How to encrypt text with a password in python?

Question

Surprisingly difficult to find a straight answer to this on Google.

I\'m wanting to collect a piece of text and a message from a user such as 1PWP7a6xgoYx81VZo

Answer 1

alpha = "abcdefghijklmnopqrstuvwxyz"
password = input("What is the password?:")
password = "".join([(str(ord(x)-96) if x.isalpha() else x) for x in list(password)])
password = int(password)

def encrypt(cleartext):
  cyphertext = ""
  for char in cleartext:
    if char in alpha:
      newpos = (alpha.find(char) + password) % 26
      cyphertext += alpha[newpos]
    else:
      cyphertext += char

  return cyphertext

def decrypt(cleartext):
  cyphertext = ""
  for char in cleartext:
    if char in alpha:
      newpos = (alpha.find(char) - password) % 26
      cyphertext += alpha[newpos]
    else:
      cyphertext += char

  return cyphertext

while True:
  cleartext = input("Cleartext:")
  cleartext = cleartext.lower()
  print(encrypt(cleartext))

  cleartext = input("Cyphertext:")
  cleartext = cleartext.lower()
  print(decrypt(cleartext))

So this is how I do it!

Answer 2

1. If you are going to use mentioned database to authorise users, you should use hashes or message digests of user's passwords, instead of 2 way encryption algorithms, that would make your data hard to use even in case of db leakage.
2. You cannot use above method to protect data that needs to be decrypted at some point, but even then you can use more secure way than just encrypting user passwords using some fixed key (which is the worst method). Take a look at OWASP's Password Storage Cheat Sheet.

As you wrote "I want to be able to encrypt/decrypt the message", I'm attaching a simple python source (tested under 2.7) for encr/decr using Blowfish.

#!/usr/bin/env python2
# -*- coding: utf-8 -*-
import os
from Crypto.Cipher import Blowfish     # pip install pycrypto

BS = 8
pad = lambda s: s + (BS - len(s) % BS) * chr(BS - len(s) % BS) 
unpad = lambda s : s[0:-ord(s[-1])]

def doEncrypt(phrase, key):
    c1  = Blowfish.new(key, Blowfish.MODE_ECB)
    return c1.encrypt(pad(phrase))

def doDecrypt(phrase, key):
    c1  = Blowfish.new(key, Blowfish.MODE_ECB)
    return unpad(c1.decrypt(phrase))

def testing123(phrase, key):
    encrypted = doEncrypt(phrase, key)
    decrypted = doDecrypt(encrypted, key)
    assert phrase == decrypted, "Blowfish ECB enc/dec verification failed"
    print ("Blowfish ECB enc/dec verified ok")
    print ('phrase/key(hex)/enc+dec: {}/{}/{}'.format(phrase, key.encode('hex'), decrypted))

if __name__== "__main__":
    phrase= 'Ala ma kota, a kot ma AIDS.'
    key= os.urandom(32)
    testing123(phrase, key)

Answer 3

Based on zwer's answers but solves a little bug when the the source is exactly a multiple of 16.

Code:

from builtins import bytes
import base64
from Crypto.Cipher import AES
from Crypto.Hash import SHA256
from Crypto import Random

def encrypt(string, password):
    """
    It returns an encrypted string which can be decrypted just by the 
    password.
    """
    key = password_to_key(password)
    IV = make_initialization_vector()
    encryptor = AES.new(key, AES.MODE_CBC, IV)

    # store the IV at the beginning and encrypt
    return IV + encryptor.encrypt(pad_string(string))

def decrypt(string, password):
    key = password_to_key(password)   

