Welcome to OStack Knowledge Sharing Community for programmer and developer-Open, Learning and Share
Welcome To Ask or Share your Answers For Others

Categories

0 votes
427 views
in Technique[技术] by (71.8m points)

ruby - How to implement a short URL like the URLs in Twitter?

If there is a long url, I want to generate a short URL like those in Twitter. Is there some way to implement this in Ruby?

See Question&Answers more detail:os

与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
Welcome To Ask or Share your Answers For Others

1 Answer

0 votes
by (71.8m points)

The easiest way is to:

  1. keep a database of all URLs
  2. when you insert a new URL into the database, find out the id of the auto-incrementing integer primary key.
  3. encode that integer into base 36 or 62 (digits + lowercase alpha or digits + mixed-case alpha). Voila! You have a short url!

Encoding to base 36/decoding from base 36 is simple in Ruby:

12341235.to_s(36)
#=> "7cik3"

"7cik3".to_i(36)
#=> 12341235

Encoding to base 62 is a bit tricker. Here's one way to do it:

module AnyBase
  ENCODER = Hash.new do |h,k|
    h[k] = Hash[ k.chars.map.with_index.to_a.map(&:reverse) ]
  end
  DECODER = Hash.new do |h,k|
    h[k] = Hash[ k.chars.map.with_index.to_a ]
  end
  def self.encode( value, keys )
    ring = ENCODER[keys]
    base = keys.length
    result = []
    until value == 0
      result << ring[ value % base ]
      value /= base
    end
    result.reverse.join
  end
  def self.decode( string, keys )
    ring = DECODER[keys]
    base = keys.length
    string.reverse.chars.with_index.inject(0) do |sum,(char,i)|
      sum + ring[char] * base**i
    end
  end
end

...and here it is in action:

base36 = "0123456789abcdefghijklmnopqrstuvwxyz"
db_id = 12341235
p AnyBase.encode( db_id, base36 )
#=> "7cik3"
p AnyBase.decode( "7cik3", base36 )
#=> 12341235

base62 = [ *0..9, *'a'..'z', *'A'..'Z' ].join
p AnyBase.encode( db_id, base62 )
#=> "PMwb"
p AnyBase.decode( "PMwb", base62 )
#=> 12341235

Edit

If you want to avoid URLs that happen to be English words (for example, four-letter swear words) you can use a set of characters that does not include vowels:

base31 = ([*0..9,*'a'..'z'] - %w[a e i o u]).join
base52 = ([*0..9,*'a'..'z',*'A'..'Z'] - %w[a e i o u A E I O U]).join

However, with this you still have problems like AnyBase.encode(328059,base31) or AnyBase.encode(345055,base31) or AnyBase.encode(450324,base31). You may thus want to avoid vowel-like numbers as well:

base28 = ([*'0'..'9',*'a'..'z'] - %w[a e i o u 0 1 3]).join
base49 = ([*'0'..'9',*'a'..'z',*'A'..'Z'] - %w[a e i o u A E I O U 0 1 3]).join

This will also avoid the problem of "Is that a 0 or an O?" and "Is that a 1 or an I?".


与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
Welcome to OStack Knowledge Sharing Community for programmer and developer-Open, Learning and Share
Click Here to Ask a Question

...