我想向文件中添加更多终端/etc/securetty
。更具体地说,我想添加pts/n
, wheren
位于范围内0-9
(如果不存在)。这可以通过sed
命令实现吗?我的内容如下/etc/securetty
:
# Local X displays (allows empty passwords with pam_unix's nullok_secure)
pts/0
pts/1
pts/2
pts/3
我尝试过类似的事情:
sudo sed '+pts/3+a pts/4' /etc/securetty
这给出了以下错误:
sed: -e expression #1, char 3: extra characters after command
答案1
当我们遇到相应的线时,我们记下分数/数字。该-p
选项将autoprint
行。当我们到达时,eof
我们提取哈希值%h
并将其通过grep
过滤器传递,以确定哪些终端没有打印,然后我们使用map
它来准备格式。
perl -lpe 'm|^pts/([0-9])$| and $h{$1}++;
END{ print for map { "pts/$_" } grep { !$h{$_} } 0 .. 9; }
' /etc/securetty
我们hold space
用数字 0 1 2 ... 9 初始化。每当我们遇到这条pts/[0-9]
线时,我们就从保留空间中剪掉它。在 处eof
,我们获取保留空间,如果找到任何数字,则应将其修改为正确的格式并打印出来。
sed -e '
# initialize the hold space with 0 1 ... 9
1{x;s|.*|'"$(echo {0..9})"'|;x}
# whatever be the line, it needs to be printed
p
# we meet a valid pts/ line
\|^pts/[0-9]$|{
# the hold space gets appended to the pattern space
G
# grab what is the pts number and search for it in the hold and
# delete itand store back the changes into hold space.
s|^pts/\([0-9]\)\n\(.*\)\1 |\2|;h
}
# weve not arrived at the eof and weve processed the input so go no further
$!d
# we are at the eof, so we bring back the hold space. just in case all
# numbers were dealt with up, we simply bail out. Else, prepend the str
# pts/ to the numbers present and simply were home
g;/[0-9]/!d;s/ //g
s|[0-9]|pts/&\n|g;s/.$//
# *TIP*: Sprinkle the l, list pattern space at various places to see
# whats going on.
' /etc/securetty
答案2
要在丢失时添加单行,可以通过删除每个出现并将其附加到末尾来完成:
sed -n '/pattern/!p;$a pattern'
但重复 10 个模式就很糟糕了。
sed '/pts\/[0-9]/d;$a pts/0 ...
如果要删除最后一行将会失败。所以反过来说,假设第一行是唯一以 开头的行#
:
sed '/#/a pts/0\
pts/1\
pts/2\
pts/3\
pts/4\
pts/5\
pts/6\
pts/7\
pts/8\
pts\9
/pts\/[0-9]/d'
可恶的。我建议在这种情况下使用不同的工具。
答案3
删除任何/所有pts/N
行,然后将它们全部添加回:
{ grep -xv '^pts/[0-9]$' /etc/securetty; printf 'pts/%d\n' {0..9}; } > /etc/securetty.new
cat /etc/securetty.new
mv /etc/securetty.new /etc/securetty
您也可以使用您最喜欢的文本处理工具一次性完成此操作,例如ed
ed -s /etc/securetty <<IN
g/^pts\/[0-9]$/d
.r ! printf pts/\%d\\\n {0..9}
,p
q
IN
(替换,p
为w
就地编辑)或sed
{ printf '%s\\\n' '$a' pts/{0..8}
printf '%s\n' 'pts/9' '/^pts\/[0-9]$/d'
} | sed -f- /etc/securetty
这与普通的几乎相同
sed '$a\
pts/0\
pts/1\
pts/2\
pts/3\
pts/4\
pts/5\
pts/6\
pts/7\
pts/8\
pts/9
/^pts\/[0-9]$/d' /etc/securetty
(使用sed
和-i
就地编辑文件)
答案4
sed
逐行处理文件,很难让它“记住”跨行的任何信息。
您可以使用grep
来查明文件是否包含给定模式;使用-f
,您可以同时提供多个模式。以下生成完整列表pts/0
.. pts/9
,然后删除给定文件中已存在的模式,并将剩余的模式添加到文件中:
#!/bin/bash
printf 'pts/%d\n' {0..9} \
| grep -vFf "$1" - >> "$1".new
mv "$1".new "$1"