使用源代码的bash脚本细化菜单

Question

我得到了这两个脚本，configScript.sh和genScript.sh，它们完全按照我的要求工作。configScript.sh设置选项，以便下次在shell中运行genScript.sh时使用。

#!/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   

---

我的问题是我想把configScript.sh中的菜单改成这样：

---

#!/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

---

我可以将所选选项写入options.sh的段中的何处？本部分：

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

如果我将它们保留在foo()和bar()中，这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*我会尽量把它说得更清楚一点。如果我还没有运行任何脚本，options.sh将为空。如果我随后运行configScript.sh并选择响应foo2的选项，然后选择响应bar1的选项，并关闭"configscript.sh . After that I run genScript.sh and I want it to echo foo2 and bar1"。

Answer 1

您的配置脚本：

#!/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