How to loop through all the files in a directory and print all the file names, like the command tree

Question

How can I mimic the tree command and iterate through all the files and subdirectories of a directory and echo all the file names?

I thought that a subdirectory within a directory is still counted as a directory, so I did:

for FILE in *; do
if [ -d $FILE ]
then echo "$FILE is a directory";
else echo "$FILE is a file"
fi
done

How can I make this recursive and loop inside the subdirectories and iterate through all the files and print their file name + path?

This is a shell script using bash. Thanks!

Answer 1

Here's an equivalent to terdon's shell function implemented as a find command:

find . -exec sh -c 'for f; do
                      [ -d "$f" ] && is_dir="" || is_dir="not "
                      printf "\"%s\" is %sa directory\n" "$f" "$is_dir"
                    done' sh {} +

Or with find & perl (indenting two spaces per directory level and using NUL as the filename separator):

find . -print0 | \
  perl -0ne '$is_dir = -d $_ ? "" : "not ";
             $f = $_; $indent = ($f =~ s=/==g);
             printf "  " x $indent . "%s is %sa directory\n", $_, $is_dir'

Most of the perl one-liner is straight-forward and obvious, but one line might require some explanation:

$f = $_; $indent = ($f =~ s=/==g);

This copies the current input record (i.e. the filename, $_) into variable $f. $f is then modified to remove all / characters while the count of changes is stored in variable $indent. $f isn't used again, it's just a throwaway variable to avoid changing $_.

The $indent variable is used with the repetition operator (x - similar to multiplication but for strings, see man perlop) to construct the printf format string.

---

To count the files and directories, with 3-digit wide numbering of each line:

$ find . -print0 | perl -0ne '
    if ( -d $_ ) {
      $is_dir = "";
      $dirs++
    } else {
      $is_dir = "not ";
      $files++
    };

    $f = $_; $indent = ($f =~ s=/==g);
    #printf "  " x $indent . "%s is %sa directory\n", $_, $is_dir;
    printf "%3i: " . "  " x $indent . "%s is %sa directory\n", $., $_, $is_dir;

    END {
      # simple method to determine singular or plural word forms
      # without using the Lingua::EN::Inflect module
      # (see https://metacpan.org/release/Lingua-EN-Inflect)
      my $d_word = ($dirs  != 1) ? "directories" : "directory";
      my $f_word = ($files != 1) ? "files" : "file";

      print "\n$dirs $d_word, $files $f_word\n";
    }'

The perl part of this one-liner has got to the point where it is no longer reasonable to edit on the command line, because dealing with absurdly long lines is a complete PITA - it's much easier to use a text editor. It should be saved to a script file somewhere in your $PATH (e.g. ~/bin/ or /usr/local/bin - add to your PATH if it's not already there) with #!/usr/bin/perl -0ne as the first line and made executable with chmod +x.

Answer 2

If you are just interested in separating directories from everything else, you can do something like this, using a recursive function:

#!/bin/bash

## Do not expand globs to themselves if they don't match
shopt -s nullglob

list_files(){
  ## if a target has been passed as an argument, use that; if not,
  ## default to '.', the current directory.
  target=${1:-.}
  for i in "$target"/*; do
    if [ -d "$i" ]; then
      list_files "$i"
    else
      echo "$i is not a directory"
    fi
  done
}

list_files "$@"

Given a directory structure like this:

$ tree
.
├── 1.txt
├── 2.txt
├── 3.txt
├── dir1
│   ├── subdir1
│   │   ├── file
│   │   ├── subsubdir1
│   │   └── subsubdir2
│   └── subdir2
│       ├── file
│       ├── subsubdir1
│       └── subsubdir2
├── dir2
│   ├── subdir1
│   │   ├── subsubdir1
│   │   └── subsubdir2
│   └── subdir2
│       ├── subsubdir1
│       └── subsubdir2
└── symlink -> 2.txt

15 directories, 6 files

The code above produces:

$ bar.sh
"./1.txt" is not a directory
"./2.txt" is not a directory
"./3.txt" is not a directory
"./dir1" is a directory
"./dir1/subdir1" is a directory
"./dir1/subdir1/file" is not a directory
"./dir1/subdir1/subsubdir1" is a directory
"./dir1/subdir1/subsubdir2" is a directory
"./dir1/subdir2" is a directory
"./dir1/subdir2/file" is not a directory
"./dir1/subdir2/subsubdir1" is a directory
"./dir1/subdir2/subsubdir2" is a directory
"./dir2" is a directory
"./dir2/subdir1" is a directory
"./dir2/subdir1/subsubdir1" is a directory
"./dir2/subdir1/subsubdir2" is a directory
"./dir2/subdir2" is a directory
"./dir2/subdir2/subsubdir1" is a directory
"./dir2/subdir2/subsubdir2" is a directory
"./symlink" is not a directory

You can also make it a little easier to read by adding an offset to indicate the depth in the directory tree, and by processing all directories first so as to have a more ordered output:

#!/bin/bash

## Do not expand globs to themselves if they don't match
shopt -s nullglob

list_files(){
  ## if a target has been passed as an argument, use that; if not,
  ## default to '.', the current directory.
  target=${1:-.}

  for i in "$target"/*; do
    offset=${2:-""}
    if [ -d "$i" ]; then
      printf '%s"%s" is a directory\n' "$offset" "$i"
      for subdir in "$i"/*/; do
        new_offset="$offset   "
        list_files "$i" "$new_offset"
        new_offset=""
      done
    else
      printf '%s"%s" is not a directory\n' "$offset" "$i" 
    fi
  done
}

list_files "$@"

This will produce the following output on the example input above:

$ foo.sh 
"./1.txt" is not a directory
"./2.txt" is not a directory
"./3.txt" is not a directory
"./dir1" is a directory
   "./dir1/subdir1" is a directory
      "./dir1/subdir1/file" is not a directory
      "./dir1/subdir1/subsubdir1" is a directory
      "./dir1/subdir1/subsubdir2" is a directory
         "./dir1/subdir1/subsubdir2/file" is not a directory
   "./dir1/subdir2" is a directory
      "./dir1/subdir2/file" is not a directory
      "./dir1/subdir2/subsubdir1" is a directory
      "./dir1/subdir2/subsubdir2" is a directory
"./dir2" is a directory
   "./dir2/subdir1" is a directory
      "./dir2/subdir1/subsubdir1" is a directory
      "./dir2/subdir1/subsubdir2" is a directory
   "./dir2/subdir2" is a directory
      "./dir2/subdir2/subsubdir1" is a directory
      "./dir2/subdir2/subsubdir2" is a directory
"./symlink" is not a directory

Note that this approach will descend into symlinked directories. For example, if I add this:

$ ln -s dir2 dirsym
$ ls -ld dirsym
lrwxrwxrwx 1 terdon terdon 4 Apr  8 16:03 dirsym -> dir2

Then dirsym will be treated as a directory and its contents will be reported just like its target, dir2, effectively repeating the dir2 output. I don't know if this is the desired behavior or not. If it isn't, use caz's find approach instead.

Answer 3

(
  find . -type f -print0 | xargs -0 printf '%s is a file\0'
  find . -type d -print0 | xargs -0 printf '%s is a directory\0'
) | sort -z | tr '\0' '\n'

I'm not sure if I consider this to be a good answer, was something of a "I wonder if I could..."

Produce a list of files and directories.

Print them with appropriate suffix.

Then sort the entire result so items appear in the right context.

Some extra fiddling needed to handle items with newline characters in their name.