Shell脚本菜单开发：如何添加返回菜单的选项？

结合你当前的脚本，我们可以通过调整循环结构、将菜单打印逻辑放入循环内，并新增返回菜单选项来实现需求。以下是修改后的完整脚本及关键说明：

while :
do
    # 将菜单打印放入循环内，每次返回都重新显示完整选项
    echo "1. I'm number one"
    echo "2. I'm number two"
    echo "3. I'm number three"
    echo "4. Return to menu"  # 新增返回菜单选项
    echo "5. Exit from menu"

    echo -n "Enter one of the following numbers: "
    read choice

    case $choice in
        1)  
            echo "You have selected the number one"
            echo "Selected number is one"
            # 执行完操作后等待用户确认，再返回菜单
            read -p "Press Enter to return to menu..."
            ;;
        2)  
            echo "You have selected the number two"
            echo "Selected number is two"
            read -p "Press Enter to return to menu..."
            ;;
        3)  
            echo "You have selected the number three"
            echo "Selected number is three"
            read -p "Press Enter to return to menu..."
            ;;    
        4)  
            echo "Returning to main menu..."
            sleep 1  # 短暂延迟提示用户，可选
            ;;
        5)  
            echo "Exiting after the information have been received by both devices"
            exit 0
            ;;
        *)  # 处理无效输入，提升脚本健壮性
            echo "Invalid choice! Please enter a number between 1-5"
            read -p "Press Enter to try again..."
            ;;
    esac
    # 清空屏幕，让菜单显示更整洁（可选）
    clear
done

关键修改点 #

• 菜单循环显示：把原来仅在开头打印一次的菜单，放到while循环内部，确保用户每次返回都能看到完整的选项列表。
• 新增返回菜单选项：添加选项4作为返回入口，执行时会提示用户并重新进入循环（即回到菜单）。
• 操作后等待确认：在1/2/3选项执行完毕后，加入read -p让用户按回车再返回菜单，避免操作信息一闪而过。
• 无效输入处理：新增*)分支，当用户输入错误选项时给出明确提示，引导重新输入。
• 选项逻辑调整：将原退出选项从4调整为5，让菜单功能划分更清晰。

内容的提问来源于stack exchange，提问作者user19873675