I have a string that is encrypted in c#.net which I have to decrypt in java using a key and a IV provided by the client. Used algorithm is AES.
I have tried few things. The key looks something like
key = "QWEEqweASDreefERTfdf45fefdWERfsdf34fedfdwn5=" //length 44 bytes
iv = "nkfghER24dfdfdf56YUIgH==" // lenght=24 bytes
When I use this with Cipher class with algorith AES/CBC/PKCS5Padding
passing the above key to Secretkeyspec class it says invalid key lenth 44 bytes
I am not able makeout whats wrong with the key. Tried all suggested solutions for few days nothing works. Can some one help please? Thank you.
Use java native for C# code.
First write C# code for decrypt the key.
and calling the code in java using native.
for reference
http://www.codeproject.com/Articles/378826/How-to-wrap-a-Csharp-library-for-use-in-Java
Related
I am calling an API which gives me back a string encrypted using AES algorithm. The result of AES algorithm is encoded using Base64. The AES algorithm uses ECB mode and PKCS5 padding.
With ruby decrypting AES is rather easy. So I tried the usual ways:
Tried the AES gem by chicks
Digged through the Cipher documentation to find a suitable algorithm
I ended up with a lot of "try and error" because it was not totally clear to me how to pass PKCS5 padding into the decryption function.
All I found is this class which refers to PKCS5 but its not totally clear how to pass my key and content in there.
So I tried many possible AES/ECB combinations like this:
decipher = OpenSSL::Cipher::AES.new(128, :ECB)
decipher.key = Base64.decode64(key)
encrypted = Base64.decode64(encrypted)
decrypted = decipher.update(encrypted) + decipher.final
but this throws:
final': wrong final block length (OpenSSL::Cipher::CipherError)
I am not an encryption specialist so I am in a dead end here. Maybe you have a hint in the right direction.
How can I decrypt AES encrypted content using my key and ECB mode with PKCS5 padding?
Thank you!
I'm currently working on a project where I have to "convert" some code from Ruby(version 1.9.3p194) to Golang(version 1.7). There is this part where Ruby uses RSA public key encryption and I always get a consistent result every time it gets executed. This is the function used:
Edit: I overlooked that after the public key encryption, there is a base 64 encoding as well
public_key = OpenSSL::PKey::RSA.new(public_encryption_key)
public_encrypted_text = public_key.public_encrypt(text, OpenSSL::PKey::RSA::NO_PADDING)
base64_encrypted_text = Base64.encode64(public_encrypted_text).gsub("\n", "")
escaped_encrypted_text = URI.escape(encrypted_key, "/+=")
However in Golang, due to the rsa library I can't get a consistent result since the function to encrypt takes a random parameter to generate different result each time. I understand why it needs to be different every time, but i can't get anything remotely similar to what ruby generates. These are the functions used in Golang:
//keyBytes is the public key as []byte
block, _ := pem.Decode(keyBytes)
key, err := x509.ParsePKIXPublicKey(block.Bytes)
if err != nil {
return nil, err
}
pubKey, ok := key.(*rsa.PublicKey)
if !ok {
return nil, errors.New("Cannot convert to rsa.PublicKey")
}
result, err := rsa.EncryptPKCS1v15(cryptorand.Reader, pubKey, text)
encryptedText := base64.URLEncoding.EncodeToString(result)
encryptedText = strings.TrimRight(encryptedText, "=")
One of the problems is that ruby can encrypt the text with no problem, and in golang I'm getting an error that the key is too short to encrypt everything.
If I encrypt something else, like "Hello". When decrypting I get from ruby the error "padding check failed". The decryption is being handle like follows:
private_key.private_decrypt(Base64.decode64(text))
EDIT: Thanks to the answer of gusto2 I now know better what is going on since I didn't have much understanding of RSA.
Now in Golang I was able to encrypt the text using PKCS1 v1.5, and just to be sure I tried to decrypt that as well, also in Golang, with no problem.
However in Ruby I still wasn't able to decrypt using the private key. So I figured that the base64 encoding used in Golang was the issue. So I changed that part to this:
encryptedText := base64.StdEncoding.EncodeToString(result)
And also I removed the last line were the equal sign was being trimmed.
With that done it worked like a charm.
I am no knowledgeable about golang, however I may know something about RSA.
The difference seems to be in the padding.
For ruby - no padding is used
For golang - PKCS1v15 padding is used
In the rubyexample you use OpenSSL::PKey::RSA::NO_PADDING is used which is VERY VERY unsafe. It is called textbook RSA and is not inteded in real-life use as it has many weaknesses and dangerous traps. So the ruby example is very dangerously unsafe because of using the textbook RSA. As well it is limited to encrypting small messages (much smaller than the keyspace).
