我正在linux中制作一个新的菜单驱动Shell脚本,我已经简化了我的表,只是hello和bye,使这个更简单,下面是我的基本菜单布局
# Menu Shell Script
#
echo ----------------
echo menu
echo ----------------
echo [1] hello
echo [2] bye
echo [3] exit
echo ----------------
基本上我有菜单,我最近一直在玩一些东西,但似乎不能得到任何工作,因为我是新的,我想下一行将是
`read -p "Please Select A Number: " menu_choice`
,但我不知道该如何处理变量和什么不。我想知道是否有人可以帮助我编写下一段代码,让它在我按1时表示hello,在按2时表示bye,在用户按3时表示exit。我已经尝试了好几天不同的方法,但似乎都没有效果,如果你能这么做,我将不胜感激。
echo...
和read
不需要这些反号
echo "----------------"
echo " menu"
echo "----------------"
echo "[1] hello"
echo "[2] bye"
echo "[3] exit"
echo "----------------"
read -p "Please Select A Number: " mc
if [[ "$mc" == "1" ]]; then
echo "hello"
elif [[ "$mc" == "2" ]]; then
echo "bye"
else
echo "exit"
fi
编辑
showMenu(){
echo "----------------"
echo " menu"
echo "----------------"
echo "[1] hello"
echo "[2] bye"
echo "[3] exit"
echo "----------------"
read -p "Please Select A Number: " mc
return $mc
}
while [[ "$m" != "3" ]]
do
if [[ "$m" == "1" ]]; then
echo "hello"
elif [[ "$m" == "2" ]]; then
echo "bye"
fi
showMenu
m=$?
done
exit 0;
示例
if [ $menu_choice -eq 1 ]
then
echo hello
elif [ $menu_choice -eq 2 ]
then
echo bye
elif [ $menu_choice -eq 3 ]
then
exit 0
fi
或使用大小写
case $menu_choice in
1) echo hello
;;
2) echo bye
;;
3) exit 0
;;
esac