Bash grep ip从行 [英] Bash grep ip from line

问题描述

我有文件ip.txt，其中包含以下内容

I have the file ip.txt which contain the following

ata001dcfe16f85.mm.ph.ph.cox.net (24.252.231.220)
220.231.252.24.xxx.com (24.252.231.220)

我做了这个bash命令来提取ips:

and I made this bash command to extract ips :

grep -Eo '(25[0-5]|2[0-4][0-9]|[01]?[0-9][0-9]?)\.(25[0-5]|2[0-4][0-9]|[01]?[0-9][0-9]?)\.(25[0-5]|2[0-4][0-9]|[01]?[0-9][0-9]?)\.(25[0-5]|2[0-4][0-9]|[01]?[0-9][0-9]?)' ip.txt | sort -u > good.txt

我想编辑代码，以便仅提取括号之间的ips.不是所有的ip都在线，因为当前代码提取了ip 220.231.252.24

I want to edit the code so it extracts the ips between the parentheses ONLY . not all the ips on the line because the current code extract the ip 220.231.252.24

推荐答案

要使IP处于括号内，您所需要做的就是将整个正则表达式包装在转义的\( \)

To get the IP within paranthesis all you need is to wrap the entire regex in an escaped \( \)

grep -Eo '\((25[0-5]|2[0-4][0-9]|[01]?[0-9][0-9]?)\.(25[0-5]|2[0-4][0-9]|[01]?[0-9][0-9]?)\.(25[0-5]|2[0-4][0-9]|[01]?[0-9][0-9]?)\.(25[0-5]|2[0-4][0-9]|[01]?[0-9][0-9]?)\)'

将输出为

(24.252.231.220)
(24.252.231.220)

如果您还希望在输出中摆脱偏执狂，环顾四周会很有用

if you want to get rid of the paranthesis as well in the output, look around would be usefull

grep -oP '(?<=\()(25[0-5]|2[0-4][0-9]|[01]?[0-9][0-9]?)\.(25[0-5]|2[0-4][0-9]|[01]?[0-9][0-9]?)\.(25[0-5]|2[0-4][0-9]|[01]?[0-9][0-9]?)\.(25[0-5]|2[0-4][0-9]|[01]?[0-9][0-9]?)(?=\))'

将产生的输出为

24.252.231.220
24.252.231.220

一个更轻巧的版本

grep -oP '(?<=\()(25[0-5]|2[0-4][0-9]|[01]?[0-9]{2}?)(\.(25[0-5]|2[0-4][0-9]|[01]?[0-9]{2}?)){3}(?=\))'

此处

[0-9]{2}匹配数字2次

(\.(25[0-5]|2[0-4][0-9]|[01]?[0-9]{2}?)){3}匹配.，后跟三位3位数

(\.(25[0-5]|2[0-4][0-9]|[01]?[0-9]{2}?)){3} matches . followed by 3 digit number three times

可以使用到uniq的管道将重复行删除为

The repeating lines can be removed using a pipe to uniq as

grep -oP '(?<=\()(25[0-5]|2[0-4][0-9]|[01]?[0-9]{2}?)(\.(25[0-5]|2[0-4][0-9]|[01]?[0-9]{2}?)){3}(?=\))' input | uniq

将输出赋予为

24.252.231.220

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