For my own understanding of Websphere LTPA keys, I'm running this Java code, which works correctly when using the author's Base64 encoded, SHA1 hashed password:

String ltpa3DESKey = "IpGJOdpSxV3J8yWAuB+UiUlYCK7sAzSfENE5MLT2q+s=";

but fails when using mine:

String ltpa3DESKey = "ExGKSrGQPGN+q0GE1GlFhK7BB8bLB0mEp9ec1Vzxv8A\=";


Invalid escape sequence (valid ones are  \b  \t  \n  \f  \r  \"  \'  \\ )

But if I escape it:

String ltpa3DESKey = "ExGKSrGQPGN+q0GE1GlFhK7BB8bLB0mEp9ec1Vzxv8A\=";

I get:

 javax.crypto.BadPaddingException: Given final block not properly padded
    at com.sun.crypto.provider.CipherCore.doFinal(CipherCore.java:991)
    at com.sun.crypto.provider.CipherCore.doFinal(CipherCore.java:847)
    at com.sun.crypto.provider.DESedeCipher.engineDoFinal(DESedeCipher.java:294)
    at javax.crypto.Cipher.doFinal(Cipher.java:2165)
    at com.me.ltpa.LtpaToken.decrypt(LtpaToken.java:79)
    at com.me.ltpa.LtpaToken.getSecretKey(LtpaToken.java:62)
    at com.me.ltpa.LtpaToken.decryptLtpaToken(LtpaToken.java:110)
    at com.me.ltpa.LtpaToken.getInstance(LtpaToken.java:118)
    at com.me.ltpa.LtpaToken.main(LtpaToken.java:139)

I've read the basics of the Base64 Wikipedia article, but apparently backslashes aren't part of the Base64 alphabet. Any thoughts on how to proceed?

  • The tool you used to base64 encode the key should not have generated a backslash. Re-encode the key with a better encoder. (see CyberChef for a good all purpose encoding tool.) – John Deters Nov 21 '17 at 19:47

There are different variants of base64 encoding; while none I know of should produce a \, you might be better off just to use a URL-compatible version (which is done in java like the first google hit says).

Obiter Dictum:

  • There is usually no good reason to use 3DES and not AES. 3DES has long been deprecated.
  • Sha1 is not a good key derivation function, please use one that is designed for this purpose, like PBKDF2.
  • I never finished this project, but the code I'm testing is old and corresponds to when 3DES was the industry standard, so I'm moving on. Thank you! – mellow-yellow Nov 22 '17 at 3:30

Guess Java tries to identify the \ as an commandin the String, like e.g. \n. There's no command like \=, so that's why the exception appears. I would recommand you to put some // instead of \ , otherwise you might not use the \ for your strings.

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