparse(s).path
'/farms/important-stuff-here'

IF you really want to do it by hand, first replace everything from "#" onwards, then replace everything from the start till "/"

$ echo "http://www.a.com/farms/important-stuff-here#ignorable-stuff" | sed 's/#.*//;s|.*\/||'
important-stuff-here

Or using just plain splits on strings

$url = "http://www.a.com/farms/important-stuff-here#ignorable-stuff";
$s = explode("#",$url,2);
$t = explode("/",$s[0]);
print end($t);