Question

我有两个脚本，configScript.sh和genScript.sh，它们完全按照我的意愿运行。 configScript.sh设置了下次在shell中运行genScript.sh时使用的选项。

#!/bin/bash -x
#configScript.sh
func()
{
echo "
Choose
1 - Option 1
2 - Option 2
"
echo -n "   Enter selection: "
read select
case $select in
            1 ) 
            echo "  Option 1 chosen"
            . ./genScript.sh one
            cat << EOF >options.sh
OPTION=$OPTION
EOF
            ;;
            2 )
            echo "  Option 2 chosen"
            . ./genScript.sh two
            cat << EOF >options.sh
OPTION=$OPTION
EOF
            ;;

esac
}
func

#!/bin/bash -x
#genScript.sh
OPTION=""
. options.sh
[ "$1" ] && OPTION=$1
func2()
{
    if [ "$OPTION" == one ] ; then
        echo "Option one"
    elif [ "$OPTION" == two ] ; then
        echo "Option two"
    else
        echo "null"
    fi
}
func2

我的问题是，我想将configScript.sh中的菜单更改为以下内容：

#!/bin/bash -x
#configScript.sh
func()
{
echo "
Choose
1 - foo
2 - bar 
"
echo -n "   Enter selection: "
read select
case $select in
            1 ) 
            echo "  foo chosen"
            foo
            ;;
            2 )
            echo "  bar chosen"
            bar
            ;;      
esac
}

foo()
{
    echo "
    Choose
    1 - foo1
    2 - foo2
    "
    echo -n "    Enter Selection: "
    read fooSelect
    case $fooSelect in
        1 )
        echo "foo1 chosen"
        ;;
        2 )
        echo "foo2 chosen"
        ;;
    esac
}

bar()
{
        echo "
    Choose
    1 - bar1
    2 - bar2
    "
    echo -n " Enter Selection: "
    read barSelect
    case $barSelect in
        1 )
        echo "bar1 chosen"
        ;;
        2 )
        echo "bar2 chosen"
        ;;
    esac
}
func

我可以在哪里放入段以将所选选项写入options.sh？这部分：

. ./genScript.sh one
cat << EOF >options.sh
OPTION=$OPTION
EOF

如果我将它们保留在foo()和bar()之内，则应该意味着如果我配置为回显foo1然后回显bar2我的genScript.sh应该只输出其中一个。我该如何解决这个问题？

更新* 我会试着让它更清楚一些。如果我没有运行任何脚本，options.sh将为空。如果我然后运行configScript.sh并选择回显foo2的选项，然后选择回显bar1的选项，然后关闭'configscript.sh . After that I run genScript.sh {{1 }} {foo2的{1}} bar1`。

Answer 1

你的配置：

#!/bin/bash -x
#configScript.sh
func()
{
    . options.sh
    echo "
    Choose
    1 - foo
    2 - bar 
    3 - update config
    "
    echo -n "   Enter selection: "
    read select
    case $select in
        1 ) echo "  foo chosen"    ; foo ;;
        2 ) echo "  bar chosen"    ; bar ;;      
        3 ) echo "  update chosen" ; update ;; 
    esac
}

update()
{
cat <<EOF >options.sh
FOO=$FOO
BAR=$BAR 
EOF

. ./genScript.sh
}

foo()
{
    echo "
    Choose
    1 - foo1
    2 - foo2
    "
    echo -n "    Enter Selection: "
    read fooSelect
    case $fooSelect in
        1 ) echo "foo1 chosen" ; FOO="foo1" ;;
        2 ) echo "foo2 chosen" ; FOO="foo2" ;;
    esac
}

bar()
{
    echo "
    Choose
    1 - bar1
    2 - bar2
    "
    echo -n " Enter Selection: "
    read barSelect
    case $barSelect in
        1 ) echo "bar1 chosen" ; BAR="bar1" ;;
        2 ) echo "bar2 chosen" ; BAR="bar2" ;;
    esac
}
func

您的genScript.sh

#!/bin/bash -x
#genScript.sh
FOO=""
BAR=""
. options.sh
[ "$1" ] && FOO=$1
[ "$2" ] && BAR=$2
func2()
{
    echo $FOO
    echo $BAR
}
func2

使用源代码的bash脚本详细说明菜单

1 个答案: