收起这个Shell脚本什么是错误的

Question

因此，我正在为Kali Linux编写一个脚本，它将我的网卡置于监视模式。但是当我试图运行它时，我遇到了第八行中的do命令的问题。下面是代码：

#!/bin/bash 
echo "Preparing to enter monitor mode."` 

检查，看看Firefox的运行

ps cax | grep firefox > /dev/null 
if [ $? -eq 0 ]; then 
echo "Firefox is running. It needs to be closed in order to properly enter monitor mode." #If it is offers to close it 
echo "Would you like me to close it for you?" 
select yn in "Yes" "No"; do 
case $yn in 
Yes) pkill firefox; echo "Firefox killed. Proceeding to enter monitor mode."; echo "Press any button when ready"; wait;; 
No) echo "Proceeding to enter monitor mode."; echo "Press any button when ready."; wait;; 
else 
echo "Proceeding to enter monitor mode." 
echo "Press any button when ready" 
wait 
fi 

询问无线接口的名称设置变量

echo "What is the name of your wireless interface?" 
read interfacevar 
echo Thank you. Entering $interfacevar into monitor mode. 
airmon-ng start $interfacevar 
airmon-ng check kill 
echo "Complete! Exiting in:" 
echo "5" 
sleep 1s 
echo "4" 
sleep 1s 
echo "3" 
sleep 1s 
echo "2" 
sleep 1s 
echo "1" 
sleep 1s 
echo "Goodbye!" 
exit 

但后来我得到的问题说，该做的在第8行是不正确的。

Answer 1

的问题是，你完成一个case块与else时，你应该esac

来完成它可以通过前右else

添加一行

esac 

修复