Bash中使用「和」 while循環

Question

好吧，實際上這是腳本的樣子：

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!" 

我想改變的是這一行：

while [ $guess != 5 ]; do 

爲了這樣的事情：

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

在Java中，我知道「和」是「& &」，但似乎並沒有工作ħ ERE。我是用正確的方式來使用while循環嗎？

編輯：更新的問題，使其在搜索更加有用..

Answer 1

的[]運營商在bash是到test一個電話，這是在man test記錄語法糖。 「或」由綴-o表達，但你需要一個「和」：

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

Answer 2

有達到你想要的2分正確的和便攜式的方式。
好老shell語法：

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

而且bash語法（如指定）：

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

Answer 3

的便攜和可靠的方法是使用case語句。如果你不習慣它，可能只需要圍繞語法來考慮一下。

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，但你實際上可以直接使用case聲明那裏。儘管如此，閱讀和維護會變得多毛。

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