Looping on empty directory content in Bash

Question

I'm writing a shell script in which I need to loop over directories and then loop over files inside them. So I've written this function:

loopdirfiles() {
    #loop over dirs
    for dir in "${PATH}"/*
    do
        for file in "${dir}"/*
            do
                echo $file
            done
    done
}

The problem is that it echoes something like path/to/dir/* on empty directories.

Is there a way to use this approach and ignore those kind of directories?

Answer 1

You can cut the * from the directory name instead of completely ignoring it:

[[ $file == *"*" ]] && file="${file/%\*/}"
#this goes inside the second loop

Or if you want to ignore the empty directory:

[[ -d $dir && $ls -A $dir) ]] || continue
#this goes inside the first loop

Another way:

files=$(shopt -s nullglob dotglob; echo "$dir"/*)
(( ${#files} )) || continue
#this goes inside the first loop

Or you can turn on the nullglob (mentioned by Etan Reisner) and dotglob altogether:

shopt -s nullglob dotglob
#This goes before first loop.

---

From Bash Manual

nullglob

If set, Bash allows filename patterns which match no files to expand to a null string, rather than themselves.

dotglob

If set, Bash includes filenames beginning with a ‘.’ in the results of filename expansion.

Note: dotglob includes hidden files (files with a . at the beginning in their names)