    # extract the IV from the beginning
    IV = string[:AES.block_size]  
    decryptor = AES.new(key, AES.MODE_CBC, IV)

    string = decryptor.decrypt(string[AES.block_size:])
    return unpad_string(string)

def password_to_key(password):
    """
    Use SHA-256 over our password to get a proper-sized AES key.
    This hashes our password into a 256 bit string. 
    """
    return SHA256.new(password).digest()

def make_initialization_vector():
    """
    An initialization vector (IV) is a fixed-size input to a cryptographic
    primitive that is typically required to be random or pseudorandom.
    Randomization is crucial for encryption schemes to achieve semantic 
    security, a property whereby repeated usage of the scheme under the 
    same key does not allow an attacker to infer relationships 
    between segments of the encrypted message.
    """
    return Random.new().read(AES.block_size)

def pad_string(string, chunk_size=AES.block_size):
    """
    Pad string the peculirarity that uses the first byte
    is used to store how much padding is applied
    """
    assert chunk_size  <= 256, 'We are using one byte to represent padding'
    to_pad = (chunk_size - (len(string) + 1)) % chunk_size
    return bytes([to_pad]) + string + bytes([0] * to_pad)
def unpad_string(string):
    to_pad = string[0]
    return string[1:-to_pad]

def encode(string):
    """
    Base64 encoding schemes are commonly used when there is a need to encode 
    binary data that needs be stored and transferred over media that are 
    designed to deal with textual data.
    This is to ensure that the data remains intact without 
    modification during transport.
    """
    return base64.b64encode(string).decode("latin-1")

def decode(string):
    return base64.b64decode(string.encode("latin-1"))

Tests:

def random_text(length):
    def rand_lower():
        return chr(randint(ord('a'), ord('z')))
    string = ''.join([rand_lower() for _ in range(length)])
    return bytes(string, encoding='utf-8')

def test_encoding():
    string = random_text(100)
    assert encode(string) != string
    assert decode(encode(string)) == string

def test_padding():
    assert len(pad_string(random_text(14))) == 16
    assert len(pad_string(random_text(15))) == 16
    assert len(pad_string(random_text(16))) == 32

def test_encryption():
    string = random_text(100)
    password = random_text(20)
    assert encrypt(string, password) != string
    assert decrypt(encrypt(string, password), password) == string

Answer 4

Here is my solution for anyone who may be interested:

from Crypto.Cipher import AES  # pip install pycrypto
import base64

def cypher_aes(secret_key, msg_text, encrypt=True):
    # an AES key must be either 16, 24, or 32 bytes long
    # in this case we make sure the key is 32 bytes long by adding padding and/or slicing if necessary
    remainder = len(secret_key) % 16
    modified_key = secret_key.ljust(len(secret_key) + (16 - remainder))[:32]
    print(modified_key)

    # input strings must be a multiple of 16 in length
    # we achieve this by adding padding if necessary
    remainder = len(msg_text) % 16
    modified_text = msg_text.ljust(len(msg_text) + (16 - remainder))
    print(modified_text)

    cipher = AES.new(modified_key, AES.MODE_ECB)  # use of ECB mode in enterprise environments is very much frowned upon

    if encrypt:
        return base64.b64encode(cipher.encrypt(modified_text)).strip()

    return cipher.decrypt(base64.b64decode(modified_text)).strip()


encrypted = cypher_aes(b'secret_AES_key_string_to_encrypt/decrypt_with', b'input_string_to_encrypt/decrypt', encrypt=True)
print(encrypted)
print()
print(cypher_aes(b'secret_AES_key_string_to_encrypt/decrypt_with', encrypted, encrypt=False))

Result:

b'secret_AES_key_string_to_encrypt'
b'input_string_to_encrypt/decrypt '
b'+IFU4e4rFWEkUlOU6sd+y8JKyyRdRbPoT/FvDBCFeuY='

b'secret_AES_key_string_to_encrypt'
b'+IFU4e4rFWEkUlOU6sd+y8JKyyRdRbPoT/FvDBCFeuY=    '
b'input_string_to_encrypt/decrypt'

