I'm a total newbie to flask let alone cryptography. I'm currently following Miguel Grinberg's flask tutorial. I'm trying to store a pbkdf2:sha512 hashed password with salt length 64 but I can't find the appropriate length to store it in the Database is it db.string(128) or 512.
Refer following pbkdf2-sha512 digest,
$pbkdf2-sha512$10001$0dr7v7eWUmptrfW.9z6HkA$w9j9AMVmKAP17OosCqDxDv2hjsvzlLpF8Rra8I7p/b5746rghZ8WrgEjDpvXG5hLz1UeNLzgFa81Drbx2b7.hg
OR
$pbkdf2.sha512.10000.4483972AD2C52E1F590B3E2260795FDA9CA0B07B96FF492814CA9775F08C4B59CD1707F10B269E09B61B1E2D11729BCA8D62B7827B25B093EC58C4C1EAC23137.DF4FCB5DD91340D6D31E33423E4210AD47C7A4DF9FA16F401663BF288C20BF973530866178FE6D134256E4DBEFBD984B652332EED3ACAED834FEA7B73CAE851D
which is ideally less than 256 character. But sometimes it will go above 256 characters also. So you should use db.string(512).
Related
I Have gone through several approaches on stack overflow but none works now.
Encrypting a text field value which is the password and saving it in the database is the requirement.
I need the approach of converting the plain text into encrypted
I have tried
https://wiki.qt.io/Simple_encryption_with_SimpleCrypt
it gives different encryptions for same text so I can not compare and validate
Simplecrypt you linked to has this piece of code in it:
//prepend a random char to the string
char randomChar = char(qrand() & 0xFF);
ba = randomChar + integrityProtection + ba;
What this means is, any piece of data can result in 256 different possible encrypted datas. This is useful in encryption, where you (among many other things) don't want an attacker to be able to see if two separate encrypted pieces of data are actually same data or not.
If you want to use SimpleCrypt, you have to compare the passwords after decrypting. You could also modify the algorithm to have a known (given by you) randomChar. But I advise against it, as that is going to extra effort to do something poorly.
You should really use something else, for example QCryptographicHash. Just remember to use salt when hashing the password (this prevents an attacker from seeing if some passwords in the database are the same).
I am referencing this gist available online.
It's using a hard-coded 128 bit AES key to encrypt/decrypt data. The key is hard-coded in this format:
wchar_t default_key[] = L"3igcZhRdWq96m3GUmTAiv9";
How is a 128 bit AES key specified in this format? And how would I, for example, specify/convert an AES 256 key in raw hex form (like shown below) in the same format:
5DD34BE0C9C344833C91E20D247C3B2988340A034C9719C7C291EC2F781F8133
Note that all keys shown here are just examples and not used to encrypt any sensitive information.
It is likely that the hardcoded key is a base64 encoded 128-bit key without padding. But it is also possible that it is arbitrary UTF8 that is padded to become the 128 bit key. I'd put my money on the prior, though.
If you'd like to take your provided key in hex and use that instead, you just need to convert from hex to base64.
I expect then that your key would look like this:
XdNL4MnDRIM8keINJHw7KYg0CgNMlxnHwpHsL3gfgTM
You may need to make further adjustments to the gist you have to ensure it knows you are using a 256-bit key now.
I'm having trouble encrypting data with node and decrypting with RNCryptor. My understanding is that RNCryptor uses a special data format?
The cryptotext is being generated as follows:
var crypto = require('crypto');
var cipher = crypto.createCipher('aes-256-cbc','InmbuvP6Z8');
var text = "123|123123123123123";
var crypted = cipher.update(text,'utf8','base64');
crypted += cipher.final('base64');
And I'm decrypting like this:
[RNDecryptor decryptData:plainText withPassword:password error:&error];
How am I supposed to do this? When I try to decrypt currently I get an empty NSData and no error.
Yes, RNCryptor outputs encrypted data in its own format. If you build this format in your own encryption code (and use the same encryption params), you can pass it to JNCryptor to decrypt.
For RNCryptor format version 3, the first 34 bytes are as follows:
byte[0] is the version (3).
byte[1] defines whether a password or a key is used (1 for password, 0 for key).
bytes[2-9] carry the encryption salt.
bytes[10-17] carry the HMAC salt.
bytes[18-33] carry the IV.
Then comes the encrypted ciphertext.
