删除所有行，在/pattern/之后的两行开始 [英] Delete all lines, starting two lines after /pattern/

问题描述

想象一下我有一个文件，以下文件:

Imagine I have a file the following file:

drink
eat
XXX
pizza
blunzn
sushi

我想从文件中删除所有行，从模式 XXX 之后的第三行开始，因此结果应类似于:

I would like to remove all lines from the file, starting with the third line after the pattern XXX, so the result should look like:

drink
eat
XXX
pizza
blunzn

删除 XXX 之后的所有行非常简单:

Removing all lines after XXX is simple enough:

sed -e '/XXX/q' -i data.txt

但是，我发现删除模式后很难跳过固定数量的行.

However, I find it hard to skip a fixed number of lines after the pattern from deletion.

到目前为止，我想到的最好的方法是:

The best I came up with so far is:

 sed -e '/XXX/ { N; N; q }' -i data.txt

有什么比添加n * N 更优雅的了(想象中，我想跳过50行)?

Is there something more elegant, than adding n * N (imagine, I would like to skip 50 lines)??

推荐答案

这可能对您有用(GNU sed):

This might work for you (GNU sed):

sed '/pattern/{:a;N;s/\n/&/2;Ta;q}' file

遇到所需的模式时，循环所需的行，然后退出.

On encountering the required pattern, loop the required lines then quit.

对于遵循所需模式的五十行，请使用:

For fifty lines following the required pattern, use:

sed '/pattern/{:a;N;s/\n/&/50;Ta;q}' file

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