如何在不使用“find”或“ls”命令的情况下递归列出 Bash 中的子目录？

Question

I know you can use the find command for this simple job, but I got an assignment not to use find or ls and do the job. How can I do that?

Answer 1

you can do it with just the shell

#!/bin/bash
recurse() {
 for i in "$1"/*;do
    if [ -d "$i" ];then
        echo "dir: $i"
        recurse "$i"
    elif [ -f "$i" ]; then
        echo "file: $i"
    fi
 done
}

recurse /path

OR if you have bash 4.0

#!/bin/bash
shopt -s globstar
for file in /path/**
do
    echo $file
done

Answer 2

尝试使用

tree -d

Answer 3

Below is one possible implementation:

# my_ls -- recursively list given directory's contents and subdirectories
# $1=directory whose contents to list
# $2=indentation when listing
my_ls() {
  # save current directory then cd to "$1"
  pushd "$1" >/dev/null
  # for each non-hidden (i.e. not starting with .) file/directory...
  for file in * ; do
    # print file/direcotry name if it really exists...
    test -e "$file" && echo "$2$file"
    # if directory, go down and list directory contents too
    test -d "$file" && my_ls "$file" "$2  "
  done
  # restore directory
  popd >/dev/null
}

# recursively list files in current
#  directory and subdirectories
my_ls .

As an exercise you can think of how to modify the above script to print full paths to files (instead of just indented file/dirnames), possibly getting rid of pushd / popd (and of the need for the second parameter $2 ) in the process.

Incidentally, note the use of test XYZ && command which is fully equivalent to if test XYZ ; then command ; fi if test XYZ ; then command ; fi if test XYZ ; then command ; fi (ie execute command if test XYZ is successful). Also note that test XYZ is equivalent to [ XYZ ] , ie the above is also equivalent to if [ XYZ ] ; then command ; fi if [ XYZ ] ; then command ; fi if [ XYZ ] ; then command ; fi . Also note that any semicolon ; can be replaced with a newline, they are equivalent.

Remove the test -e "$file" && condition (only leave the echo ) and see what happens.

Remove the double-quotes around "$file" and see what happens when the directory whose contents you are listing contains filenames with spaces in them. Add set -x at the top of the script (or invoke it as sh -x scriptname.sh instead) to turn on debug output and see what's happenning in detail (to redirect debug output to a file, run sh -x scriptname.sh 2>debugoutput.txt ).

To also list hidden files (eg .bashrc ):

...
for file in * .?* ; do
  if [ "$file" != ".." ] ; then
    test -e ...
    test -d ...
  fi
done
...

Note the use of != (string comparison) instead of -ne (numeric comparison.)

Another technique would be to spawn subshells instead of using pushd / popd :

my_ls() {
  # everything in between roundbrackets runs in a separatly spawned sub-shell
  (
    # change directory in sub-shell; does not affect parent shell's cwd
    cd "$1"
    for file in ...
      ...
    done
  )
}

Note that on some shell implementations there is a hard limit (~4k) on the number of characters which can be passed as an argument to for (or to any builtin, or external command for that matter.) Since the shell expands, inline, * to a list of all matching filenames before actually performing for on it, you can run into trouble if * is expanded inside a directory with a lot of files (same trouble you'll run into when running, say ls * in the same directory, eg get an error like Command too long .)

Answer 4

Since it is for bash, it is a surprise that this hasn't been already said:
(globstar valid from bash 4.0+)

shopt -s globstar nullglob dotglob
echo **/*/

That's all.
The trailing slash / is there to select only dirs.

Option globstar activates the ** (search recursivelly). Option nullglob removes an * when it matches no file/dir. Option dotglob includes files that start wit a dot (hidden files).

Answer 5

The du command will list subdirectories recursively.

I'm not sure if empty directories get a mention, though

Answer 6

Like Mark Byers said you can use echo * to get a list of all files in the current directory.

The test or [] command/builtin has an option to test if a file is a directory.

Apply recursion and you're done.

Answer 7

$ function f { for i in $1/*; do if [ -d $i ]; then echo $i; f $i; fi; done }
$ mkdir -p 1/2/3 2/3 3
$ f .
./1
./1/2
./1/2/3
./2
./2/3
./3

Answer 8

Technically, neither find norls are used by find2perl | perl or File::Find directly.

$ find2perl -type d | perl
$ perl -MFile::Find -e'find(sub{-d&&print"$File::Find::name\n"},".")'

Answer 9

Based on this answer ; use shell options for the desired globbing behaviour:

• enable ** with globstar (Bash 4.0 or newer)
• include hidden directories with dotglob
• expand to the empty string instead of **/*/ if there is no match with nullglob

and then useprintf with the %q formatting directive to quote directory names with special characters in them:

shopt -s globstar dotglob nullglob
printf '%q\n' **/*/

so if you have directories like has space or even containing a newline, you'd get output like

$ printf '%q\n' **/*/
$'has\nnewline/'
has\ space/

with one directory per line.

Answer 10

I wrote a another solution iterative instead of recursive.

iterativePrintDir() {    
    dirs -c;
    currentd=$1
    if [ ! -d $currentd ];then
            echo "Diretorio não encontrado. \"$currentd\""
    fi
    pushd $(readlink -f $currentd)

    while popd 2>&-; do
            echo $currentd
            for d in $(dir $currentd); do
                    test -d "${currentd}/${d}" && pushd  "${currentd}/${d}"
            done
            currentd=$(dirs -l +0)
    done
}