How to extract only 7 characters using grep

Question

I am using a regular expression with grep. I want to extract exactly 7 character passwords, but I am getting 7 and more than 7 characters as a result.

grep '[a-zA-Z0-9]\{7\}' /usr/share/wordlists/rockyou.txt

grep '[a-zA-Z0-9]\{7,7\}' /usr/share/wordlists/rockyou.txt

Answer 1

Use extended grep:

grep  -E '^[a-zA-Z0-9]{7}$' /usr/share/wordlists/rockyou.txt

or your own version like:

grep '^[a-zA-Z0-9]\{7\}$' /usr/share/wordlists/rockyou.txt

or even:

egrep '^.{7}$' /usr/share/wordlists/rockyou.txt

Answer 2

Any line that contains more than 7 characters also contains a substring of 7 characters (which will match your expression).

You can force only complete matches by anchoring the expression to the start and end of the line:

grep '^[a-zA-Z0-9]\{7\}$' /usr/share/wordlists/rockyou.txt

or specify whole-line matching using the -x option

grep -x '[a-zA-Z0-9]\{7\}' /usr/share/wordlists/rockyou.txt

From man grep:

-x, --line-regexp
       Select  only  those  matches  that exactly match the whole line.
       For a regular expression pattern, this  is  like  parenthesizing
       the pattern and then surrounding it with ^ and $.

Answer 3

you can cd to the directory and try the below command:

grep ^.......$ *

• . --> any single character (so seven . is seven characters)
• * (asterisk) --> all the files in the current directory (except, usually, those whose names begin with a literal .). You can give the file name(s) too if preferred.