There are two padding types used with RSA:
PKCS1 v1 (commonly referred as PKCS1) - this is a deterministic padding (the output is always the same), many cryptographers consider this option obsolete as some weaknesses has been found when not used properly, but it is still in use and not considered broken.
PKCS1 v2 (commonly refered as OAEP or PSS) which is stochastic (randomized) padding. You can distinguish the last two as the output of OAEP is always different.
One of the problems is that ruby can encrypt the text with no problem, and in golang I'm getting an error that the key is too short to encrypt everything
You've provided only a small part of the golang example, so here I may only assume many things.
As you claim the golang example outputs randomized output and according to the parameters PKCS1 v1.5 is used, I'd assume the implementation is doing hybrid encryption which is good and much safer way to encrypt data with RSA (using symmetric encryption with random key and wrap/encrypt the key with RSA).
I am encrypting my data using AES-128 CBC. However it is decrypted by another software CodeIgnitor, which actually converts password into 2 elements. HMAC key and encryption_key using hkdf and SHA-512 digest.
How can i implement this thing in my code. (I am using VB 6)
Re-phrasing my question: How is password converted into 2 elements HMAC_key and encryption_key?
Let's see if I can explain further..
Encryption software used is CryptoSys while decryptionis done by CodeIgnitor..which is not working
So the difference I can see between the two is that codeignitor is expecting an HMAC authmessage when decrypting while the encryption does not prepend it the final message. So how can i do that? I am using VB6 for encryption.
So far I have come to the conclusion that there could be 2 key-derivation function a pb-kdf or a h-kdf.
The h-kdf authentication in particular works as follows:
A password is enterd by user
Based on this password and a random IV, an HMAC key is generated using HKDF and SHA-512 digest algo.
This HMAC key is saved and also expanded and then using IV/salt encrypts the plaintext message.
The IV is pre-pended to the plain-text message and base64 encoded.
The HMAC key is then pre-pended to the final message gereated in step 4.
So basically hkdf does not convert a key into 2 elements but only uses the basic key generated to generate other keys (and HMAC key is the original key used to generate other keys) now my question becomes...
Is there any vb6 library that uses hkdf and sha-512? As far as i have researched, Chilkat and CryptoSys APIs are only pbkdf based..Am I correct?
I think I started getting this error when I switched from MySQL to PostgreSQL.
I had written code to encrypt decrypt model attributes containing sensitive data and I had it working until the db switch.
I have the following code:
#pbk = OpenSSL::PKey::RSA.new File.read("#{RAILS_ROOT}/cert/pb_sandwich.pem")
#pvk = OpenSSL::PKey::RSA.new File.read("#{RAILS_ROOT}/cert/tuna_salad.pem"), 'pass45*'
model.sendata = Base64.encode64 #pbk.public_encrypt(model.sendata)
I run that code on save. I've also tried with and with out first using Base64.
Then when I try to read:
#pvk.private_decrypt Base64.decode64(model.sendata)
I get this error:
OpenSSL::PKey::RSAError: data greater than mod len
I never got that before when I used MySQL. I can't really remember what datatype the sendata column was in MySQL but in my current PostgreSQL setup that column is datatype bytea
I'm assuming that is the problem since it used to work fine with MySQL. What datatype should the column be if I wanted to skip having to do that extra step to Base64 encode/decode? If that is the problem that is.
Another thing of note is that I've tried generating the private key with mod lengths: 2048, 4096, and 5120 and I always get the same error. Also, the sendata field isn't very long before encoding, it's under 40 chars.
I'm stumped right now, any ideas?
You are probably not storing the keys properly in the Database. There's probably some field that is being truncated.
The message you are getting probably means that the data is too long to be encrypted with such a small key. If this is the case, you should encrypt the data with AES and encrypt the AES key with RSA. Then send both the encryted data and the encrypted key.
I am attempting to decrypt a number encrypted by another program that uses the BouncyCastle library for Java.
In Java, I can set the key like this: key = Hex.decode("5F3B603AFCE22359");
I am trying to figure out how to represent that same step in Ruby.
To get Integer — just str.hex. You may get byte array in several ways:
str.scan(/../).map(&:hex)
[str].pack('H*').unpack('C*')
[str].pack('H*').bytes.to_a
See other options for pack/unpack and examples (by codeweblog).
For a string str:
"".tap {|binary| str.scan(/../) {|hn| binary << hn.to_i(16).chr}}