Question

具有以下shell，其中（while true）循环内部（while true）循环。我试图用“休息”打破内循环，但事实并非如此。我想打破内循环＆amp;显示用户第一循环的选项;建议请。

#!/bin/ksh
    sub_menu() {
    echo "~~~~~~~~~~~~~~~~~~~~~~~~~"    
    echo "    S U B - M E N U "
    echo "~~~~~~~~~~~~~~~~~~~~~~~~~"
    echo "1. Display properties"
    echo "2. Back"
    }

    read_sub_options(){

    echo "Please select option" 
    read option
    case $option in
        1) sub_menu ;;
        ***2) break ;;***
        *) echo "Please insert options 1 ~ 2";;
        esac
    }

showSubMenu(){
    while true
    do
        sub_menu
        read_sub_options
done    
}

read_options(){ 
echo "Please select option "
    read option
    case $option in
        1) showSubMenu ;;
        2) exit 0;;
        *) echo "Please insert options ";;
    esac
}

show_menus() {
    echo "~~~~~~~~~~~~~~~~~~~~~"    
    echo "   M A I N - M E N U "
    echo "~~~~~~~~~~~~~~~~~~~~~"
    echo "1. Sub Menu"
    echo "2. Exit"
}

# -----------------------------------
#           MAIN 
# ------------------------------------
while true
do
    show_menus
    read_options
done

以下是输出：

~~~~~~~~~~~~~~~~~~~~~
   M A I N - M E N U
~~~~~~~~~~~~~~~~~~~~~
1. Sub Menu
2. Exit
Please select option
1
~~~~~~~~~~~~~~~~~~~~~~~~~
    S U B - M E N U
~~~~~~~~~~~~~~~~~~~~~~~~~
1. Display properties
2. Back
Please select option
2
~~~~~~~~~~~~~~~~~~~~~~~~~
    S U B - M E N U
~~~~~~~~~~~~~~~~~~~~~~~~~
1. Display properties
2. Back
Please select option
2

Answer 1

以下脚本适用于此。

如果read_sub_options被调用，则sub_menu返回2，当选择返回选项时返回1，然后检查showSubMenu中最后一次运行命令的返回值，如果返回则返回地选择。

有关更多KSH脚本参考信息，请参阅KSH script basics和Learning the Korn shell。

有关您的问题评论中提到的其他菜单选择方法的信息，请参阅select和ksh手册页。

#!/bin/ksh
sub_menu() {
echo "~~~~~~~~~~~~~~~~~~~~~~~~~"    
echo "    S U B - M E N U "
echo "~~~~~~~~~~~~~~~~~~~~~~~~~"
echo "1. Display properties"
echo "2. Back"
return 2;
}

read_sub_options(){

echo "Please select option" 
read option
case $option in
    1) sub_menu ;;
    2) return 1;;
    *) echo "Please insert options 1 ~ 2";;
    esac
}

showSubMenu(){
while true
do
    sub_menu
    read_sub_options
    if [[ $? == 1 ]] ; then
    break
    fi
done    
}

read_options(){ 
echo "Please select option "
    read option
    case $option in
        1) showSubMenu;;
        2) exit 0;;
        *) echo "Please insert options ";;
    esac
}

show_menus() {
    echo "~~~~~~~~~~~~~~~~~~~~~"    
    echo "   M A I N - M E N U "
    echo "~~~~~~~~~~~~~~~~~~~~~"
    echo "1. Sub Menu"
    echo "2. Exit"
}

# -----------------------------------
#           MAIN 
# ------------------------------------
while true
do
    show_menus
    read_options
done

Answer 2

shell“始终”支持：

break 2

打破两个级别的shell循环。即使是第7版UNIX™Bourne shell也是如此。

Answer 3

您正试图从该函数内部的函数外部循环。我不完全清楚这是否应该有效，但我发现它不适用于ksh93，mksh和Heirloom Bourne shell，而它适用于bash，ash（如dash和FreeBSD sh）和zsh。 / p>

您可以返回一个特殊状态，并在循环中执行相同的功能，就像在其他一个答案中一样。

Shell break while-true循环

3 个答案: