Question

CONFIRM=""

echo "Do you want to backup your home directory? Please enter y|Y or n|N"

read CONFIRM;
if [[ "$CONFIRM" != "Y" ]] || [[ "$CONFIRM" != "y" ]]; then
  echo "The backup will not run. Aborting now..."
else
  echo "The backup will run now. Backing up to $DEST"
  # more code
fi

每当我输入“y”或“Y”时，应执行else下的条件。相反，无论我输入什么，if下的条件都会被执行。我尝试了多种方括号和引用的组合，但是没有任何效果。为什么else分支没有正确执行？

Answer 1

此：

[[ "$CONFIRM" != "Y" ]] || [[ "$CONFIRM" != "y" ]]

表示＆＃34; $CONFIRM不是Y 或 $CONFIRM不是y＆＃34; 。这总是保证是真的：

如果$CONFIRM是Y，那么它不是y，所以第二个选项是真的
如果$CONFIRM为y，那么它不是Y，因此第一个选项为true
如果$CONFIRM是其他任何内容，那么两个选项都是真的

要解决此问题，您需要使用&&（＆＃34;和＆＃34;）代替||（＆＃34;或＆＃34;）：

[[ "$CONFIRM" != Y ]] && [[ "$CONFIRM" != y ]]

，De Morgan's laws，相当于：

! { [[ "$CONFIRM" = Y ]] || [[ "$CONFIRM" = y ]] ; }

Answer 2

如果我们仔细查看提示＆＃34;请输入 y | Y或n | N ＆＃34; ，然后缺少第三种情况，处理无效输入。与@cdarke的答案一起，这会导致例如在以下短期条件下：

CONFIRM=""

echo "Do you want to backup your home directory? Please enter y|Y or n|N"

read CONFIRM;
if [[ "$CONFIRM" = [Yy] ]]; then
  echo "The backup will run now. Backing up to $DEST"
  # more code
elif [[ "$CONFIRM" = [Nn] ]]; then
  echo "The backup will not run. Aborting now..."
else
  echo "Wrong Input. Please enter y|Y or n|N."
fi

Answer 3

另一个例子，我确保在特定y / n / q作为输入之前我们不会继续。使用-p -n1无需按Enter键。总结：一个case语句，读取一个按键，小写检查等...这个例子允许是/否或退出。

typeset confirm=''
while true; do
    read -p "Select (y/n/q) ?" -n1 confirm
    case ${confirm,,} in
        y|n|q) break;;
        *) echo "Answer y for yes / n for no  or q for quit.";;
    esac
done
echo -e "\nAnswer = $confirm" ;sleep 2

if [[ "${confirm,,}" == "q" ]] ; then
        echo "OK Quitting.."
        exit 0
fi

if [[ "${confirm,,}" == "y" ]] ; then
        echo "Continuing ..."
else
        echo "No? Do something else etc.. "
fi

测试不平等不起作用

3 个答案: