counting files, directories and subdirectories in a given path

Question

I am trying to figure how to run a simple script "count.sh" that is called together with a path, eg:

count.sh /home/users/documents/myScripts

The script will need to iterate over the directories in this path and print how many files and folders (including hidden) in each level of this path.

For example:

1. 7
2. 8
3. 9
4. 10

(myScripts - 7, documents - 8, users -9, home - 10)

And by the way, can I run this script using count.sh pwd ?

Answer 1

More or less something like that:

#!/bin/sh

P="$1"

while [ "/" != "$P" ]; do
    echo "$P `find \"$P\" -maxdepth 1 -type f | wc -l`"
    P=`dirname "$P"`;
done
echo "$P `find \"$P\" -maxdepth 1 -type f | wc -l`"

You can use it from the current directory with script.sh `pwd`

Answer 2

You could try the following

#!/bin/bash

DIR=$(cd "$1" ; pwd)
PREFIX=
until [ "$DIR" = / ] ; do 
    echo -n "$PREFIX"$(basename "$DIR")" "$(ls -Ab "$DIR" | wc -l)
    DIR=$(dirname "$DIR")
    PREFIX=", "
done
echo

( ls -Ab lists all files and folders except . and .. and escapes special characters so that only one line is printed per file even if filenames include newline characters. wc -l counts lines.)

You can invoke the script using

count.sh `pwd`

Answer 3

Another approach is to handle separation or tokenizing the path using an array and controlling word-splitting with the internal field separator ( IFS ). You can include the root directory if desired (you would need to trim the additional leading '/' in the printout in that case)

#!/bin/bash

[ -z "$1" -o ! -d "$1" ] && {
    printf "error: directory argument required.\n"
    exit 1
}

p="$1"  ## remove two lines to include /
[ "${p:0:1}" = "/" ] && p="${p:1}" 

oifs="$IFS"     ## save internal field separator value
IFS=$'/'        ## set to break on '/'
array=( $p )    ## tokenize given path into array
IFS="$oifs"     ## restore original IFS

## print out path level info using wc
for ((i=0; i<${#array[@]}; i++)); do
    dirnm="${dirnm}/${array[i]}"
    printf "%d. %s -- %d\n" "$((i+1))" "$dirnm" $(($(ls -Al "$dirnm" | wc -l)-1))
done

Example Output

$ bash filedircount.sh /home/david/tmp
1. /home -- 5
2. /home/david -- 132
3. /home/david/tmp -- 113

As an alternative, you could use a for loop to loop through and count the items in the directory at each level instead of using wc if desired.