March 10th, 2011 04:29 PM
John the ripper charset
Hey all, I have a question about John the Ripper.
I know that a password that is 7 characters long and used only 4 keys. I know what the three keys are, but I don't know how to make a custom charset. I have edited the john.conf file to have the following:
Now i just don't know how to make the custom.chr so that it only contains the 4 keys 'asdf'
File = custom.chr
MinLen = 7
MaxLen = 7
CharCount = 3
Last edited by Phalse; March 10th, 2011 at 04:37 PM.
March 10th, 2011 06:47 PM
I am confused
If you have 4 characters, why have you set "CharCount = 3" ?
Also, don't you need to declare a, s, d, f as your character variables?
If you cannot do someone any good: don't do them any harm....
As long as you did this to one of these, the least of my little ones............you did it unto Me.
What profiteth a man if he gains the entire World at the expense of his immortal soul?
March 10th, 2011 06:57 PM
Yes, sorry nihil that was a typo. I have CharCount=4.
And i guess my question is HOW do i declare my variables as a,s,d,f? From what I can gather the character sets are locked up in the .chr files, but they aren't ascii so I can't just make one.
There is the -make-charset flag, but i must not be using it properly because it expects a file.
By sploiterwannabe in forum Newbie Security Questions
Last Post: May 23rd, 2005, 05:34 PM
By !mitationRust in forum Cosmos
Last Post: August 2nd, 2004, 04:42 PM
By lpaulgib in forum Cryptography, Steganography, etc.
Last Post: April 15th, 2004, 01:42 AM
By steve.milner in forum Tech Humor
Last Post: July 26th, 2003, 09:09 PM
By I am a cracker in forum The Security Tutorials Forum
Last Post: February 24th, 2002, 10:01 PM