一. 需求

?有如下文件，現要求

• 刪除含有xuecheng關鍵字的行
• 刪除含有192.168.1.1關鍵字的行
• 也就是說，最終只會留下127.0.0.1 license.sublimehq.com

127.0.0.1 www.xuecheng.com
127.0.0.1 img.xuecheng.com
192.168.1.1 www.test.com
127.0.0.1 video.xuecheng.com
127.0.0.1 teacher.xuecheng.com
127.0.0.1 system.xuecheng.com
127.0.0.1 ucenter.xuecheng.com
127.0.0.1 license.sublimehq.com

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二. grep -v 排除

?方式1

• -E：用來指定正則表達式
• -v：用來排除

grep -E "xuecheng|192.168.1.1" -v ./hosts

# 查看hosts文件內容
fengyehong@ubuntu:~/jmw_work_space/20240625$ cat hosts
127.0.0.1 www.xuecheng.com
127.0.0.1 img.xuecheng.com
192.168.1.1 www.test.com
127.0.0.1 video.xuecheng.com
127.0.0.1 teacher.xuecheng.com
127.0.0.1 system.xuecheng.com
127.0.0.1 ucenter.xuecheng.com
127.0.0.1 license.sublimehq.com# 排除指定的行
fengyehong@ubuntu:~/jmw_work_space/20240625$ grep -E "xuecheng|192.168.1.1" -v ./hosts
127.0.0.1 license.sublimehq.com

?方式2

grep -v "xuecheng" ./hosts | grep -v "192.168.1.1"

# 查看hosts文件內容
fengyehong@ubuntu:~/jmw_work_space/20240625$ cat hosts
127.0.0.1 www.xuecheng.com
127.0.0.1 img.xuecheng.com
192.168.1.1 www.test.com
127.0.0.1 video.xuecheng.com
127.0.0.1 teacher.xuecheng.com
127.0.0.1 system.xuecheng.com
127.0.0.1 ucenter.xuecheng.com
127.0.0.1 license.sublimehq.com# 先使用-v排除掉xuecheng關鍵字之后,再使用-v排除192.168.1.1關鍵字
fengyehong@ubuntu:~/jmw_work_space/20240625$ grep -v "xuecheng" ./hosts | grep -v "192.168.1.1"
127.0.0.1 license.sublimehq.com

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三. 創建新文件

grep -v "xuecheng" ./hosts | grep -v "192.168.1.1" > new_hosts