Question

我正在尝试编写一个脚本并使其尽可能干净明亮。总之，我想呈现一个菜单，让一个人选择一个与他们喜欢的颜色相对应的数字。从那里我想获取菜单中带有数字的颜色名称，并将其用作将放置在其他地方的脚本中的变量。我的目标是在菜单后面有一个语法，但会使用颜色变量。这是让我回归的一件事。下面是一个片段..有什么想法吗？

color_pref=
while [ -z "$color_pref" ] ;
do
echo
echo
echo "  ***************************************"
echo " What is your favorite color? "
echo "  1 - red"
echo "  2 - blue"
echo "  3 - green"
echo "  4 - orange"
echo "    Select 1, 2, 3, or 4 :" \n
echo "  ***************************************"
printf "    Enter Selection> "; read color_pref
echo [[[whatever variable is for color selected]]]

Answer 1

您可以使用case语句将set变量等于基于所选数字的颜色。

case $color_pref in
    1) color=red ;;
    2) color=blue ;;
    3) color=green ;;
    4) color=blue ;;
    *) printf "Invalid color choice: %s" "$color_pref" >&2
       exit;
esac

您可能需要查看select命令，该命令负责处理大部分菜单显示和选择。

Answer 2

您也可以使用关联数组：

declare -A colors=( [1]=red [2]=blue [3]=green [4]=orange )

示例：

declare -A colors=( [1]=red [2]=blue [3]=green [4]=orange )
color_pref=
while [ -z "$color_pref" ]
do
echo
echo
echo "  ***************************************"
echo " What is your favorite color? "
echo "  1 - red"
echo "  2 - blue"
echo "  3 - green"
echo "  4 - orange"
echo "    Select 1, 2, 3, or 4 :" \n
echo "  ***************************************"
printf "    Enter Selection> "; read color_pref
echo ${colors[$color_pref]}
done

或索引数组：

declare -a colors=('invalid' 'red' 'blue' 'green' 'orange' )

用法：

echo ${colors[$color_pref]}

Answer 3

您可以将COLORS存储在类似

的数组中

COLORS=('red' 'blue' 'green' 'orange')

然后您可以使用类似

的内容回显所选值

echo $color_pref ${COLORS[color_pref-1]}

你需要添加一个

done

结束循环。总之类似的事情，

#!/usr/bin/env bash

COLORS=('red' 'blue' 'green' 'orange')
color_pref=
while [ -z "$color_pref" ] ;
do
        echo
        echo
        echo "  ***************************************"
        echo " What is your favorite color? "
        echo "  1 - red"
        echo "  2 - blue"
        echo "  3 - green"
        echo "  4 - orange"
        echo "    Select 1, 2, 3, or 4 :" \n
        echo "  ***************************************"
        printf "    Enter Selection> "; read color_pref
        echo $color_pref ${COLORS[color_pref-1]}
done

Linux变量基于用户选择

3 个答案: