日期:2008-10-10  浏览次数:20944 次

在.net中,由于.net framework 封装了常见的加密算法,因此实现标准的MD5算法只需要短短几行代码即可实现:

 public string Md5(string strPassword)

{



MD5CryptoServiceProvider hashmd5;

  hashmd5 = new MD5CryptoServiceProvider();

return BitConverter.ToString(hashmd5.ComputeHash(Encoding.Default.GetBytes(strPassword))).Replace("-","").ToLower();

}

,但是当加密字符串含有中文或者其它双字节字符时,这种算法的结果与目前网上流行的ASP写的MD5算法的结果却不一致,这主要是由于目前网上流行的ASP写的MD5加密算法,存在一个缺陷,它使用了mid函数,取出的是“字符”,而正确的做法应该是取出字节,因此当加密字符串有双字节字符时,结果会与标准的MD5算法不一致.。但是由于在ASP向ASP.net的系统进行升级的过程中,已经向数据库内写入了大量以前ASP算法加密的密码,为了使新系统能够与原来的系统完全兼容,因此只有在.net 环境下实现与原来ASP算法完全一致的MD5算法。我的实现思路如下:

1、先用正则表达式对要加密的字符串进行判断,是否含有双字节字符。

2、如果含有双字节字符,调用实现与ASP算法一致的非托管COM组件进行处理。(这个组件只需要将原ASP加密算法代码进行适当修改,然后加入VB6中进行编译即可得到)。

3、否则直接调用.net的标准MD5加密算法。

可能有的朋友会问到,为什么不都直接调用COM组件来加密呢?这主要是从性能角度考虑的,由于调用非托管COM组件,.net需要生成一个代理类与之交互,所以在性能方面要远远低于.net内置的函数。

下面给出所有代码:

原ASP加密代码:

<%



Private Const BITS_TO_A_BYTE = 8

Private Const BYTES_TO_A_WORD = 4

Private Const BITS_TO_A_WORD = 32

Private m_lOnBits(30)

Private m_l2Power(30)



m_lOnBits(0) = CLng(1)

m_lOnBits(1) = CLng(3)

m_lOnBits(2) = CLng(7)

m_lOnBits(3) = CLng(15)

m_lOnBits(4) = CLng(31)

m_lOnBits(5) = CLng(63)

m_lOnBits(6) = CLng(127)

m_lOnBits(7) = CLng(255)

m_lOnBits(8) = CLng(511)

m_lOnBits(9) = CLng(1023)

m_lOnBits(10) = CLng(2047)

m_lOnBits(11) = CLng(4095)

m_lOnBits(12) = CLng(8191)

m_lOnBits(13) = CLng(16383)

m_lOnBits(14) = CLng(32767)

m_lOnBits(15) = CLng(65535)

m_lOnBits(16) = CLng(131071)

m_lOnBits(17) = CLng(262143)

m_lOnBits(18) = CLng(524287)

m_lOnBits(19) = CLng(1048575)

m_lOnBits(20) = CLng(2097151)

m_lOnBits(21) = CLng(4194303)

m_lOnBits(22) = CLng(8388607)

m_lOnBits(23) = CLng(16777215)

m_lOnBits(24) = CLng(33554431)

m_lOnBits(25) = CLng(67108863)

m_lOnBits(26) = CLng(134217727)

m_lOnBits(27) = CLng(268435455)

m_lOnBits(28) = CLng(536870911)

m_lOnBits(29) = CLng(1073741823)

m_lOnBits(30) = CLng(2147483647)



m_l2Power(0) = CLng(1)

m_l2Power(1) = CLng(2)

m_l2Power(2) = CLng(4)

m_l2Power(3) = CLng(8)

m_l2Power(4) = CLng(16)

m_l2Power(5) = CLng(32)

m_l2Power(6) = CLng(64)

m_l2Power(7) = CLng(128)

m_l2Power(8) = CLng(256)

m_l2Power(9) = CLng(512)

m_l2Power(10) = CLng(1024)

m_l2Power(11) = CLng(2048)

m_l2Power(12) = CLng(4096)

m_l2Power(13) = CLng(8192)

m_l2Power(14) = CLng(16384)

m_l2Power(15) = CLng(32768)

m_l2Power(16) = CLng(65536)

m_l2Power(17) = CLng(131072)

m_l2Power(18) = CLng(262144)

m_l2Power(19) = CLng(524288)

m_l2Power(20) = CLng(1048576)

m_l2Power(21) = CLng(2097152)

m_l2Power(22) = CLng(4194304)

m_l2Power(23) = CLng(8388608)

m_l2Power(24) = CLng(16777216)

m_l2Power(25) = CLng(33554432)

m_l2Power(26) = CLng(67108864)

m_l2Power(27) = CLng(134217728)

m_l2Power(28) = CLng(268435456)

m_l2Power(29) = CLng(536870912)

m_l2Power(30) = CLng(1073741824)

Private Function LShift(lValue, iShiftBits)

If iShiftBits = 0 Then

LShift = lValue

Exit Function

ElseIf iShiftBits = 31 Then

If lValue And 1 Then

LShift = &H80000000

Else

LShift = 0

End If

Exit Function

ElseIf iShiftBits < 0 Or iShiftBits > 31 Then

Err.Raise 6

End If

If (lValue And m_l2Power(31 - iShiftBits)) Then

LShift = ((lValue And m_lOnBits(31 - (iShiftBits + 1))) * m_l2Power(iShiftBits)) Or &H80000000

Else

LShift = ((lValue And m_lOnBits(31 - iShiftBits)) * m_l2Power(iShiftBits))