Bash – How to remove all files in a directory except hidden ones

In BASH you can use the trailing slash (I think it should work in any POSIX shell):

rm -R -- */

Note the -- which separates options from arguments and allows one to remove entries starting with a hyphen - otherwise after expansion by the shell the entry name would be interpreted as an option by rm (the same holds for many other command line utilities).

Add the -f option if you don't want to be prompted for confirmation when deleting non-writeable files.

Note that by default, hidden directories (those whose name starts with .) will be left alone.

An important caveat: the expansion of */ will also include symlinks that eventually resolve to files of type directory. And depending on the rm implementation, rm -R -- thelink/ will either just delete the symlink, or (in most of them) delete the content of the linked directory recursively but not that directory itself nor the symlink.

If using zsh, a better approach would be to use a glob qualifier to select files of type directory only:

rm -R -- *(/) # or *(D/) to include hidden ones

or:

rm -R -- *(-/)

to include symlinks to directories (but because, this time, the expansion doesn't have trailing /s, it's the symlink only that is removed with all rm implementations).

With bash, AT&T ksh, yash or zsh you can do:

set -- */
rm -R -- "${@%/}"

to strip the trailing /.

With the file zz containing the list of file names, this works, so just replace cat zz.

cat zz | grep -vF -f <(cat zz|sort -r|uniq -w11)

e.g. echo *.sql | grep -vF -f <( echo *.sql | sort -r | uniq -w11 ) | xargs rm

As is, it won't work if spaces in file names, and very fragile to filename length.