如果條件錯誤，bash腳本

Question

這是我的第一個bash腳本，用於將我的內容同步到服務器。我想通過要同步的文件夾作爲參數。但是，它不能按預期工作。

這是我的腳本（保存爲sync.sh）：

echo "STARTING SYNCING... PLEASE WAIT!" 
var="$1" ; 
echo "parameter given is $var" 

if [ "$var"=="main" ] || [ "$var"=="all" ] ; then 
    echo "*** syncing main ***" ; 
    rsync -Paz /home/chris/project/main/ user@remote_host:webapps/project/main/ 
fi 

if [ "$var"=="system" ] || [ "$var"=="all" ] ; then 
    echo "*** syncing system ***" ; 
    rsync -Paz /home/chris/project/system/ user@remote_host:webapps/project/system/ 
fi 

if [ "$var"=="templates" ] || [ "$var"=="all" ] ; then 
    echo "*** syncing templates ***" ; 
    rsync -Paz /home/chris/project/templates/ user@remote_host:webapps/project/templates/ 
fi 

這我的輸出：

chris@mint-desktop ~/project/ $ sh ./sync.sh templates 
STARTING SYNCING... PLEASE WAIT! 
parameter given is templates 
*** syncing main *** 
^Z 
[5]+ Stopped     sh ./sync.sh templates 

雖然我給的 「模板」 作爲一個參數，它忽略它。爲什麼？

Answer 1

基於下面的意見，這是我的建議修正後的腳本：

#!/bin/bash 
echo "STARTING SYNCING... PLEASE WAIT!" 
var="$1" ; 
echo "parameter given is $var" 

if [ "$var" == "main" -o "$var" == "all" ] ; then 
    echo "*** syncing main ***" ; 
    rsync -Paz /home/chris/project/main/ user@remote_host:webapps/project/main/ 
fi 

if [ "$var" == "system" -o "$var" == "all" ] ; then 
    echo "*** syncing system ***" ; 
    rsync -Paz /home/chris/project/system/ user@remote_host:webapps/project/system/ 
fi 

if [ "$var" == "templates" -o "$var" == "all" ] ; then 
    echo "*** syncing templates ***" ; 
    rsync -Paz /home/chris/project/templates/ user@remote_host:webapps/project/templates/ 
fi 

Answer 2

您需要在==運算符兩邊的空間。將其更改爲：

if [ "$var" == "main" ] || [ "$var" == "all" ] ; then 
    echo "*** syncing main ***" ; 
    rsync -Paz /home/chris/project/main/ user@remote_host:webapps/project/main/ 
fi