如何使用sed删除与模式匹配的行及其后的行? [英] How to delete the line that matches a pattern and the line after it with sed?

问题描述

我有一个看起来像这样的文件:

I have a file that looks something like:

good text
good text
FLAG bad text
bad text
good text
good text
good test
bad Text FLAG bad text
bad text
good text

我需要删除任何包含"FLAG" 的行，并且我总是也需要删除紧接在"FLAG"行之后的一行.

I need to delete any line containing "FLAG" and I always need to delete the one line immediately following the "FLAG" line too.

"FLAG"行不规则地出现，以至于我不能依靠任何种类的行号策略.

"FLAG" lines come irregularly enough that I can't rely on any sort of line number strategy.

有人知道如何用sed做到这一点吗?

Anyone know how to do this with sed?

推荐答案

使用

Using an extension of the GNU version of sed:

sed -e '/FLAG/,+1 d' infile

它产生:

good text
good text
good text
good text
good test
good text

这篇关于如何使用sed删除与模式匹配的行及其后的行?的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！