Answer 5

Here's how to do it properly in CBC mode, including PKCS#7 padding:

import base64
from Crypto.Cipher import AES
from Crypto.Hash import SHA256
from Crypto import Random

def encrypt(key, source, encode=True):
    key = SHA256.new(key).digest()  # use SHA-256 over our key to get a proper-sized AES key
    IV = Random.new().read(AES.block_size)  # generate IV
    encryptor = AES.new(key, AES.MODE_CBC, IV)
    padding = AES.block_size - len(source) % AES.block_size  # calculate needed padding
    source += bytes([padding]) * padding  # Python 2.x: source += chr(padding) * padding
    data = IV + encryptor.encrypt(source)  # store the IV at the beginning and encrypt
    return base64.b64encode(data).decode("latin-1") if encode else data

def decrypt(key, source, decode=True):
    if decode:
        source = base64.b64decode(source.encode("latin-1"))
    key = SHA256.new(key).digest()  # use SHA-256 over our key to get a proper-sized AES key
    IV = source[:AES.block_size]  # extract the IV from the beginning
    decryptor = AES.new(key, AES.MODE_CBC, IV)
    data = decryptor.decrypt(source[AES.block_size:])  # decrypt
    padding = data[-1]  # pick the padding value from the end; Python 2.x: ord(data[-1])
    if data[-padding:] != bytes([padding]) * padding:  # Python 2.x: chr(padding) * padding
        raise ValueError("Invalid padding...")
    return data[:-padding]  # remove the padding

It's set to work with bytes data, so if you want to encrypt strings or use string passwords make sure you encode() them with a proper codec before passing them to the methods. If you leave the encode parameter to True the encrypt() output will be base64 encoded string, and decrypt() source should be also base64 string.

Now if you test it as:

my_password = b"secret_AES_key_string_to_encrypt/decrypt_with"
my_data = b"input_string_to_encrypt/decrypt"

print("key:  {}".format(my_password))
print("data: {}".format(my_data))
encrypted = encrypt(my_password, my_data)
print("\nenc:  {}".format(encrypted))
decrypted = decrypt(my_password, encrypted)
print("dec:  {}".format(decrypted))
print("\ndata match: {}".format(my_data == decrypted))
print("\nSecond round....")
encrypted = encrypt(my_password, my_data)
print("\nenc:  {}".format(encrypted))
decrypted = decrypt(my_password, encrypted)
print("dec:  {}".format(decrypted))
print("\ndata match: {}".format(my_data == decrypted))

your output would be similar to:

key:  b'secret_AES_key_string_to_encrypt/decrypt_with'
data: b'input_string_to_encrypt/decrypt'

enc:  7roSO+P/4eYdyhCbZmraVfc305g5P8VhDBOUDGrXmHw8h5ISsS3aPTGfsTSqn9f5
dec:  b'input_string_to_encrypt/decrypt'

data match: True

Second round....

enc:  BQm8FeoPx1H+bztlZJYZH9foI+IKAorCXRsMjbiYQkqLWbGU3NU50OsR+L9Nuqm6
dec:  b'input_string_to_encrypt/decrypt'

data match: True

Proving that same key and same data still produce different ciphertext each time.

Now, this is much better than ECB but... if you're going to use this for communication - don't! This is more to explain how it should be constructed, not really to be used in a production environment and especially not for communication as its missing a crucial ingredient - message authentication. Feel free to play with it, but you should not roll your own crypto, there are well vetted protocols that will help you avoid the common pitfalls and you should use those.

Answer 6

You can do it by using two of the built-in functions on the standard Python library. The first one is the function ord( ), which takes a unicode string character as single input parameter and converts it to its corresponding unicode code (an integer). Two simple examples of the usage of this function are provided:

>>> ord('a')
    97

>>> ord('b')
    98

Then, you also have the inverse function of ord(): chr( ). And as you can imagine it works all the way around: it has a unicode code as an input (integer) and gets the corresponding unicode character (string):

>>> chr(97)
    'a'

>>> chr(98)
    'b'

Then you can do a simple encription by adding or substracting by some arbitrary integer... in this case, the number 2:

