Shell script to check if a directory contains more than 4 files-CodePudding

I need a shell script that checks if a directory contains more than 4 files and change directory owner.

The below script only checks if a directory is empty or not , if the directory is empty it changes the directory owner to lohith and if directory is not empty it changes owner to satha .

Here what needs to be added to script to check if a directory contains more than 4 files it needs to change owner to satha and if directory contains less than 4 files or if directory is empty it needs to change owner to lohith.

#!/bin/bash
FILE=""
DIR="/home/ec2-user/test1"
# init
# look for empty dir 
if [ "$(ls -A $DIR)" ]; then
     chown satha $DIR
else
     chown lohith $DIR
fi

CodePudding user response：

You generally do not want to parse the output of the ls command in a script. A non-ls example might look like:

#!/bin/bash
dir_files=("$DIR"/*)
if [[ "${#dir_files[@]}" -gt 4 ]] ; then
 #More than 4 files
elif [[ -e "${dir_files[0]}" ]] ; then
 #non-empty
else
 #empty
fi

This puts the list of files in an array, and then checks in order: If the array has more than 4 elements (thus more than 4 files), then checks if the directory is non-empty by confirming the first element of the array exists (which proves the glob matched something), then falls into the final case where the directory was empty.

CodePudding user response：

Here is an example that does use ls. wc -l <filename> will return the number of lines in a file, so if we pipe the output of ls to wc, we can get the number of lines in the file. Since $(pwd) gets the current working directory, you can replace it with any other directory.

#!/bin/bash

DIR=$(pwd)

count=$(ls -a $DIR | wc -l)
if (( $count > 4 )); then
     echo "more than four"
else
     echo "less than four"
fi