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Classic ASP

Duplicate PHP RSA Encryption

See more RSA Examples

Demonstrates how to duplicate the following PHP function.

Chilkat Classic ASP Downloads

Classic ASP
<html>
<head>
<meta http-equiv="Content-Type" content="text/html; charset=utf-8">
</head>
<body>
<%
success = 0

' This example requires the Chilkat API to have been previously unlocked.
' See Global Unlock Sample for sample code.

' Duplicate the following PHP code:
' 
'    public function encryptRSA($plainText,$rsaMOD,$pubKEY){
'         $rsa  = new RSA();
'         $rsa->setEncryptionMode(RSA::ENCRYPTION_PKCS1);
'         $publicKey = [
'             'e' => new BigInteger($pubKEY,16),
'             'n' => new BigInteger($rsaMOD,16)
'         ];
' 		
'         $rsa->loadKey($publicKey);
'         $ciphertext = $rsa->encrypt($plainText);
'         return bin2hex($ciphertext);
'     }
' 
'     $plainText="key=abcdefghijkmnopq&iv=abcdefghijkmnopq&h=12345678&s=12345678"
'     $rsaMOD="F0946D8F05604809E24B8CFFD30349CEA9E5F4D320BFD9E9AA1B088863F02C43E7997D37A3E27B4F8F359F1744DB6B20A437067C0D325A80660D12FF56A57673"
'     $pubKEY="010001"

' We have the RSA modulus in hex
rsaMOD = "F0946D8F05604809E24B8CFFD30349CEA9E5F4D320BFD9E9AA1B088863F02C43E7997D37A3E27B4F8F359F1744DB6B20A437067C0D325A80660D12FF56A57673"

' The RSA exponent in hex is "010001", which is 65537 in decimal.  It's typically the exponent that is always used.
rsaEXP = "010001"

' Get the RSA modulus and exponent in base64.
set bdMod = Server.CreateObject("Chilkat.BinData")
set bdExp = Server.CreateObject("Chilkat.BinData")
success = bdMod.AppendEncoded(rsaMOD,"hex")
success = bdExp.AppendEncoded(rsaEXP,"hex")

' Build the XML representation of the RSA public key
set xml = Server.CreateObject("Chilkat.Xml")
xml.Tag = "RSAPublicKey"
xml.UpdateChildContent "Modulus",bdMod.GetEncoded("base64")
xml.UpdateChildContent "Exponent",bdExp.GetEncoded("base64")

' Load the RSA public key into a Chilkat public key object.
set pubkey = Server.CreateObject("Chilkat.PublicKey")
success = pubkey.LoadFromString(xml.GetXml())

' Setup the RSA object for encryption and do it..
set rsa = Server.CreateObject("Chilkat.Rsa")
rsa.VerboseLogging = 1
success = rsa.UsePublicKey(pubkey)

' Use PKCSv1.5 padding
rsa.PkcsPadding = 1

' Encrypt and return the string as hex.
rsa.EncodingMode = "hex"
plainText = "key=abcdefghijkmnopq&iv=abcdefghijkmnopq&h=12345678&s=12345678"
cipherText = rsa.EncryptStringENC(plainText,0)
If (rsa.LastMethodSuccess = 0) Then
    Response.Write "<pre>" & Server.HTMLEncode( rsa.LastErrorText) & "</pre>"
    Response.End
End If

' Note: The PKCSv1_5 padding incorporates random bytes.  Therefore, the RSA encryption will produce different results each time -- all of which are valid 
' and decrypt correctly to the same original text.
Response.Write "<pre>" & Server.HTMLEncode( cipherText) & "</pre>"

%>
</body>
</html>