如何执行忽略注释掉的行的grep？ [英] How to perform a grep which ignores commented out lines?

问题描述

我想知道如何执行一个'grep'搜索，它不会显示任何在行前有注释标记//的行，但也会忽略//标记前面的空格

I was wondering how to perform a 'grep' search which doesn't display any lines that have the comment marks "//" in the front of the line, but also ignore whitespace in front of the // marks?

我正在使用grep*，但我不知道如何不使用-v标志来搜索某些内容，我尝试了但没有成功。 / p>

I am using grep "" *, but I don't know how to not search for something using the -v flag, I tried but it didn't work.

推荐答案

 grep 'your_search_pattern' where_to_look | grep -v '^//'

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