Shell script to list out the contents of a file

This shell script will list out the contents of a file using MENU DRIVEN, considering the following:
- The user enters the file name (may enter a path)
- The script reads in the file name
- It checks if it exists and it is a file
- If it does not, it warns the user
- Otherwise it should list the contents

# !/bin/bash
# MENU DRIVEN BASH SCRIPT CREATED BY Nathan Joseph (http://NJ180degree.net)
# CREATED ON APRIL 30th 2010
# ISSUED UNDER THE GNU GPL (GNU GPL)
echo "Hello $USER, Welcome to THE CONTROL MENU"
echo "The Date & Time is:"
date

function FILE {
	echo -e "Please enter a file name"
	read FILE
	if [ ! -f $FILE ]
	then
		echo -e "Error: The file $FILE does NOT exists!"
	else
		cat "$FILE"
		echo -e ""

fi
}

echo -e "Please choose from the following options by entering 0 or 1"
echo -e "1. List out the contents of a file!"
echo -e "0. Exit"

read option
while [ "$option" != "0" ]
	do
		case $option in
			1) FILE ;;
			*) echo -e "Invaild choice, please choose from 0 or 1 only" ;;
			0) echo " "
			echo -e "0.Exit the program"
			exit 0
			 ;;
		esac
		echo -e "Please press return to go back to the main menu"
		read RETURN
		echo -e "Please choose from the following options by entering 0 or 1"
		echo -e "1. List out the contents of a file!"
		echo -e "0. Exit"
			read option
	done
echo -e "Thanks for using my script.. you may change or modify it to suit your purposes :)"
exit

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