Dictionary attack on SHA1 hash

Question

The following is my code. I have been trying to fix the code to perform dictionary attack on a hash (SHA-1) and I am getting the following result. P.S. I am a beginner to coding.

import hashlib
import random
#plug in the hash that needs to be cracked
hash_to_crack = "5baa61e4c9b93f3f0682250b6cf8331b7ee68fd8"
#direct the location of the dictionary file
dict_file = "C:/Users/kiran/AppData/Local/Programs/Python/Python37/dictionary.txt"

def main():
    with open(dict_file) as fileobj:
        for line in fileobj:
            line = line.strip()
            if hashlib.sha1(line.encode()).hexdigest() == hash_to_crack:
                print ("The password is %s") % (line);
                return ""
    print ("Failed to crack the hash!")
    return ""


if __name__ == "__main__":
    main()

The result:

RESTART: C:/Users/kiran/AppData/Local/Programs/Python/Python37/Codes/datest1.py
The password is %s
Traceback (most recent call last):
  File "C:/Users/kiran/AppData/Local/Programs/Python/Python37/Codes/datest1.py", line 20, in <module>
    main()
  File "C:/Users/kiran/AppData/Local/Programs/Python/Python37/Codes/datest1.py", line 13, in main
    print ("The password is %s") % (line);
TypeError: unsupported operand type(s) for %: 'NoneType' and 'str'

Please, don't use capslock for writing question. Also, take some time to explain what you are trying to do. To learn how to ask better questions, please go over [how to ask](https://stackoverflow.com/help/how-to-ask) and [on-topic](https://stackoverflow.com/help/on-topic). This will help others to help you. If you have questions, provide your code as [mvce](https://stackoverflow.com/help/mcve). — TwistedSim, Apr 17 '18 at 23:53
You can use [hashcat](https://github.com/hashcat/hashcat) instead of trying to program this yourself. — Nick, Apr 17 '18 at 23:54

score 2 · Answer 1 · answered Apr 17 '18 at 23:56

You're using Python 3, where print is a function. The line:

print ("The password is %s") % (line)

calls the function print with the argument "The password is %s". The function returns None. Then None % (line) gives the error message you see.

Most idiomatic is to write the line this way instead:

print("The password is", line)

Other ways that would work:

print("The password is %s" % line)
print(("The password is %s") % (line))

TwistedSim · Accepted Answer · 2018-04-18T00:09:46.340

-1

This line is not valid python3 syntax: (EDIT: it's a valid syntax ! But not what you want :) )

print ("The password is %s") % (line);

Use this instead:

print ("The password is %s" % line)

Also, accordig to the error message, line might be None.

edited Apr 18 '18 at 00:09

answered Apr 17 '18 at 23:55

TwistedSim

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See my answer: the syntax is fine in Python 3 - it just doesn't at all do what they hoped it would do ;-) If the syntax were wrong, they would gotten a `SyntaxError ` instead. The problem is that that `print()` function returns `None`, and _then_ `None % a_string` makes no sense (which _is_ the error message they got). – Tim Peters Apr 18 '18 at 00:01

Dictionary attack on SHA1 hash

2 Answers2