Bash - command is executing but do nothing without complaining

Shell Programming and Scripting
1

Hello.
This part of script do nothing but no complain.
Here the directory content :

The script run by root

cd /root
CMD="rmdir -p -v --ignore-fail-on-non-empty  .ssh"
echo $CMD
$CMD

The ouput

linux:~/bin # ./test_05
rmdir -p -v --ignore-fail-on-non-empty .ssh
rmdir: removing directory, `.ssh'
linux:~/bin # ls /root/.ssh
config  list_user
linux:~/bin #
2

its doing exactly what it should be doing, the rmdir command wont delete the directory because it has files/directories inside it.

The error message is being suppresed by the option "--ignore-fail-on-non-empty".

Remove the option and you will see the error output, use rm -rf and the directory will be deleted.

3

I suppose that ignore-fail-on-non-empty just suppresses the error message and sets the exit status

`--ignore-fail-on-non-empty'
     Ignore each failure to remove a directory that is solely because
     the directory is non-empty.

`-p'
`--parents'
     Remove DIRECTORY, then try to remove each component of DIRECTORY.
     So, for example, `rmdir -p a/b/c' is similar to `rmdir a/b/c a/b
     a'.  As such, it fails if any of those directories turns out not
     to be empty.  Use the `--ignore-fail-on-non-empty' option to make
     it so such a failure does not evoke a diagnostic and does not
     cause `rmdir' to exit unsuccessfully.

As you probably already know, if you intend to remove the .ssh directory and its content, you should use rm instead.

4

Thank you for helping.

I come from ms world ( nobody is perfect ) and it is not the first time I made this kind of error.