我在网上下了一段rsa加密程序,看不大懂,请高手进来给看看(asp实现)

ebombsuhocom 2003-08-23 12:16:22
rsatest.asp
<%
rem 文章标题:在asp中通过vbs类实现rsa加密与解密
rem 收集整理:yanek
rem 联系:aspboy@263.net

%>
<%Option Explicit%>
<!--#INCLUDE FILE="clsRSA.asp"-->
<%

Dim LngKeyE
Dim LngKeyD
Dim LngKeyN
Dim StrMessage
Dim ObjRSA
If Not Request.Form = "" Then

LngKeyE = Request.Form("KeyE")
LngKeyD = Request.Form("KeyD")
LngKeyN = Request.Form("KeyN")
StrMessage = Request.Form("Message")

Set ObjRSA = New clsRSA

Select Case Request.Form("Action")
Case "Generate Keys"
Call ObjRSA.GenKey()
LngKeyE = ObjRSA.PublicKey
LngKeyD = ObjRSA.PrivateKey
LngKeyN = ObjRSA.Modulus
Case "Encrypt"
ObjRSA.PublicKey = LngKeyE
ObjRSA.Modulus = LngKeyN
StrMessage = ObjRSA.Encode(StrMessage)
Case "Decrypt"
ObjRSA.PrivateKey = LngKeyD
ObjRSA.Modulus = LngKeyN
StrMessage = ObjRSA.Decode(StrMessage)
End Select

Set ObjRSA = Nothing

End If
%>
<HTML>
<HEAD>
<TITLE>RSA Cipher Demonstration</TITLE>
</HEAD>
<BODY>
<H1>RSA Cipher Demonstration</H1>
<P>
You will first need to generate your public/privage key-pair
before you can encrypt/decrypt messages.
</P>
<FORM method="post">
<TABLE>
<TR>
<TD>Public Key</TD>
<TD><INPUT name="KeyE" value="<%=Server.HTMLEncode(LngKeyE)%>"></TD>
<TD rowspan="3">
<INPUT type="Submit" name="Action" value="Generate Keys">
</TD>
</TR>
<TR>
<TD>Private Key</TD>
<TD><INPUT name="KeyD" value="<%=Server.HTMLEncode(LngKeyD)%>"></TD>
</TR>
<TR>
<TD>Modulus</TD>
<TD><INPUT name="KeyN" value="<%=Server.HTMLEncode(LngKeyN)%>"></TD>
</TR>
<TR>
<TD colspan="3">
Test Message:<BR>
<TEXTAREA name="Message" cols="50" rows="7"><%=Server.HTMLEncode(StrMessage)%></TEXTAREA>
</TD>
</TR>
<TR>
<TD align="right" colspan="3">
<INPUT type="Submit" name="Action" value="Encrypt">
<INPUT type="Submit" name="Action" value="Decrypt">
</TD>
</TR>
</TABLE>
</FORM>
</BODY>
</HTML>




clsrsa.asp
<%
rem 实现rsa加密与解密的vbs类文件
rem 文章标题:在asp中通过vbs类实现rsa加密与解密
rem 收集整理:yanek
rem 联系:aspboy@263.net

' RSA Encryption Class
'
' .PrivateKey
' Your personal private key. Keep this hidden.
'
' .PublicKey
' Key for others to encrypt data with.
'
' .Modulus
' Used with both public and private keys when encrypting
' and decrypting data.
'
' .GenKey()
' Creates Public/Private key set and Modulus
'
' .Crypt(pLngMessage, pLngKey)
' Encrypts/Decrypts message and returns
' as a string.
'
' .Encode(pStrMessage)
' Encrypts message and returns in double-hex format
'
' .Decode(pStrMessage)
' Decrypts message from double-hex format and returns a string
'
Class clsRSA

Public PrivateKey
Public PublicKey
Public Modulus

Public Sub GenKey()
Dim lLngPhi
Dim q
Dim p

Randomize

Do
Do

' 2 random primary numbers (0 to 1000)
Do
p = Rnd * 1000 \ 1
Loop While Not IsPrime(p)
Do
q = Rnd * 1000 \ 1
Loop While Not IsPrime(q)
'p,p是两个1 到1000的素数

' n = product of 2 primes
Modulus = p * q \ 1

' random decryptor (2 to n)
PrivateKey = Rnd * (Modulus - 2) \ 1 + 2

lLngPhi = (p - 1) * (q - 1) \ 1
PublicKey = Euler(lLngPhi, PrivateKey)

Loop While PublicKey = 0 Or PublicKey = 1

' Loop if we can't crypt/decrypt a byte
Loop While Not TestCrypt(255)

End Sub

Private Function TestCrypt(ByRef pBytData)
Dim lStrCrypted
lStrCrypted = Crypt(pBytData, PublicKey)
TestCrypt = Crypt(lStrCrypted, PrivateKey) = pBytData
End Function

Private Function Euler(ByRef pLngPHI, ByRef pLngKey)

