如何获取从最后匹配到文件结尾的行? [英] How to get lines from the last match to the end of file?

问题描述

需要在最后一次匹配到文件末尾之后打印行.比赛的数量可以是任意的，而不是确定的.我有一些文字，如下所示.

Need to print lines after the last match to the end of file. The number of matches could be anything and not definite. I have some text as shown below.

MARKER
aaa
bbb
ccc
MARKER
ddd
eee
fff
MARKER
ggg
hhh
iii
MARKER
jjj
kkk
lll

期望的输出是

jjj
kkk
lll

我可以将awk与RS和FS一起使用以获得所需的输出吗?

Do I use awk with RS and FS to get the desired output?

推荐答案

实际上，您可以使用 awk (gawk)进行操作，而无需使用任何管道.

You can actually do it with awk (gawk) without using any pipe.

$ awk -v RS='(^|\n)MARKER\n' 'END{printf "%s", $0}' file
jjj
kkk
lll

说明:

• 您可以通过 RS ='(^ | \ n)MARKER \ n'将记录分隔符定义为(^ | \ n)MARKER \ n ，默认情况下是是 EOL char
• 'END {printf％s"，$ 0}' =>在文件末尾，您将整行打印出来，因为 RS 设置为(^ | \ n)MARKER \ n ， $ 0 将包括所有行，直到EOF.

• You define your record separator as (^|\n)MARKER\n via RS='(^|\n)MARKER\n', by default it is the EOL char
• 'END{printf "%s", $0}' => at the end of the file, you print the whole line, as RS is set at (^|\n)MARKER\n, $0 will include all the lines until EOF.

---

另一种选择是使用 grep (GNU):

$ grep -zoP '(?<=MARKER\n)(?:(?!MARKER)[^\0])+\Z' file
jjj
kkk
lll

说明:

• -z 使用ASCII NUL字符作为分隔符
• -o 仅打印匹配项
• -P 激活perl模式
• PCRE正则表达式:(?< = MARKER \ n)(?:( ?! MARKER)[^ \ 0])+ \ Z 在这里 https://regex101.com/r/RpQBUV/2/

• -z to use the ASCII NUL character as delimiter
• -o to print only the matching
• -P to activate the perl mode
• PCRE regex: (?<=MARKER\n)(?:(?!MARKER)[^\0])+\Z explained here https://regex101.com/r/RpQBUV/2/

---

最后但并非最不重要的是，还可以使用以下 sed 方法:

sed -n '/^MARKER$/{n;h;b};H;${x;p}' file
jjj
kkk
lll

说明:

• n 跳至下一行
• h 用当前行替换保留空间
• H 进行相同的操作，但不要替换，而是添加
文件交换结束时
• $ {x; p} ( x )保留空间和模式空间并打印( p )

• n jump to next line
• h replace the hold space with the current line
• H do the same but instead of replacing, append
• ${x;p} at the end of the file exchange (x) hold space and pattern space and print (p)

可以变成:

tac file |  sed -n '/^MARKER$/q;p' | tac

如果我们使用 tac .

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