如何在指定的行号后grep字符串?

问题描述

我有一个很大的文本文件，我想在这个文本文件中看到包含花费的时间"的行，我使用:

I have a large text file, I want to see the lines containing "time spent" in this text file, I use:

grep -in "time spent" myfile.txt

但我只对 50000 之后的行感兴趣.在输出中，我想看到 50000 之后的行并且包含花费的时间".有没有办法做到这一点?

But I'm interested only in the lines after 50000. In the output I want to see lines after 50000 and that contain "time spent". Is there a way to do this?

推荐答案

你可以拖尾，然后 grep:

You can tail it, then grep:

tail -n +50000 myfile.txt | grep -in "time spent"

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