CODEWITHSUNDEEP
python
Adel Zana
Adel Zana

Reputation: 3

MD5 decrypt script

__author__ = 'Zane'
import hashlib
import sys

if (len(sys.argv)!=2 ) or (len(sys.argv[1])!= 32):

    print("[---] md5cracker.py & hash")
    sys.exit(1)


    crackedmd5 = sys.argv[1]

    # open a file and read its contents

    f = open('file.txt')

    lines = f.readline()

    f.close()


    for line in lines:

        cleanline = line.rstrip()

        hashobject = hashlib.md5(cleanline)

        if (hashobject==crackedmd5):

            print('Plain text password for ' + crackedmd5 + "is "  + hashobject + '\n')

I get no error with exit code 1 and i do not know where i get it wrong

Upvotes: 0

Views: 931

Answers (2)

Germar
Germar

Reputation: 446

Pythons code structure is based on indent of lines. For now your whole code is part of the if (len(sys.argv)!=2 ) or (len(sys.argv[1])!= 32): condition.

You need to unindent all lines with one tab starting from crackedmd5 = sys.argv[1]

EDIT

You also used lines = f.readline() which will read only one line and so for line in lines will iterate over every single char in that line and not over multiple lines. You need to use lines = f.readlines() instead.

Upvotes: 0

dsh
dsh

Reputation: 12234

Your program exits with status code one because you told it so (roughly on line 8):

sys.exit(1)

Upvotes: 2

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