问题
Given any valid HTTP/HTTPS string, I would like to parse/transform it such that the end result is exactly the root of the string.
So given URLs:
http://foo.example.com:8080/whatsit/foo.bar?x=y
https://example.net/
I would like the results:
http://foo.example.com:8080/
https://example.net/
I found the documentation for URI::Parser not super approachable.
My initial, naïve solution would be a simple regex like:
/\A(https?:\/\/[^\/]+\/)/
(That is: Match up to the first slash after the protocol.)
Thoughts & solutions welcome. And apologies if this is a duplicate, but my search results weren't relevant.
回答1:
With URI::join:
require 'uri'
url = "http://foo.example.com:8080/whatsit/foo.bar?x=y"
baseurl = URI.join(url, "/").to_s
#=> "http://foo.example.com:8080/"
回答2:
Use URI.parse and then set the path to an empty string and the query to nil:
require 'uri'
uri = URI.parse('http://foo.example.com:8080/whatsit/foo.bar?x=y')
uri.path = ''
uri.query = nil
cleaned = uri.to_s # http://foo.example.com:8080
Now you have your cleaned up version in cleaned. Taking out what you don't want is sometimes easier than only grabbing what you need.
If you only do uri.query = '' you'll end up with http://foo.example.com:8080? which probably isn't what you want.
回答3:
You could use uri.split() and then put the parts back together...
WARNING: It's a little sloppy.
url = "http://example.com:9001/over-nine-thousand"
parts = uri.split(url)
puts "%s://%s:%s" % [parts[0], parts[2], parts[3]]
=> "http://example.com:9001"
来源：https://stackoverflow.com/questions/6755919/ruby-rails-3-1-given-a-url-string-remove-path