Bash script instead of rm to prompt with number of files about to delete (a safer rm)

[OS/4]

Background
I always review my rm command line carefully before executing, so I thought I'd be fine. The other day while using the command history, I recalled and executed a previous command, thinking it's "ls some files", but didn't notice it's "rm some files". I executed that in a different dir before, but the same files happened to be in the current dir and got deleted - oops.

I don't want to start using rm -i, because I hate that it prompts for every single file - annoying when you're deleting many!
What I really want is a single "are you sure" prompt, so this can't happen again. I'd also like it to tell me how many files I'm about to delete, because that would act as a sanity check (especially when using -R and accidentally being in a directory higher up ;-)

I came across the idea to count the files using grep on the output of "ls". It looked promisingly simple, but then I tested it with wildcards and files with spaces and it got a bit more complicated than I would have liked:

rmi.sh

#!/bin/bash
# Writes out how many files and then prompts for deletion
if [ "$#" -eq 0 ] ; then
    echo "usage: rmi [-R] file ..."
    exit 1
fi

RECURSIVE=""
for PARAM in "$@" ; do
    if [ "$PARAM" = "-R" ] ; then
        RECURSIVE="-R "
    fi
done

if [ -z "$RECURSIVE" ] ; then
    NUMBER=0
    for PARAM in "$@" ; do
        if [ -f "$PARAM" ] ; then
            ((NUMBER+=$(ls -1Ap "$PARAM" | grep -c '[^:]$')))
        fi
    done
else
    NUMBER=$(ls -1Ap $RECURSIVE "$@" | grep -c '[^:]$')
fi

if [ $NUMBER = 0 ] ; then
    echo "rmi: $@: No such file or directory"
else
    read -n 1 -p "Deleting $NUMBER file(s), Are you sure (Y/N)? " yn
    echo
    if [ "$yn" != "${yn#[Yy]}" ] ; then
        # rm "$@"
    fi
fi

I would greatly appreciate if someone with more UNIX/shell skills could point ot any flaws in this.

1. What I'm not happy with, is that I need the if [ -z "$RECURSIVE" ] ... it feels like a hack.
I would like a simple ls -1Ap "$@", but if I call the script with a single * wildcard, the shell expands it and passes all files and directories in the current directory and then the ls command also picks up the files in the first level of sub-directories.

2. I don't mind if it only works in bash. I don't mind if the only option that works is -R (which happens to work the same for rm and ls).

3. I don't mind that it's wrong with "." or ".." as args, rm doesn't work with those anyway.

4. I've added "-p" to the "ls" because I thought I might filter out directories (from the number of file(s) reported) based on ending with a slash later, but I don't really mind.

5. I would train myself to use an alias called "rmi" (short for "rm -i" ;-) rather than set up an "rm" alias that lures me into safety and might be disastrous if the alias isn't there (e.g. on a different computer).

I have commented out the only call to actual "rm" so I can test it safely.

Many thanks to anyone who bothers to look at my script ;-)

[OS/4]

I found out about the -d option on ls, which avoid listing directory contents if a dir is specified (or expanded from *). That makes the script simpler:

#!/bin/bash
# Writes out how many files and then prompts for deletion
if [ "$#" -eq 0 ] ; then
    echo "usage: rmi [-f | -i] [-PRrvW] file ..."
    exit 1
fi

FLAGS="d" # don't go into specified/expanded directories unless recursive flag (below)
for PARAM in "$@" ; do
    if [ "$PARAM" = "-r" ] || [ "$PARAM" = "-R" ] ; then
        FLAGS="R"
    fi
done

NUMBER=$(ls -1APp$FLAGS "$@" 2>/dev/null | grep -c '[^/:]$') # don't count directories or directory headings
if [ $NUMBER = 0 ] ; then
    echo "rmi: $@: No such file or directory"
else
    read -n 1 -p "Deleting $NUMBER file(s), Are you sure (Y/N)? " yn
    echo
    if [ "$yn" != "${yn#[Yy]}" ] ; then
        echo rm "$@"
    fi
fi

So now the only questions I would have for someone experienced with UNIX is whether there are any big dangers or flaws in this script. Thanks!