bash 有没有更简洁的方法来获取每行的最后 N 个字符？
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Is there a cleaner way of getting the last N characters of every line?
提问by merlin2011
To simplify the discussion, let N = 3.
为了简化讨论，让N = 3。
My current approach to extracting the last three characters of every line in a file or stream is to use sedto capture the last three characters in a group and replace the entire line with that group.
我当前提取文件或流中每一行的最后三个字符的方法是使用sed捕获组中的最后三个字符并将整行替换为该组。
sed 's/^.*\(.\{3\}\)//'
It works but it seems excessively verbose, especially when we compare to a method for getting the firstthree characters in a line.
它有效但似乎过于冗长，尤其是当我们与获取一行中的前三个字符的方法进行比较时。
cut -c -3
Is there a cleaner way to extract the last N characters in every line?
有没有更简洁的方法来提取每行的最后 N 个字符？
回答by Tiago Lopo
It's very simple with grep -o '...$':
这非常简单grep -o '...$'：
cat /etc/passwd | grep -o '...$'
ash
/sh
/sh
/sh
ync
/sh
/sh
/sh
Or better yer:
或者更好：
N=3; grep -o ".\{$N\}$" </etc/passwd
ash
/sh
/sh
/sh
ync
/sh
/sh
That way you can adjust your Nfor whatever value you like.
这样你就可以调整N你喜欢的任何值。
回答by Cyrus
rev /path/file | cut -c -3 | rev
回答by Ed Morton
Why emphasize brevity when it's a tiny command either way? Generality is much more important:
当它是一个很小的命令时，为什么要强调简洁？一般性更重要：
$ cat file
123456789
abcdefghijklmn
To print 3 characters starting from the 4th character:
从第 4 个字符开始打印 3 个字符：
$ awk '{print substr($ awk '{print substr($ awk '{print substr($ while read -r in; do echo "${in: -3}"; done
hello
llo
$
,(length($ sed 's,.*\(.\{3\}\)$,,'
hallo
llo
$
)-3)/2,3)}' file
345
efg
,length(##代码##)-3,3)}' file
678
klm
,4,3)}' file
456
def
To print 3 characters starting from the 4th-last character:
从倒数第 4 个字符开始打印 3 个字符：
##代码##To print 3 characters from [around] the middle of each line:
要从每行中间的 [周围] 打印 3 个字符：
##代码##回答by lx42.de
Pure bash solution:
纯 bash 解决方案：
##代码##sed
sed
##代码##