“阅读” bash 脚本中的命令被跳过

Question

如果我需要擦除硬盘或者需要在另一台机器上安装 Linux，我正在创建一个脚本来更新我的 Linux 发行版。所以这个脚本基本上安装了我通常需要的所有程序。一开始有一个“读取”命令，询问我是否要自动安装所有软件包。如果我选择不，对于每个未找到的程序，它应该询问我是否要安装它并使用此代码

if [[ $installall == "yes" ]]; then
    echo " Installing $sciprog..."
    sudo apt-get install -y $sciprog >/dev/null
    {
        scitest=`dpkg -s $sciprog | grep Status`
    } 2>${HOME}/musthave.errorlog
    if [[ $scitest != "Status: install ok installed" ]]; then
        echo " I've encountered problems installing $sciprog that I can't resolve. "
        echo " Consider installing $sciprog manually. "
        {
            echo "=========="
            echo " $sciprog"
        } >>${HOME}/musthave.notinstalled
    else
        echo " $sciprog installed correctly!"
        {
            echo "=========="
            echo " $sciprog"
        } >>${HOME}/musthave.installed
    fi
else
    echo " Seems like $sciprog is not installed... Do you want to download it?"
    echo " Type 'y' for yes."

    read secondyn ### THIS IS THE GUILTY COMMAND ###

    if [[ $secondyn == "y" ]]; then
        echo " Installing $sciprog ..."
        sudo apt-get install -y $sciprog >/dev/null
        {
            checkinstall=`dpkg -s $sciprog | grep Status`
        } 2>>${HOME}/musthave.errorlog
        if [[ $checkinstall != "Status: install ok installed" ]]; then
            echo " I've encountered problems installing $sciprog that I can't resolve. "
            echo " Consider installing $sciprog manually. "
            {
                echo "=========="
                echo " $sciprog"
            } >>${HOME}/musthave.notinstalled
        else
            echo " $sciprog installed correctly!"
            {
                echo "=========="
                echo " $sciprog"
            } >>${HOME}/musthave.installed
        fi
    else
        echo " Skipping $sciprog ..."
        {
            echo "=========="
            echo " $sciprog"
        } >>${HOME}/musthave.notinstalled
    fi
### some more code which works as expected. All the code above is inside a 
### while...do...done loop which reads line by line the file at the end
done <${HOME}/file.list

但是如果我运行脚本，它会跳过 else 子句中的“read”命令并假设它是“n “...

我不明白为什么，if...then...else...fi 循环中还有其他读取函数，并且它们按预期工作...

任何想法？

Answer 1

代码的相关部分仍然不完整，但根据注释，我猜测您的 while 循环看起来像

while read -r ... ; do 
    # do stuff ...

    # read user input
    read -r var

done < file

这样，问题立即显而易见：内部 read 正在从以下位置获取输入与外循环相同的位置，即从 file 重定向的 stdin，而不是用户。对于不依赖于 /dev/tty 的内核级支持的稍微更便携的替代方案，只需在 while 循环中使用 stdin 以外的不同文件描述符即可。

while read -r ... <&9; do
    # loop stuff

    # stdin still attached to the terminal untouched, 
    # so this reads from the terminal as expected
    read -r var 

done 9< file

请注意，此示例使用 fd 9 作为文件，仅保留 fd 0 (stdin)。请查看 BashFAQ 089 了解更多详细信息。

Answer 2

尝试从控制终端设备读取：

read secondyn </dev/tty

Answer 3

read secondyn < /proc/${PPID}/fd/0

这将查看父级的输入，该输入仍然应该是标准输入。