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(Visual Basic 6.0) Duplicate PHP RSA EncryptionDemonstrates how to duplicate the following PHP function.
' 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 Dim rsaMOD As String rsaMOD = "F0946D8F05604809E24B8CFFD30349CEA9E5F4D320BFD9E9AA1B088863F02C43E7997D37A3E27B4F8F359F1744DB6B20A437067C0D325A80660D12FF56A57673" ' The RSA exponent in hex is "010001", which is 65537 in decimal. It's typically the exponent that is always used. Dim rsaEXP As String rsaEXP = "010001" ' Get the RSA modulus and exponent in base64. Dim bdMod As New ChilkatBinData Dim bdExp As New ChilkatBinData Dim success As Long success = bdMod.AppendEncoded(rsaMOD,"hex") success = bdExp.AppendEncoded(rsaEXP,"hex") ' Build the XML representation of the RSA public key Dim xml As New ChilkatXml 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. Dim pubkey As New PublicKey success = pubkey.LoadFromString(xml.GetXml()) ' Setup the RSA object for encryption and do it.. Dim rsa As New ChilkatRsa rsa.VerboseLogging = 1 success = rsa.ImportPublicKeyObj(pubkey) ' Use PKCSv1.5 padding rsa.OaepPadding = 0 ' Encrypt and return the string as hex. rsa.EncodingMode = "hex" Dim plainText As String plainText = "key=abcdefghijkmnopq&iv=abcdefghijkmnopq&h=12345678&s=12345678" Dim cipherText As String cipherText = rsa.EncryptStringENC(plainText,0) If (rsa.LastMethodSuccess = 0) Then Debug.Print rsa.LastErrorText Exit Sub 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. Debug.Print cipherText |
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