请教如何从文本中取一个关键词后的字符串然后输出呢

mudfr0g

请教一下，如何用shell脚本如何从一个txt文本中取出特定关键词后的字符串并且输出呢

这是个自定义软件输出日志，格式很变态。


比如内容是这样：

[host "oa.cop.com"] [ip 192.168.4.13] [ver "main 1.0"] test备注信息test111
bbbbb报错信息bbbbb [host "oa1.cop.com"] [ip 192.168.4.12] [ver "main 1.0"] test备注信息test111
[host "oa2.cop.com"] [ip 192.168.4.11] [ver "main 1.0"] test备注信息test111



要求输出的是
"oa.cop.com","192.168.4.13","main 1.0"
"oa1.cop.com","192.168.4.12","main 1.0"

这样的
思考再三，感觉似乎只有提取关键词后的字符串，并且输出，否则用域分割直接输出会出错

水平太差，求帮忙 谢谢。

haooooaaa

1. awk -vFPAT='"[^"]+"|[0-9.]{7,15}' '{$2="\""$2"\""}1' OFS=, a
   
2. "oa.cop.com","192.168.4.13","main 1.0"
   
3. "oa1.cop.com","192.168.4.12","main 1.0"
   
4. "oa2.cop.com","192.168.4.11","main 1.0"
   

复制代码

jason680

回复 1# mudfr0g

$ awk -F']|[[][a-z]+ ' -vOFS=, '{print $2,"\""$4"\"",$6}' FILE
"oa.cop.com","192.168.4.13","main 1.0"
"oa1.cop.com","192.168.4.12","main 1.0"
"oa2.cop.com","192.168.4.11","main 1.0"

$ awk '{gsub("] +\\[","#");if(match($0,"host +([^#]+)#ip +([^#]+)#ver +([^]]+)",m))print m[1]",\""m[2]"\","m[3]}' FILE
"oa.cop.com","192.168.4.13","main 1.0"
"oa1.cop.com","192.168.4.12","main 1.0"
"oa2.cop.com","192.168.4.11","main 1.0"

mudfr0g

非常感谢 的确思路很重要