bash - Bash while 语句 - 无限循环

Question

#!/bin/bash
CHECKUSER=$(grep "$USER" /var/log/string.log)
if [[ $CHECKUSER == "" ]];
   then
        echo  "Please enter y or n? (y/n)?"
        read string
        if [ "$string" = "y" -o "$string" = "n" ];
           then
              {
              echo "$USER - $string" >> /var/log/string.log
              }
           else
              while [ "$string" != "y" -o "$string" != "n" ];
                 do
                 echo "'$string' is an invalid option, please enter y or n: "
                 read string
              done
        fi
elif [[ $CHECKUSER == "$USER - n" ]];
   then
        echo "User selected n"
elif [[ $CHECKUSER == "$USER - y" ]];
   then
        echo "You've already said that you would like your account backed up."
else echo "User entered something other than y or n"
fi

这一切都很好！但是，如果您输入的不是 y|n，您就会陷入无限循环。

有任何想法吗？

Answer 1

你必须改变

while [ "$string" != "y" -o "$string" != "n" ];

到

while [ "$string" != "y" -a "$string" != "n" ];

因为否则它总是正确的：

if string="y" ----------> [ false -o true  ] == [ true ]
if string="n" ----------> [ true  -o false ] == [ true ]
if string="whatever" ---> [ true  -o true  ] == [ true ]

Answer 2

在这个 while 循环中有两个问题：

while [ "$string" != "y" -o "$string" != "n" ];
do
    echo "'$string' is an invalid option, please enter y or n: "
    read backup
done

正如 fedorqui 指出的那样，您的测试始终是正确的，但即使您修复了该问题，您也将循环，因为您正在读取backup而不是string。将内容更改为：

while [ "$string" != "y" -a "$string" != "n" ];
do
    echo "'$string' is an invalid option, please enter y or n: "
    read string
done