Bash or condition in IF

Question

爲什麼OR條件在下面的bash腳本中不起作用？

read number 
if [ "$number" != 1 -o "$number" != 2 ] ; then echo not 1 or 2 fi 
if [ "$number" == 1 -o "$number" == 2] ; then echo 1 or 2 fi 
exit 

Answer 1

我想這和它的作品：

#!/bin/bash 
read number 
if [ "$number" != 1 -o "$number" != 2 ];then 
    echo not 1 or 2 
fi 
if [ "$number" == 1 -o "$number" == 2 ];then 
    echo 1 or 2 
fi 
exit 

輸出：

[~]$./test1.sh 
2 
not 1 or 2 
1 or 2 

反正按照正確的語法和你的第一個輸出消息，你應該這樣做：

#!/bin/bash 
read number 
if [ "$number" -ne 1 ] && [ "$number" -ne 2 ];then 
    echo not 1 or 2 
fi 
if [ "$number" -eq 1 ] || [ "$number" -eq 2 ];then 
    echo 1 or 2 
fi 
exit 

輸出：

[~]$./test1.sh 
1 
1 or 2 
[~]$./test1.sh 
2 
1 or 2 
[~]$./test1.sh 
3 
not 1 or 2 
[~]$./test1.sh 
4 
not 1 or 2 

問候 克勞迪奧