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shellunixls
Yoo SungKyung
Yoo SungKyung

Reputation: 21

Setting directory as variable in Shell script

What I am trying to do is to count all the files in a directory using shell script.

For example, when execute the program,

./test.sh project

it should count all the files in the folder called "project".

But I am having trouble with the directory part.

What I have done so far is,

#!/bin/bash

directory=$1
count=ls $directory | wc -l
echo "$folder has $count files"

but it does not work... Can anyone blow up my confusion please?

Thanks!

Upvotes: 1

Views: 822

Answers (2)

Guru
Guru

Reputation: 148

#!/bin/bash

directory=$1
count=`ls $directory | wc -l`

echo "$folder has $count files"

Upvotes: 0

Inian
Inian

Reputation: 85895

You have an incorrect syntax while setting the count, for running nested commands in bash you need to use command-substitution using $(..) which runs the commands in a sub-shell and returns the restult

count=$(ls -- "$directory" | wc -l)

But never parse ls output in scripts for any purpose, use the more general purpose find command

find "$1" -maxdepth 1 -type f  | wc -l

Check more about what $(..) form Wiki Bash Hackers - Command substitution

Upvotes: 1

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