11 - Extract the domain name from a URL

Don't say so much, just coding...

Instruction

Write a function that when given a URL as a string, parses out just the domain name and returns it as a string. For example:

  domain_name("http://github.com/carbonfive/raygun") == "github" 
  domain_name("http://www.zombie-bites.com") == "zombie-bites"
  domain_name("https://www.cnet.com") == "cnet"

Ruby

Init

  def domain_name(url)
    # return just the domain name
  end

Sample Testing

  Test.assert_equals(domain_name("http://google.com"), "google")
  Test.assert_equals(domain_name("http://google.co.jp"), "google")
  Test.assert_equals(domain_name("www.xakep.ru"), "xakep")
  Test.assert_equals(domain_name("https://youtube.com"), "youtube")

Javascript

Init

  function domainName(url){
    //your code here
  }

Sample Testing

  Test.assertEquals(domainName("http://google.com"), "google");
  Test.assertEquals(domainName("http://google.co.jp"), "google");
  Test.assertEquals(domainName("www.xakep.ru"), "xakep");
  Test.assertEquals(domainName("https://youtube.com"), "youtube");

Thinking

想法(1): 原本其實是從 URI 的方式寫完的，不過發現要處理的東西有點多，後來覺得 start_with 來解決就可以，也不用 regex 去 metch 所有條件（regex 就爛）
想法(2): 用 gsub 把所有要篩選的條件寫進來也不錯

圖片來源：Unsplash Marvin Meyer

Hint & Reference

• Ruby
  • Ruby - apidock Start_with?
  • Ruby - apidock Gsub
  • Ruby - rubydocs URI
• JavaScript
  • JS - MDN String.prototype.replace()

Solution

Ruby

  # Solution 1
  def domain_name(url)
    url = url.start_with?('www.') ? url.split('www.').last : url.split('//').last
    url.split('.').first
  end
  
  # Solution 2
  def domain_name(url)
    url.start_with?('www.') ? url.split('www.').last.split('.').first
                            : url.split('//').last.split('.').first
  end
  
  # Solution 3
  def domain_name(url)
    url.gsub('http://', '')
       .gsub('https://', '')
       .gsub('www.', '')
       .split(".")[0]
  end

Javascript

  // Solution 1 gsub to replace XDDD
  function domainName(url){
    return url.replace('http://', '')
              .replace('https://', '')
              .replace('www.', '')
              .split('.')[0];
  }