Dim lLngR(3)
Dim lLngP(3)
Dim lLngQ(3)

Dim lLngCounter
Dim lLngResult

Euler = 0

lLngR(1) = pLngPHI: lLngR(0) = pLngKey
lLngP(1) = 0: lLngP(0) = 1
lLngQ(1) = 2: lLngQ(0) = 0

lLngCounter = -1

Do Until lLngR(0) = 0

lLngR(2) = lLngR(1): lLngR(1) = lLngR(0)
lLngP(2) = lLngP(1): lLngP(1) = lLngP(0)
lLngQ(2) = lLngQ(1): lLngQ(1) = lLngQ(0)

lLngCounter = lLngCounter + 1

lLngR(0) = lLngR(2) Mod lLngR(1)
lLngP(0) = ((lLngR(2)\lLngR(1)) * lLngP(1)) + lLngP(2)
lLngQ(0) = ((lLngR(2)\lLngR(1)) * lLngQ(1)) + lLngQ(2)

Loop

lLngResult = (pLngKey * lLngP(1)) - (pLngPHI * lLngQ(1))

If lLngResult > 0 Then
Euler = lLngP(1)
Else
Euler = Abs(lLngP(1)) + pLngPHI
End If

End Function

Public Function Crypt(pLngMessage, pLngKey)
On Error Resume Next
Dim lLngMod
Dim lLngResult
Dim lLngIndex
If pLngKey Mod 2 = 0 Then
lLngResult = 1
For lLngIndex = 1 To pLngKey / 2
lLngMod = (pLngMessage ^ 2) Mod Modulus
' Mod may error on key generation
lLngResult = (lLngMod * lLngResult) Mod Modulus
If Err Then Exit Function
Next
Else
lLngResult = pLngMessage
For lLngIndex = 1 To pLngKey / 2
lLngMod = (pLngMessage ^ 2) Mod Modulus
On Error Resume Next
' Mod may error on key generation
lLngResult = (lLngMod * lLngResult) Mod Modulus
If Err Then Exit Function
Next
End If
Crypt = lLngResult
End Function

Private Function IsPrime(ByRef pLngNumber)
Dim lLngSquare
Dim lLngIndex
IsPrime = False
If pLngNumber < 2 Then Exit Function
If pLngNumber Mod 2 = 0 Then Exit Function
lLngSquare = Sqr(pLngNumber)
For lLngIndex = 3 To lLngSquare Step 2
If pLngNumber Mod lLngIndex = 0 Then Exit Function
Next
IsPrime = True
End Function

Public Function Encode(ByVal pStrMessage)
Dim lLngIndex
Dim lLngMaxIndex
Dim lBytAscii
Dim lLngEncrypted
lLngMaxIndex = Len(pStrMessage)
If lLngMaxIndex = 0 Then Exit Function
For lLngIndex = 1 To lLngMaxIndex
lBytAscii = Asc(Mid(pStrMessage, lLngIndex, 1))
lLngEncrypted = Crypt(lBytAscii, PublicKey)
Encode = Encode & NumberToHex(lLngEncrypted, 4)
Next
End Function

Public Function Decode(ByVal pStrMessage)
Dim lBytAscii
Dim lLngIndex
Dim lLngMaxIndex
Dim lLngEncryptedData
Decode = ""
lLngMaxIndex = Len(pStrMessage)
For lLngIndex = 1 To lLngMaxIndex Step 4
lLngEncryptedData = HexToNumber(Mid(pStrMessage, lLngIndex, 4))
lBytAscii = Crypt(lLngEncryptedData, PrivateKey)
Decode = Decode & Chr(lBytAscii)
Next
End Function

Private Function NumberToHex(ByRef pLngNumber, ByRef pLngLength)
NumberToHex = Right(String(pLngLength, "0") & Hex(pLngNumber), pLngLength)
End Function

Private Function HexToNumber(ByRef pStrHex)
HexToNumber = CLng("&h" & pStrHex)
End Function

End Class




%>
...全文
46 3 打赏 收藏 转发到动态 举报
写回复
用AI写文章
3 条回复
切换为时间正序
请发表友善的回复…
发表回复
寻开心 2003-09-03
  • 打赏
  • 举报
回复
http://soft.winzheng.com/InfoView/Article_122.htm
http://soft.winzheng.com/InfoView/Article_123.htm
这两个文章是介绍RSA的,供参考。
LLnju 2003-09-03
  • 打赏
  • 举报
回复
这也能算是 RSA 加密吗? p , q 怎么也得 512b 吧 , 你这个就跟没有加密一样
WYlslrt 2003-09-02
  • 打赏
  • 举报
回复
不懂,帮你顶吧

33,008

社区成员

发帖
与我相关
我的任务
社区描述
数据结构与算法相关内容讨论专区
社区管理员
  • 数据结构与算法社区
加入社区
  • 近7日
  • 近30日
  • 至今
社区公告
暂无公告

试试用AI创作助手写篇文章吧