[SOLVED] How to create argument variable in bash script

Issue

I am trying to write a script such that I can identify number of characters of the n-th largest file in a sub-directory. I was trying to assign n and the name of sub-directory into arguments like $1, $2.

Current directory: Greetings
   Sub-directory: language_files, others
     Sub-directory: English, German, French
          Files: Goodmorning.csv, Goodafternoon.csv, Goodevening.csv ….

I would be at directory “Greetings”, while I indicating subdirectory (English, German, French), it would show the nth-largest file in the subdirectory indicated and calculate number of characters as well.

For instance, if I am trying to figure out number of characters of 2nd largest file in English, I did:

langs=$1
n=$2
for langs in language_files/;
 Do count=$(find language_files/$1 name "*.csv" | wc -m | head -n -1 | sort -n -r | sed -n $2(p))
Done | echo "The file has $count bytes!"

The result I wanted was:

$ ./script1.sh English 2
The file has 1100 bytes!

The main problem of all the issue is the fact that I don't understand how variables and looping work in bash script.

Solution

no need for looping

find language_files/"$1" -name "*.csv" | xargs wc -m | sort -nr | sed -n "$2{p;q}"

for byte counting you should use -c, since -m is for char counting (it may be the same for you).

You don't use the loop variable in the script anyway.

Answered By - karakfa