RijndaelManaged Decryption From SQL Server

Jan 17, 2008

Subject:RijndaelManaged decryption from SQL Server

Issue: I am trying to move the decryption code from C# to SQL Server 2005.


Web.config
...

<symmetricCryptoProviders>


<add algorithmType="System.Security.Cryptography.RijndaelManaged, mscorlib, Version=2.0.0.0, Culture=neutral, PublicKeyToken=xxxxxx" protectedKeyFilename="C:Keysfffff.key" protectedKeyProtectionScope="LocalMachine" type="Microsoft.Practices.EnterpriseLibrary.Security.Cryptography.SymmetricAlgorithmProvider, Microsoft.Practices.EnterpriseLibrary.Security.Cryptography, Version=2.0.0.0, Culture=neutral, PublicKeyToken=null" name="RijndaelManaged"/>

</symmetricCryptoProviders>
...

.cs file
....

decString = Cryptographer.DecryptSymmetric("RijndaelManaged", encString);
....

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Some thing like this..

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OUPPUT
--------
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How to Decrypt ( âUÙN_{�� ) using the below function.

CREATE FUNCTION dbo.fnEncDecRc4
(
@Pwd VARCHAR(256),
@Text VARCHAR(8000)
)
RETURNSVARCHAR(8000)
AS

BEGIN
DECLARE@Box TABLE (i TINYINT, v TINYINT)

INSERT@Box
(
i,
v
)
SELECTi,
v
FROMdbo.fnInitRc4(@Pwd)

DECLARE@Index SMALLINT,
@i SMALLINT,
@j SMALLINT,
@t TINYINT,
@k SMALLINT,
@CipherBy TINYINT,
@Cipher VARCHAR(8000)

SELECT@Index = 1,
@i = 0,
@j = 0,
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WHILE @Index <= DATALENGTH(@Text)
BEGIN
SELECT@i = (@i + 1) % 256

SELECT@j = (@j + b.v) % 256
FROM@Box b
WHEREb.i = @i

SELECT@t = v
FROM@Box
WHEREi = @i

UPDATEb
SETb.v = (SELECT w.v FROM @Box w WHERE w.i = @j)
FROM@Box b
WHEREb.i = @i

UPDATE@Box
SETv = @t
WHEREi = @j

SELECT@k = v
FROM@Box
WHEREi = @i

SELECT@k = (@k + v) % 256
FROM@Box
WHEREi = @j

SELECT@k = v
FROM@Box
WHEREi = @k

SELECT@CipherBy = ASCII(SUBSTRING(@Text, @Index, 1)) ^ @k,
@Cipher = @Cipher + CHAR(@CipherBy)

SELECT@Index = @Index +1
END

RETURN@Cipher
END

GO

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v
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UPDATE@Box
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SELECT@Index = @Index + 1
END

RETURN
ENDANd this function does the encrypt/decrypt partCREATE FUNCTION dbo.fnEncDecRc4
(
@Pwd VARCHAR(256),
@Text VARCHAR(8000)
)
RETURNSVARCHAR(8000)
AS

BEGIN
DECLARE@Box TABLE (i TINYINT, v TINYINT)

INSERT@Box
(
i,
v
)
SELECTi,
v
FROMdbo.fnInitRc4(@Pwd)

DECLARE@Index SMALLINT,
@i SMALLINT,
@j SMALLINT,
@t TINYINT,
@k SMALLINT,
@CipherBy TINYINT,
@Cipher VARCHAR(8000)

SELECT@Index = 1,
@i = 0,
@j = 0,
@Cipher = ''

WHILE @Index <= DATALENGTH(@Text)
BEGIN
SELECT@i = (@i + 1) % 256

SELECT@j = (@j + b.v) % 256
FROM@Box b
WHEREb.i = @i

SELECT@t = v
FROM@Box
WHEREi = @i

UPDATEb
SETb.v = (SELECT w.v FROM @Box w WHERE w.i = @j)
FROM@Box b
WHEREb.i = @i

UPDATE@Box
SETv = @t
WHEREi = @j

SELECT@k = v
FROM@Box
WHEREi = @i

SELECT@k = (@k + v) % 256
FROM@Box
WHEREi = @j

SELECT@k = v
FROM@Box
WHEREi = @k

SELECT@CipherBy = ASCII(SUBSTRING(@Text, @Index, 1)) ^ @k,
@Cipher = @Cipher + CHAR(@CipherBy)

SELECT@Index = @Index +1
END

RETURN@Cipher
END

Peter Larsson
Helsingborg, Sweden

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This is related to post :
http://www.sqlteam.com/forums/topic.asp?TOPIC_ID=78552

got a issue with this one..im not sure why..

My results are as follows:

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Code Block
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ExternalAccountID Encrypted ID Number Decrypted ID Number

-------------------------------------------------- ------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------ ------------------------------

12345654444 0x007B9CA28582AC42B36EBD8DFF91434701000000E267255EC20E27E5B4AC03E68C508ADCFA2EB815C234116D54FEE639685D7ECF49BB200140952E45A246AFE83BF53444 ㈱�㘵��4

12312312312 0x007B9CA28582AC42B36EBD8DFF91434701000000DB3838A7C459A2F031B1FFECF2499647C7DE10B7A20171EFDEA68320F7996CCC1C6A45011E93361F9A93494F49311CB9 ㈱ㄳ㌲㈱ㄳ2

12345678901 0x007B9CA28582AC42B36EBD8DFF9143470100000057EEEC0B1E223819EBF8186F0CBFEF8FDF4BFF4C0DCE322FAB8735DC8EA144B5937FFCC745A7FC427282C493DFECD901 ㈱�㘵㠷〹1



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