Linux shell腳本程序在29行後退出？

Question

我正在編寫一個shell腳本來創建一個用戶並將它們添加到一個組中。我遇到了第17行的障礙，當我有一個錯誤，可以在這裏看到http://imgur.com/XhNGIeW ，但一半的代碼似乎是完美的（就添加用戶而言），但任何人都可以幫助我這個謝謝。

#!/bin/bash 
let repeat=1 
let counter=0 
while [ $repeat -eq 1 ]; 
     do 
     echo "Please enter the username for the created user" 
     read username 
     sudo useradd -m $username 
     echo "" 
     sudo passwd $username 
     let counter=$counter+1 
     while [ $repeat -eq 1 ]; 
       do 
       echo "please enter the name of the group to put the user into" 
       read group 
       if [ $(getent group $group) ]; then #line 17 
         sudo usermod -G $group,sudo $username 
         let repeat=0 
       else 
         echo "The group "$group$" does not exist on our system" 
         echo "Do you want to create it as a new group (y)" 
         read input 
         if [ $input == "y" ]; then 
           sudo groupadd $groups 
           sudo usermod -G $group,sudo $username 
           let repeat=0 

         fi 

       fi 

     done 
     echo $username" has beem added to the group "$group 
    done 

    echo "You have now created "$counter" new user(s) and have added them to their new group(s)" 

Answer 1

如果組的名稱是無效的，getent group $group將不返回任何 輸出，使線17將等同於：

if [ != '' ]; then 

的!=操作前，操作後比較字符串。如果 所需的字符串缺少一個，殼報告錯誤：

[: !=: unary operator expected 

如果你想檢查一個有效的組，你可以簡單地檢查一些輸出 返回：

if [ $(getent group $group) ]; then