Help on Simple Menu Script (PLEASE HELP!!!)

[kriminul1982]

This code is then ran give me an error saying there is something wrong with the syntax at the "done"

begin code

#! /bin/bash

#start a loop
while [ choice != "0" ]
do
clear

#display menu
echo "UNIX Helper Utility


l -- list file names in current directory
o -- sign on this session
f -- sign off this session
d -- display the time and date
b -- background job
s -- interactive shell


q -- quit"

#get variable
read choice


#if else statements for your choices
if [ "$choice" = "l" ]
#list files
then ls
else if [ "$choice" = "o" ]
#sign on
then signon
else if [ "$choice" = "f" ]
#sign off
then signoff
else if [ "$choice" = "d" ]
#display date
then date
else if [ "$choice" = "b" ]
#run background job
then sleep 500 &
else if [ "$choice" = "s" ]
#run interactive shell
then good
else [ "$choice" = "q" ]
#quit
exit
fi

#display menu again
echo "press RETURN for menu"
read key

done
exit 0

end code

[Freston]

Hi, your usage of the 'else if' statement is causing this. In bash, you use elif. If you adjust it, it'll run.

A tip on style, for this sort of menus a case statement may be more readable.

case $choice in
     option1)
          command1
     ;;
     option2)
          command2
     ;;
     option3)
          command3
     ;;
     option4|option5|option6)  # more than one option is possible
          command456
     ;;
     *)                        # all the rest
          command-x
     ;;
esac

[kriminul1982]

ok now I need to be able to return to the beginning of the script and redisplay the menu until the user chooses "q" for quit. What I have isnt working.