Then the last 32 bytes hold the (SHA256) HMAC for the ciphertext.
The spec is here: https://github.com/RNCryptor/RNCryptor-Spec/blob/master/RNCryptor-Spec-v3.md
RNCryptor is more than just encryption, it is an entire secure "stack" including authentication, password key derivation, versioning and random iv. Since your encryption is just the encryption portion it will be incompatible.
Either use RNCryptor for the encryption or change to a simple decryption method. Since the question has a commoncrypto tag and the encryption code look like it might be Swift (there is no language tag) just consider using Common Crypto in Objective-C for the decryption. See the SO answer for example code.
I just want to know the exact formula (or algorithm) used for generating the pseudo random values used in encrypting the zip file. I am trying to create a password hacker(for zip files) and I also require to know how to verify if the random password generated by my program is correct. I have tried searching for an answer to this in Google but I could't find a direct solution.
I am trying to program this zip hacker in c++.
note: by formula (or algorithm) I meant: key derivation function.
I just want the necessary information as quick as possible, that's why I posted it here!
Different versions of zip-files do it differently, but basically you have an encryption-header specifying what encryption is used according to the zip-file specification.
For example, the strong encryption header looks like below and specifies the encryption algorithm in the AlgID-field.
4.5.12 -Strong Encryption Header (0x0017):
Value Size Description
----- ---- -----------
0x0017 2 bytes Tag for this "extra" block type
TSize 2 bytes Size of data that follows
Format 2 bytes Format definition for this record
AlgID 2 bytes Encryption algorithm identifier
Bitlen 2 bytes Bit length of encryption key
Flags 2 bytes Processing flags
CertData TSize-8 Certificate decryption extra field data
(refer to the explanation for CertData
in the section describing the
Certificate Processing Method under
the Strong Encryption Specification)
7-zip uses AES-256 encrpytion for 7z/zip archives.(see here)
7-Zip also supports encryption with AES-256 algorithm. This algorithm uses cipher key with length of 256 bits. To create that key 7-Zip uses derivation function based on SHA-256 hash algorithm. A key derivation function produces a derived key from text password defined by user. For increasing the cost of exhaustive search for passwords 7-Zip uses big number of iterations to produce cipher key from text password.
Also, keep in mind that Brute force attacks are a waste of time. I won't go into details why, I will instead direct you to Jeff Atwood's blog, he has an excellent post.
I need to use current time because later on, another program will need to know when I generated the pin and whats the username and password.
Summary:
Username + password + current time = pin [6 digits]
Reverse, generated pin, I need to know which particular, and check if its already pass 1 minute.
I am not asking for direct code, but I need to know the best, not say best, good way/algorithm for it. Thanks (btw, I am cpp beginner)
EDIT:
I am sorry for not making things clear. Actually, I dun need the OTP, after generated the pin, the other program will need to run like this: validate {username} {password} {pin}
999999 minutes, gives ~694 days, ~1.9 years. So if you used all the entropy available just for recording the current time, you'd cycle the value in less than 2 years.
And things will be much worse if you want to include the username and password, and avoid easy guessing.
With 6 decimal digits, you can store about 19 bits of data. So you'll have to make sure that you have strong anti-brute-force protection on your server end, otherwise it'll be trivial to try all possible combinations.
One-time passwords do not have internal decode-able structure, they are typically used in addition to a normal password to act as a second factor in authentication. Then can be based on the time as you are suggesting, but are not reversible - the other end also has the secret key, and can generate the possible list itself.
So, for example, as well as entering a username and (normal) password, the user enters the value from a token, which is generated as AES(secretkey, currentminute), and the server computes AES(secretkey, currentminute) and AES(secretkey, currentminute-1) etc, to compare the value against. It might also record which token matches, so that it records an estimate of the token's clock accuracy, which allows some drift of the token's clock, as long as it's used frequently enough. To work out how to best use the 19 bits you have in a 6-digit pin, you'll need someone who is a real cryptographer - as I would guess that taking a simple truncation might be in-secure.
Promoted my earlier comment to an answer, as an afterthought:
I'm pretty sure it is deceiving to call this 'otp'. OTP is purely random and secret.
What you describe is a simple hash.
You could
MD5(username+password_hash+(seconds_past_1970%60))
I'm sure besides using public key encryption instead of the password, this is more or less how RSA keys do it.
Edit Oh yes: it will be pretty trivial to generate 6 digits from the resulting hash :)