Suppose a user submits a url: The New York Times - Breaking News, US News, World News and Videos. How can this be shortened to a cleaner url, like nyt.com?
require 'uri'
uri = URI.parse("http://www.ruby-lang.org/"\) p uri # => #<URI::HTTP:0x202281be URL:http://www.ruby-lang.org/> p uri.scheme # => "http" p uri.host # => "www.ruby-lang.org"
Sean,
Are you saving this url to the model? I'm not sure what you mean by shorten it up?
--Tom
tomrossi7 wrote:
Sean,
Are you saving this url to the model? I'm not sure what you mean by shorten it up?
--Tom
I want a user to submit a link to a webpage. After they submit the link I want to parse it to the homepage url. Like if the article is www.nyt.com/sports/lakers/123. I want to have this shortened to the homepage link nyt.com. You can try it in the interpreter as someone posted above.
Yeah, URI is the library to work with. You could tighten it up and just do:
homepage = URI.parse("The New York Times - Breaking News, US News, World News and Videos).host
On the model, you could put "require 'uri' " but I think its loaded with Rails by default.
--Tom
I don't think
I have @article.url and I want to parse that url. When I tried: @homepage = URI.parse(@article.url).host in the controller I simply got a blank output.
Can you post a sample @article.url that results in blank .host output from URI.parse?
