Sort and copy files from several directories

Question

I have a directory containing several sub-directories, containing files.

I would like to sort the files in each of these directories by name alphabetically and copy the first file from each directory to a given new directory.

• I would like to ignore hidden files.
• If there is a file with the same name in the destination directory, both files should be kept.

Sample:

parentdir
->dir1
    ->afile
    ->bfile
    ->cfile
->dir2
    ->dfile
    ->efile
    ->ffile
->dir3
    ->afile
    ->hfile
    ->ifile

I would like to run a command that would populate my new directory like so:

newdir
->afile
->afile.1
->dfile

How can this be accomplished?

Answer 1

I guess the straightforward way is to use ls to list the files, which it does in alphabetical order (and omitting hidden files) by default, take the first one from each directory with head, and then have a bit of shell code to figure out what filename they should be copied to. As in this bash script:

for d in ./*/; do
    src="$(ls "$d" | head -n 1)"
    dstbase="newdir/$(basename "$src")"
    if [[ ! -f "$dstbase" ]]; then
        cp "$src" "$dstbase"
    else
        suffix=1
        while [[ -f "$dstbase.$suffix" ]]; do
            let suffix++
        done
        cp "$src" "$dstbase.$suffix"
    fi
done

Obviously you will have to adjust the file paths as needed for your actual situation. This is also fairly fragile in the sense that it may have problems if there are any directories under dir1, dir2, etc. which you would want to exclude from being copied. It also has a potential race condition in the loop that looks for an unused filename; that probably won't be a problem, but don't do anything like running multiple copies of this in parallel.

Answer 2

This can be achieved with the following one line:

ls -ltr /parent_directory_path/ |
awk '$1~"^drw" {print $9}' |
while read line; do
  ls -l $line |
  awk '$1 ~ "^-r" {print $9}' |
  head -1 |
  awk -v line="$line" '{print "cp -rvfp"  "  " line"/"$1 " " "/newdir/" }';
done |
sh

Answer 3

With zsh and GNU cp:

cd parentdir &&
  for dir (*(/)) (cp --backup=numbered -- $dir/*(.[1]) newdir/)

the --backup=numbered implements a backup scheme where files are named file.~1~, file.~2~... So the first copy will make will be file.~max~ and the last one file (process the list backward (*(/On) instead of *(/)) if you want the order reversed).

POSIXly, you could do:

cp_numbered() (
  suffix=
  n=0
  while [ -e "$2$suffix" ] || [ -L "$2$suffix" ]; do
    suffix=.$((n += 1))
  done
  exec cp -- "$1" "$2$suffix"
)
cd parentdir &&
  for dir in */; do
    for file in "$dir"/*; do
      if [ -f "$file" ]; then
        cp_numbered "$file" "newdir/${file#*/}"
        continue 2
      fi
    done
  done