Question

好的，基本上这就是脚本的样子：

echo -n "Guess my number: "
read guess

while [ $guess != 5 ]; do
echo Your answer is $guess. This is incorrect. Please try again.
echo -n "What is your guess? "
read guess
done

echo "That's correct! The answer was $guess!"

我想改变的是这一行：

while [ $guess != 5 ]; do

对于这样的事情：

while [ $guess != 5 and $guess != 10 ]; do

在Java中我知道“和”是“＆amp;＆amp;”但这似乎不适用于此。我是否使用while循环以正确的方式进行此操作？

编辑：更新了问题，使其在搜索中更有用..

Answer 1

有两种正确且便携的方法可以达到你想要的效果良好的旧shell语法：

while [ "$guess" != 5 ] && [ "$guess" != 10 ]; do

bash语法（如您所指定）：

while [[ "$guess" != 5 && "$guess" != 10 ]]; do

Answer 2

bash中的[]运算符是test调用的语法糖，man test中对此进行了记录。 “或”由中缀-o表示，但您需要“和”：

while [ $guess != 5 -a $guess != 10 ]; do

Answer 3

可移植且可靠的方法是使用case语句。如果你不习惯它，可能需要一些看起来只是围绕语法。

while true; do
    case $guess in 5 | 10) break ;; esac
    echo Your answer is $guess. This is incorrect. Please try again.
    echo -n "What is your guess? "
    read guess  # not $guess
done

我使用while true但您实际上可以直接使用case语句。但是，阅读和维护会变得毛茸茸。

while case $guess in 5 | 10) false;; *) true;; esac; do ...

在Bash while循环中使用“和”

3 个答案: