使用否定排除包含`#`的行

Question

假设我有这样一个文件：

    $ head -n 15 foo.md
    # User Database
    # 
    # Note that this file is consulted directly only when the system is running
    # in single-user mode.  At other times this information is provided by
    # Open Directory.
    #
    # See the opendirectoryd(8) man page for additional information about
    # Open Directory.
    ##
    nobody:*:-2:-2:Unprivileged User:/var/empty:/usr/bin/false
    root:*:0:0:System Administrator:/var/root:/bin/sh
    daemon:*:1:1:System Services:/var/root:/usr/bin/false
    _uucp:*:4:4:Unix to Unix Copy Protocol:/var/spool/uucp:/usr/sbin/uucico
    _taskgated:*:13:13:Task Gate Daemon:/var/empty:/usr/bin/false

我想看看不包括“#”的内容，

它可以通过反转选项来实现，比如

$ grep -v "^#“< foo.md i意图与否定^交替

    $ grep -E "[^#]" < foo.md
    # User Database
    # 
    # Note that this file is consulted directly only when the system is running
    # in single-user mode.  At other times this information is provided by

和

    $ grep -Eo "[^#]+" < foo.md
     User Database

     Note that this file is consulted directly only when the system is running
     in single-user mode.  At other times this information is provided by
     Open Directory.
     See the opendirectoryd(8) man page for additional information about
     Open Directory.
    nobody:*:-2:-2:Unprivileged User:/var/empty:/usr/bin/false

命令没有输出我想要的结果。

如何用否定来解决这样的问题？

Answer 1

您的文件在#之前有空格，因此您需要对其进行说明。否定在这里是不必要的，并且使一切变得复杂--尤其是当试图进行逆匹配时，所以不要去理会它。

grep -v "^\s*#" file.txt

输出

nobody:*:-2:-2:Unprivileged User:/var/empty:/usr/bin/false
root:*:0:0:System Administrator:/var/root:/bin/sh
daemon:*:1:1:System Services:/var/root:/usr/bin/false
_uucp:*:4:4:Unix to Unix Copy Protocol:/var/spool/uucp:/usr/sbin/uucico
_taskgated:*:13:13:Task Gate Daemon:/var/empty:/usr/bin/false

要得到相反的结果，只需失去反向匹配(-v)选项：

grep "^\s*#" 

输出

# User Database
# 
# Note that this file is consulted directly only when the system is running
# in single-user mode.  At other times this information is provided by
# Open Directory.
#
# See the opendirectoryd(8) man page for additional information about
# Open Directory.
##

使用命令grep -Eo "[^#]+"，只告诉grep (-o)与行中没有#的部分匹配，因此显然它将输出除#字符之外的所有内容。

POSIX版

grep '^[[:blank:]]*#'

Answer 2

如果我是对的，你希望只有非评论行，最后5行，而不以“#”开头？

对于我来说，在Mac或-v上使用grep ^# "^#“非常好，输出中只有最后5行。

选项-v的意思是反转匹配，字符“^”表示“行的开头”