在每行匹配后使用 grep 获取下一个 WORD [英] Using grep to get the next WORD after a match in each line

问题描述

我想从我的服务器日志中获取GET"查询.

I want to get the "GET" queries from my server logs.

例如，这是服务器日志

1.0.0.127.in-addr.arpa - - [10/Jun/2012 15:32:27] code 404, message File not fo$
1.0.0.127.in-addr.arpa - - [10/Jun/2012 15:32:27] "GET /hello HTTP/1.1" 404 -   
1.0.0.127.in-addr.arpa - - [10/Jun/2012 15:41:57] code 404, message File not fo$
1.0.0.127.in-addr.arpa - - [10/Jun/2012 15:41:57] "GET /ss HTTP/1.1" 404 -

当我尝试使用简单的 grep 或 awk 时，

When I try with simple grep or awk,

Adi:~ adi$ awk '/GET/, /HTTP/' serverlogs.txt

它发出

1.0.0.127.in-addr.arpa - - [10/Jun/2012 15:32:27] "GET /hello HTTP/1.1" 404 -
1.0.0.127.in-addr.arpa - - [10/Jun/2012 15:41:57] "GET /ss HTTP/1.1" 404 -

我只想显示:hello 和 ss

有什么办法可以做到吗?

Is there any way this could be done?

推荐答案

假设你有 gnu grep，你可以使用 perl-style regex 做一个积极的回顾:

Assuming you have gnu grep, you can use perl-style regex to do a positive lookbehind:

grep -oP '(?<=GETs/)w+' file

如果您没有 gnu grep，那么我建议您只使用 sed:

If you don't have gnu grep, then I'd advise just using sed:

sed -n '/^.*GET[[:space:]]{1,}/([-_[:alnum:]]{1,}).*$/s//1/p' file

如果你碰巧有 gnu sed，那可以大大简化:

If you happen to have gnu sed, that can be greatly simplified:

sed -n '/^.*GETs+/(w+).*$/s//1/p' file

这里的底线是，您当然不需要管道来完成此操作.grep 或 sed 就足够了.

The bottom line here is, you certainly don't need pipes to accomplish this. grep or sed alone will suffice.

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