grep 行不以 # 或空行开头 [英] grep lines NOT starting with # or empty lines

问题描述

当我使用 grep (GNU grep) 2.22 打印时，not 以 # 开头或not强>空我需要做的

When I use grep (GNU grep) 2.22 to print that do not start by # or that are not empty I need to do

$ grep -v '^#' fileNameIGrepFor |grep -v '^$'

我觉得这很丑，有没有更聪明的方法(请使用grep)

I think this is ugly, is there a smarter way (using grep please)

推荐答案

grep -v '^#' fileNameIGrepFor | grep -v '^$'

可以简化为:

grep -v '^#\|^$' fileNameIGrepFor

要删除丑陋的\，您可以使用grep -E，或等效地egrep:

To remove the ugly \ you can use grep -E, or equivalently egrep:

egrep -v '^#|^$' fileNameIGrepFor

然后您可以通过对术语进行分组来澄清这一点:

You could then clarify this a bit by grouping the terms:

egrep -v '^(#|$)' fileNameIGrepFor

然后通过在 # 之前检查空格使其更加健壮:

And then make it a little more robust by including a check for whitespace before the #:

egrep -v '^(\s*#|$)' fileNameIGrepFor

也许您还想排除所有空行(仅包含空格)?在这种情况下，更改也很简单:

Maybe you'll also want to exclude all blank lines (only contain whitespace)? In which case, again, the change is simple:

egrep -v '^\s*(#|$)' fileNameIGrepFor

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