Count lines with different characters

This could be your command. It's a script that takes a filename on the command line and outputs all lines in it that has exactly seven characters:

#!/bin/sh
grep -c '^.......$' "$1"

You have to make the above script executable (with something like chmod +x script.sh), and then run it with ./script.sh file where file is the name of the file that you'd like to examine.

The command

grep -c '^.......$'

will output the number of lines in the given file that contains exactly seven characters (including spaces).

This may be shortened to

grep -cEx '.{7}'

The -E in necessary because the expression is now an extended regular expression. The -x flag will force the expression to match a full line.

If you don't want to count spaces and you're only interested in "word characters" (alphanumerics), you may use

grep -cExw '.{7}'

The -w flag to grep will force the match to be a "word" (a string of alphanumeric characters delimited by non-alphanumeric characters).

If you want to count how many unique lines there are that fulfill this criteria (this is how I interpret "different lines"), then pipe the result of grep without the -c flag through uniq and wc -l. This will count the number of unique lines with exactly seven word characters.

grep -Exw '.{7}' | uniq | wc -l

In the script, this becomes

grep -Exw '.{7}' "$1" | uniq | wc -l

perl -l -0777pe '$_ =()= /^.{7}$/gm' "$1"

awk 'length == 7' "$1" | wc -l

sed -Ee '/.{8}/d; /.{7}/!d' "$1" | grep -c .