NOTE: Watch out in not utting very big values or you'll get an error id you reach a negative nber, for example.

def encrypt(message):
    newS=''
    for car in message:
        newS=newS+chr(ord(car)+2)
    return newS


print(encrypt('hello world'))

And getting as a result:

jgnnq"yqtnf

Now you can copy and past the same function and generate the decrypt function. In this case, it requires, obviously, to substract by 2:

def decrypt(message):
    newS=''
    for car in message:
        newS=newS+chr(ord(car)-2)
    return newS


print(decrypt('jgnnq"yqtnf'))

And the result will be the original message again:

'hello world'

This would be a great way to encrypt messages to non programmers. However, anyone with a little of programming knowledge could write a program that varied the integer we used until they found we have just added (2) to the unicode characters to encrypt the code...

In order to avoid that, I would propose two more complex alternatives.

1. The first one is the simplest: it consists in applying a different sum value to the chr function depending on the position of the character (for example, adding 2 to each unicode code when it occupies an even position in the string and substracting 3 when sits on an odd position).

2. The second one will generate the maximum security. It will consist on adding or substracting every unicode code for a number that will be randomly generated for each character. It will require to store an array of values to decript back the message. Make sure, then, this array of values is not available to third parties.

There it goes a possible solution for 1.:

def encryptHard(message):
newS=''
for i in range(len(message)):
  if i%2==0:
    newS=newS+chr(ord(message[i])+2)
  else:
    newS=newS+chr(ord(message[i])-3)
return newS


print(encryptHard('hello world'))

And the result would be:

jbniqyltif

With the information hereby privided the decrypting script is obvious, so I won't bother you with coping, pasing and changing two values.

Finally, let's go into an in-depth-analysis of the second more complex alternative. With this one we can say that the encription will be almost indefitable. The idea is to vary the value we add or substract to each unicode code by a random number comprized between 0 and 255 (this is the range of numbers the chr( ) function admits, so do not try to play with other numbers o you will definitely get an error).

In this case, my proposal also randomizes the operation (sum or subtract), and avoids that the final number be a 0 (i.e. we would get an original character). Finally, its also returns a list with the numers it has been subtracted to, something you will need in order to decrypt the message back.

The chances that you get the same encrypted message if you call two times this function using the same message of length n are somewhat near to 255^n... So don't worry (I say somewhat, as the algorithm created would actually generate more repeated values on the low-end or high-end range of values, for example, in case the most frequent characters were not centered in this distrubution unicode caracrer set (from 0 to 255), which is the case. However, the program, though not perfect, works flawlessly and protects the information.

import random as r
def encryptSuperHard(message):
  newS=''
  l_trans=[]
  for car in message:
    code=ord(car)
    add_subtract=r.choice([True,False])
    if add_subtract:
      transpose=r.randint(0,code-1)
      newS=newS+chr(code-transpose)
      l_trans=l_trans+[-transpose]
    else:
      transpose=r.randint(code+1,255)
      newS=newS+chr(code+transpose)
      l_trans=l_trans+[transpose]
  return newS, l_trans

print(encryptSuperHard('hello world'))

In this case, this random encrypting script I've made has returned this two value tuple, where the first value is the encrypted message and the second one is the value that has "transposed" every character in order of apearance.

('A0ŤłY\x10řG;,à', [-39, -53, 248, 214, -22, -16,     226, -40, -55, -64, 124])

Decrypting, in this case would need to take the encrypred message and the list and proceed as follows:

def decryptSuperHard(encriptedS,l):
  newS=''
  for i in range(len(l)):
    newS=newS+chr(ord(encriptedS[i])-l[i])
  return newS

print(decryptSuperHard('A0ŤłY\x10řG;,à', [-39,-53,248,214,-22,-16,226,-40,-55,-64,124]))

And the results goes back to:

hello world

print(deccryptSuperHard('A0ŤłY\x10řG;,à', [-39, -53, 248, 214, -22, -16,     226, -40, -55, -64, 124])