如何执行"grep"搜索，以排除以"//"开头的行及其前面的空格? [英] How to perform a 'grep' search that excludes line starting with "//"and its preceding spaces?

问题描述

如何执行"grep"搜索，该搜索不显示在行首带有注释标记"//"的任何行，而忽略"//"标记前面的空格? >

我尝试了grep your_search_pattern' where_to_look | grep -v '^//，但是它不会忽略在"//"之前有空格的行.

解决方案

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

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?

I attempted grep your_search_pattern' where_to_look | grep -v '^//, but it does not ignore lines that have spaces before "//".

解决方案

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

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