Check if a directory contains another directory - bash

Question

How do I check if a given directory contains another directory in shell. I want to pass 2 full path directories. (I know this is stupid, but just for learning purposes). Then I want to see if any one of those 2 paths is contained in the other one.
parent=$1
child=$2
if [ -d $child ]; then
echo "YES"
else
echo "NO"
fi
this however makes no use of the parent directory. Only checks if the child exists.

Answer 1

You can use find to see if one name is contained within another:
result=$(find "$parent" -type d -name "$child")
if [[ -n $result ]]
then echo YES
else echo NO
fi

Answer 2

Create a file (ex: dircontains.sh) with this code:
#!/bin/bash
function dircontains_syntax {
local msg=$1
echo "${msg}" >&2
echo "syntax: dircontains <parent> <file>" >&2
return 1
}
function dircontains {
local result=1
local parent=""
local parent_pwd=""
local child=""
local child_dir=""
local child_pwd=""
local curdir="$(pwd)"
local v_aux=""
# parameters checking
if [ $# -ne 2 ]; then
dircontains_syntax "exactly 2 parameters required"
return 2
fi
parent="${1}"
child="${2}"
# exchange to absolute path
parent="$(readlink -f "${parent}")"
child="$(readlink -f "${child}")"
dir_child="${child}"
# direcory checking
if [ ! -d "${parent}" ]; then
dircontains_syntax "parent dir ${parent} not a directory or doesn't exist"
return 2
elif [ ! -e "${child}" ]; then
dircontains_syntax "file ${child} not found"
return 2
elif [ ! -d "${child}" ]; then
dir_child=`dirname "${child}"`
fi
# get directories from $(pwd)
cd "${parent}"
parent_pwd="$(pwd)"
cd "${curdir}" # to avoid errors due relative paths
cd "${dir_child}"
child_pwd="$(pwd)"
# checking if is parent
[ "${child_pwd:0:${#parent_pwd}}" = "${parent_pwd}" ] && result=0
# return to current directory
cd "${curdir}"
return $result
}
Then run these commands
. dircontains.sh
dircontains path/to/dir/parent any/file/to/test
# the result is in $? var
# $1=0, <file> is in <dir_parent>
# $1=1, <file> is not in <dir_parent>
# $1=2, error
Obs:
- Tested only in ubuntu 16.04/bash
- In this case, the second parameter can be any Linux file

Answer 3

Pure bash, no external commands used:
#!/bin/bash
parent=$1
child=$2
[[ $child && $parent ]] || exit 2 # both arguments must be present
child_dir="${child%/*}" # get the dirname of child
if [[ $child_dir = $parent && -d $child ]]; then
echo YES
else
echo NO
fi

Answer 4

Works for sub-directories too:
parent=$1
child=$2
if [[ ${child/$parent/} != $child ]] ; then
echo "YES"
else
echo "NO"
fi

Answer 5

Similar to Barmar's answer but far more reliable than name comparisons:
if find "$parent" -samefile "$child" -printf 'Y\n' -quit | grep -qF Y; then
echo "contains '$child'"
fi
To be even safer, you can also follow symlinks to ensure that there's no way a recursive operation on $parent could damage $child or anything in it:
if find -L "$parent" -samefile "$child" -printf 'Y\n' -quit | grep -qF Y; then
echo "contains '$child' or link thereto"
fi

Answer 6

You can accomplish this in pure bash. Loop over every file in $1 and see if "$1/$2" is a dir, like so:
parent=$1
child=$(basename $2)
if [ -d $parent ] && [ -d $child ]; then
for child in $parent; do
if [ -d "$parent/$child" ]; then
echo "Yes"
else
echo "No"
fi
done
fi