删除所有行，在/ pattern /之后开始两行(Delete all lines, starting two lines after /pattern/)

删除所有行，在/ pattern /之后开始两行(Delete all lines, starting two lines after /pattern/)

想象一下，我有一个文件如下文件：

drink eat XXX pizza blunzn sushi

我想删除文件中的所有行，从模式XXX之后的第三行开始，因此结果应如下所示：

drink eat XXX pizza blunzn

删除XXX后的所有行很简单：

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

但是，我发现在删除模式后很难跳过固定数量的行。

到目前为止我想出的最好的是：

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

有没有比添加n * N更优雅的东西（想象一下，我想跳过50行）？

Imagine I have a file the following file:

drink eat XXX pizza blunzn sushi

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

Removing all lines after XXX is simple enough:

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

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

The best I came up with so far is:

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

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

最满意答案

这可能适合你（GNU sed）：

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

在遇到所需的模式时，循环所需的行然后退出。

对于遵循所需模式的50行，使用：

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

This might work for you (GNU sed):

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

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

For fifty lines following the required pattern, use:

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