这个Bash脚本菜单有什么问题？

Question

I'm tired of typing in long commands to do simple things so I've written this short Bash menu script to speed things up:

#!/usr/bin/env bash
PS3="Nginx? "
options=("start", "stop", "status", "exit")
select opt in "${options[@]}"
do
    case $opt in
        "start")
            sudo service nginx start
            ;;
        "stop")
            sudo service nginx stop
            ;;
        "status")
            sudo service nginx status
            ;;
        "exit")
            echo "Exiting"
            exit 1
            ;;
          *) echo "Invalid selection";;
     esac
done

It's displaying the menu correctly:

1) start
2) stop
3) status
4) exit

But only option 4, exit, works correctly. When I hit 1, 2, or 3, I get "Invalid selection". Am I entering the sudo commands incorrectly? Thank you.

Answer 1

The actual menu displayed by this code is not what you gave in the question, but instead:

1) start,
2) stop,
3) status,
4) exit

This difference -- commas on the end of the first three items -- is instructive.

---

This is an array-syntax declaration issue. Take out the commas:

options=( "start" "stop" "status" "exit" )

...by the way, since your contents are all single-word literals with no glob characters, the quotes don't do anything; it would be just as correct to write:

options=( start stop status exit )

The original code accepts not start but start, ; not stop , but stop, .