am new to ruby
using regular expression .how can i remove https and http and www from a string
server= http://france24.miles.com
server= https://seloger.com
from these sites i want to remove all http ,https and www
france24.miles.com
seloger.com
i used following code but it is not woking for me
server = server.(/^https?\:\/\/(www.)?/,'')
server = server.(/^https?\:\/\/(www.)?/,'')
This didn't work, because you aren't calling a method of the string server. Make sure you call the sub method:
server = server.sub(/^https?\:\/\/(www.)?/,'')
Example
> server = "http://www.stackoverflow.com"
> server = server.sub(/^https?\:\/\/(www.)?/,'')
stackoverflow.com
As per the requirement if you want it to work with the illegal format http:\\ as well, use the following regex:
server.sub(/https?\:(\\\\|\/\/)(www.)?/,'')
http:\\ is simply wrong. You are doing your users a disservice if you accept such a URL.Std-lib URI is dedicated for such kind of work. Using this would be simpler and may be more reliable
require 'uri'
uri = URI.parse("http://www.ruby-lang.org/")
uri.host
=> "www.ruby-lang.org"
uri.host.sub(/\Awww\./, '')
=> "ruby-lang.org"
uri = URI.parse(server) uri.host content = Net::HTTP.get(uri.host.gsub('\Awww\.', ''), '/status') but getting bad URI(is not URI?): http:\\ingen.twosmiles.com
Net::HTTP.get expect a full url with protocol http or https etc. If this is your usage, you should not remove 'http'www.ruby-lang.org string back unless I do uri.host.sub(/\Awww\./, '')See the String#sub(...) method.
Also, consider using the %r{...} literal notation for Regexp objects so that forward-slashes (/) are easier to recognize:
def trim_url(str)
str.sub %r{^https?:(//|\\\\)(www\.)?}i, ''
end
trim_url 'https://www.foo.com' # => "foo.com"
trim_url 'http://www.foo.com' # => "foo.com"
trim_url 'http://foo.com' # => "foo.com"
trim_url 'http://foo.com' # => "foo.com"
Here is what each part of the regular expression means:
%r{^https?:(//|\\\\)(www\.)?}
# │├──┘├┘│├───────┘ ├─┘├┘ └── everything in the group (...), or nothing.
# ││ │ ││ │ └── the period character "."
# ││ │ ││ └── the letters "www".
# ││ │ │└── the characters "//" or "\\".
# ││ │ └── the colon character ":".
# ││ └── the letter "s", or nothing.
# │└── the letters "http".
# └── the beginning of the line.
def strip_url(url)
return url.to_s.sub!(/https?(\:)?(\/)?(\/)?(www\.)?/, '') if url.include?("http")
url.to_s.sub!(/(www\.)?/,'') if url.include?("www")
end
This will change in place the provided url, stripped of any leading http(s) or www. It covers the following formats:
http://www.example.comhttp:/www.example.comhttp:www.example.comhttps://www.example.comhttps:/www.example.comhttps:www.example.comhttp://example.comhttp:/example.comhttp:example.comhttps://example.comhttps:/example.comhttps:example.comwww.example.comexample.comYou'll end up with example.com using this method.
With this regex: server\s*=\s*\Khttps?://(?:www\.)?
In Ruby 2.0+
result = subject.gsub(/server\s*=\s*\Khttps?:\/\/(?:www\.)?/, '')
In the demo, see the replacements at the bottom.
Hang tight for explanation. :)
Explanation
server\s*=\s* matches server= with optional spaces, to make sure we are looking at the right strings\K tells the engine to drop what was matched so far from the final matchhttps? matches http with an optional s:// matches these literal characters(?:www\.)? matches an optional www.Earlier Versions of Ruby
\K is only supported from Ruby 2.0+. Earlier versions have to use a lookbehind:
result = subject.gsub(/(?:(?<=server=)|(?<=server= ))https?:\/\/(?:www\.)?/, '')
