问题

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im tryng to run this command to grep the exploit MS-* from metasploit exploit's dir

我尝试运行此命令来从metasploit exploit目录中查找MS-*的漏洞

locate -r ".rb$" | xargs grep "MSB" | grep metasploit | grep exploit | grep smb | awk '{print $3,$4}'

找到 -r ".rb$" | xargs grep "MSB" | grep metasploit | grep exploit | grep smb | awk '{print $3,$4}'

i got this result :

我得到了这个结果：

> grep: /usr/share/doc/ruby-http-parser.rb: Is a directory

> grep：/usr/share/doc/ruby-http-parser.rb：是一个目录

> 'MS17-010'],

> 'MS17-010'],

> 'MSB', 'MS03-049'

> 'MSB', 'MS03-049'

> 'MSB', 'MS04-007'],

> 'MSB', 'MS04-007'],

> 'MSB', 'MS04-011'

> 'MSB', 'MS04-011'

> 'MSB', 'MS04-031'],

> 'MSB', 'MS04-031'],

> 'MSB', 'MS05-039'

> 'MSB', 'MS05-039'

> 'MSB', 'MS06-025'

> 'MSB', 'MS06-025'

> 'MSB', 'MS06-040'

> 'MSB', 'MS06-040'],

> 'MSB', 'MS06-066'],

> 'MSB', 'MS06-066'],

> 'MSB', 'MS06-070'

> 'MSB', 'MS06-070'],

> 'MS07-029']

> 'MS07-029']

> MS08-067),

> MS08-067),

> 'MSB', 'MS09-050'

> 'MSB', 'MS09-050'

how i can remove this ('MSB', ' ) and ('])from each line

如何删除每行中的（'MSB'，'）和（']）？

i need output like this :

我需要类似这样的输出：

> MS09-050

> MS09-050

> MS08-067

> MS08-067

> MS06-070

> MS06-070

and also i need to remove the first line ( grep: /usr/share/doc/ruby-http-parser.rb: Is a directory )

我还需要删除第一行（grep：/usr/share/doc/ruby-http-parser.rb：是一个目录）

locate -r "\.rb$" | xargs grep "MSB" | grep metasploit | grep exploit | grep smb | awk '{print $3,$4}'

答案1

得分: 0

locate -r "\.rb$" | xargs grep -s "MSB" | grep metasploit | grep exploit | grep smb | awk '{print $3,$4}' | sed "s/[,' )]//g" | sed "s/MSB //g" | sed "s/]//g"

locate -r "\.rb$" | xargs grep -s "MSB" | grep metasploit | grep exploit | grep smb | awk '{print $3,$4}'|sed "s/[,')]//g" |sed "s/MSB //g"|sed "s/]//g"