Bash中使用“和” while循环

Question

好吧，实际上这是脚本的样子：

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 

为了这样的事情：

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

在Java中，我知道“和”是“& &”，但似乎并没有工作ħ ERE。我是用正确的方式来使用while循环吗？

编辑：更新的问题，使其在搜索更加有用..

Answer 1

的[]运营商在bash是到test一个电话，这是在man test记录语法糖。 “或”由缀-o表达，但你需要一个“和”：

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

Answer 2

有达到你想要的2分正确的和便携式的方式。
好老shell语法：

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

而且bash语法（如指定）：

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

Answer 3

的便携和可靠的方法是使用case语句。如果你不习惯它，可能只需要围绕语法来考虑一下。

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 

我以前while true，但你实际上可以直接使用case声明那里。尽管如此，阅读和维护会变得多毛。

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