Phr*_*ogz 61
最简单的方法是:
在Ruby中编码到基础36 /从基础36解码很简单:
12341235.to_s(36)
#=> "7cik3"
"7cik3".to_i(36)
#=> 12341235
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对基础62的编码有点过时.这是一种方法:
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
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......在这里它正在行动:
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
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编辑
如果您想避免碰巧是英文单词的网址(例如,四个字母的发誓单词),您可以使用一组不包含元音的字符:
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
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然而,有了这个,你仍然有像AnyBase.encode(328059,base31)或AnyBase.encode(345055,base31)或等问题AnyBase.encode(450324,base31).因此,您可能希望避免使用类似元音的数字:
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
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这也可以避免"是0还是O?"的问题.并且"那是1还是我?".
e3m*_*eus 12
我使用了有点宝石.这很简单.
gem install bitly
# Use api version 3 or get a deprecation warning
Bitly.use_api_version_3
# Create a client
bitly = Bitly.new(username, api_key)
# Call method shorten
bitly.shorten('http://www.google.com').short_url
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