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从php到ruby的加密算法(vignere变体)

code-translation ruby php

J. Martin · 技术社区 · 15 年前

我有点受不了。我必须与一个API接口,该API使用的加密算法版本似乎是从AriKuorikoski编写的typo3中提取出来的。

我需要创建一个Ruby库来与他们的API接口,所以必须将他们的算法移植到Ruby中,而在加密方面,我有点不知所措。

这是代码:

private function keyED($txt) { 
$encrypt_key = md5($this->encrypt_key); 
$ctr=0; 
$tmp = ""; 
for ($i=0;$i<strlen($txt);$i++) { 
   if ($ctr==strlen($encrypt_key)) $ctr=0; 
   $tmp.= substr($txt,$i,1) ^ substr($encrypt_key,$ctr,1); 
   $ctr++; 
} 
return $tmp; 
} 

private function encrypt($txt){ 
srand((double)microtime()*1000000); 
$encrypt_key = md5(rand(0,32000)); 
$ctr=0; 
$tmp = ""; 
for ($i=0;$i<strlen($txt);$i++){ 
   if ($ctr==strlen($encrypt_key)) $ctr=0; 
   $tmp.= substr($encrypt_key,$ctr,1) . 
       (substr($txt,$i,1) ^ substr($encrypt_key,$ctr,1)); 
   $ctr++; 
} 
return base64_encode($this->keyED($tmp)); 
}

让我头疼的是,我必须为Ruby1.8.6编写它,因为这是它将要使用的服务器。字符串没有XOR。如果有的话,我也不会理解。

任何帮助、指点、想法都会受到赞赏。

附录:

我意识到,我没有添加任何代码,唯一的困难实际上是XOR问题,但这里是我迄今为止的代码:

def xor(s1,s2)
        if s2.empty? then
            return s1
        else
            a1 = s1.unpack("c*")
            a2 = s2.unpack("c*")
            a2 *= 2 while a2.length < a1.length
            return a1.zip(a2).collect {|c1,c2| c1 ^ c2}.pack("c*")
        end
    end
    def keyED(str)
        encrypt_key = Digest::MD5.digest(@key)
        ctr = 0
        tmp = ''
        for i in 0...str.length do
             ctr = 0 if ctr == encrypt_key.length
             tmp << xor(str.slice(i,1), encrypt_key.slice(ctr,1)).to_s
            ctr = ctr + 1
        end
        return tmp
    end

    # === Ported Code
    # This code was ported from Ari's Typo 3 Session Encryption
    def encrypt(str)
        encrypt_key = Digest::MD5.digest(rand(32000).to_s)
        ctr = 0
        tmp = ''
        for i in 0...str.length do
            ctr=0 if ctr==encrypt_key.length 
            tmp << encrypt_key.slice(ctr,1) << xor(str.slice(i,1), encrypt_key.slice(ctr,1))
            ctr = ctr + 1
        end
        return Base64.encode64(keyED(tmp))
    end
    def decrypt(str)
        txt = keyED(str)
        tmp = ''
        for i in 0...txt.length do 
            md = txt.slice(i,1)
            i = i + 1
            tmp << xor(txt.slice(i,1),md)
        end
        puts "Decrypte string:#{Base64.decode64(tmp)}EOSTRING"
    end

2 回复 | 直到 9 年前

Community CDub 7 年前

从 Erik Veenstra :

class String
  def xor(other)
    if other.empty?
      self
    else
      a1        = self.unpack("c*")
      a2        = other.unpack("c*")
      a2 *= 2   while a2.length < a1.length
      a1.zip(a2).collect{|c1,c2| c1^c2}.pack("c*")
    end
  end
end

然后你的代码变成

tmp << str.slice(i,1).xor(encrypt_key.slice(i,1))

替代的实现 String#xor ,根据建议 jolierouge 和 David Garamond :

class String
  def xor(other)
    raise ArgumentError, "Can't bitwise-XOR a String with a non-String" unless other.kind_of? String
    raise ArgumentError, "Can't bitwise-XOR strings of different length" unless self.length == other.length
    (0..self.length-1).collect { |i| self[i] ^ other[i] }.pack("C*")
  end
end

JasonMArcher 9 年前

编者按:最后的工作代码从问题转移到了自己的答案。

最后的工作来源,基于詹姆斯的有用答案,主要道具!还有大卫·加拉蒙德。

def xor(s1,s2)
    raise ArgumentError, "Can't bitwise-XOR a String with a non-String" unless s2.kind_of? String
    raise ArgumentError, "Can't bitwise-XOR strings of different length" unless s1.length == s2.length
    (0..s1.length-1).collect { |i| s1[i] ^ s2[i] }.pack("C*")
end

def keyED(txt,key)
    ctr,tmp = 0,''
    key = Digest::MD5.hexdigest(key)
    for i in 0...txt.length do
        ctr = 0 if ctr == key.length
        str = xor(txt.slice(i,1),key.slice(ctr,1))
        tmp << str
        ctr = ctr + 1
    end

    return tmp
end
def encrypt(txt,key)
    ctr,tmp = 0,''
    ekey = Digest::MD5.hexdigest(rand(32000).to_s)
    for i in 0...txt.length do
        ctr = 0 if ctr == ekey.length
        str = xor(txt.slice(i,1), ekey.slice(ctr,1))
        tmp << ekey.slice(ctr,1) << str
        ctr = ctr + 1
    end
    return Base64.encode64(keyED(tmp,key))
end