linux - 打破 Bash 中的嵌套函数循环

Question

我正在使用嵌套函数来分区并为连接到新 Linux 机器的驱动器制作文件系统。我在尝试打破所有循环时遇到一个奇怪的问题。

我正在跟踪嵌套循环索引并使用“break n”。当用户对“Do you have any additional drives to partition?”的问题回答“n”时我希望跳出所有嵌套循环并继续执行脚本，但结果是问题再次被问到。

你能帮我解决这个问题吗？

INIT_STARTED=0
chooseDisks()
{
    INIT_STARTED=$((INIT_STARTED+1))    
    # Choosing which drive to work on
    read -p "Please type the name of the disk you want to partition: " DISK

    while true; do
        read -p "Are you sure you want to continue ? y (partition)/n (choose another drive) /x (continue) " ynx
        case $ynx in
            [Yy]* )
                containsElement "$DISK"
                if [ $? == 1 ]; then
                        initializeDisk $DISK 

                        # remove element from found disk to prevent trying to partition it again.
                        delete=($DISK)
                        FOUNDDISKS=( "${FOUNDDISKS[@]/$delete}" )                       
                else
                    echo "${red}$DISK is not a valid choice, please select a valid disk.${reset}"
                    chooseDisks
                fi

                break;;
            [Nn]* ) 
                chooseDisks
                break $((INIT_STARTED));;
            [Xx]* ) 
                return
                break;;
            * ) echo "Please answer y or n. x to continue the script.";;
        esac
    done

    # Any additional disks to partition?
    while true; do
        read -p "Do you have any additional drives to partition ? y/n " yn
        case $yn in
            [Yy]* )
                #chooseDisks $FOUNDDISKS
                chooseDisks
                break $((INIT_STARTED));;
            [Nn]* )
                return
                break $((INIT_STARTED));;
            * ) echo "Please answer y or n";;
        esac
    done

}

我希望这样:

break $((INIT_STARTED));;

结束第n次循环并退出函数。

Answer 1

不要玩嵌套逻辑 break，只需使用一些变量，如 $userStop 而不是 while true;做 放

userStop = false
while[!${userStop}]
do
#...
# replace break $((INIT_STARTED));; by
# userStop = true