bash script to compare file names in one directory with file names in another while ignoring file extensions

Question

I am trying to compare the files in one directory with those in another. Let's take directories test 1 and test2. If test 1 contains a file named 123.JPG and if test 2 contains a file named 123.JPG.gpg I would then take no course of action: 123.JPG has already been encrypted. However if no match was successful I would then run gpg to encrypt 123.JPG.

I found this script elsewhere which I tried to modify to achieve the above but to no avail:

cd source
for x in *; do
  set -- "…/dest/${x%.*}".*
  if [ $# -eq 1 ] && ! [ -e "$1" ]; then
    echo "$x has not been converted"
  elif [ $# -gt 1 ]; then
    echo "$x has been converted to more than one output file: " "$@"
  else
    echo "$x has been converted to $1"
  fi
done

Can anyone please help?

Thank you

Answer 1

If you have GNU's find and you can use the printf option, then the follwing command:

comm -23 <(find A -type f -printf '%f.gpg\n'|sort) <(find B -type f -printf '%f\n'|sort)

will give you a list of missing files:

If directory A is:

A
├── 123.JPG
├── 456.JPG
└── 789.JPG

And directory B is:

B
├── 123.JPG.gpg
└── 456.JPG.gpg

The output would be 789.JPG.gpg which is the missing encrypted file - it does not exist in directory B, and a file with a similar name but without the extension, does exist in directory A.

You can then use the result list to encrypt the missing files.

Please notice that this command works recursively.