Ubuntu – Using shell find all sub-directories that contain certain files

command linescripts

I am trying to use shell to find all sub-directories in any directory. What I would want is to have a .sh file (shell script file) that can receive as a parameter the name of the directory I'm interested in and the list of files I want to find (NOTE: I want only sub-directories that have all these files).

I know I can use this:

find $D -perm -u=rx -type f

Where D is the directory, -u is the user, r is the users right to read and x is the right to modify I believe, but uhm I have no idea how to make the file accept parameters and I have no idea how to use -u=rx

EDIT: I now understand how to use parameters for a shell script file, so that's ok. I still don't get most of the rest.

I would love it if someone could either explain the code I mentioned or … give an alternative ?

I'm also ok with a partial answer, I just need some help.

Best Answer

If I understand correctly what you want to do, this is the solution:

#!/bin/sh

USAGE="Usage: $0 dir file1 file2 ... fileN\nto find all subdirectories of dir that contain all the given files.\n"

if [ "$#" == "0" ]; then
    printf "$USAGE"
    exit 1
fi

ARG=""
DIR=$1
shift

while (( "$#" )); do
  ARG="$ARG -exec test -e \"{}/$1\" \; "
  shift
done

cmd="find $DIR -type d $ARG -print"
eval $cmd

What it does is this:

The use find ... -type d to find all subdirectories (including the directory given as first parameter). The test -e command checks if a file exists. So for a given directory we have to check all the files given in the command line: test -e /path/to/directory/file1 test -e /path/to/directory/file2 test -e /path/to/directory/file3 ... The /path/to/directory is {} - a single result of find. Then the find-parameter -exec can be used to check for a single file. To check for all files several -exec test parameters are needed. So while loop build a list of there parameters, then this list is put together in a single command and evaluated.

Have fun ...

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