bash 如何grep最后一次出现的线条图案
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How to grep the last occurrence of a line pattern
提问by user3702858
I have a file with contents
我有一个包含内容的文件
x
a
x
b
x
c
I want to grep the last occurrence,
我想 grep 最后一次出现，
x
c
when I try
当我尝试
sed -n "/x/,/b/p" file
it lists all the lines, beginning xto c.
它列出了所有的线，开始x对c。
回答by steffen
I'm not sure if I got your question right, so here are some shots in the dark:
我不确定我的问题是否正确，所以这里有一些黑暗中的镜头：
Print last occurence of
x(regex):grep x file | tail -1Alternatively:
tac file | grep -m1 xPrint file from first matching line to end:
awk '/x/{flag = 1}; flag' filePrint file from last matching line to end (prints all lines in case of no match):
tac file | awk '!flag; /x/{flag = 1};' | tac
打印
x（正则表达式）的最后一次出现：grep x file | tail -1或者：
tac file | grep -m1 x从第一个匹配行到结尾打印文件：
awk '/x/{flag = 1}; flag' file从最后一个匹配行到结尾打印文件（在不匹配的情况下打印所有行）：
tac file | awk '!flag; /x/{flag = 1};' | tac
回答by potong
This might work for you (GNU sed):
这可能对你有用（GNU sed）：
sed '/x/h;//!H;$!d;x' file
Saves the last xand what follows in the hold space and prints it out at end-of-file.
将最后一个x和后面的内容保存在保留空间中，并在文件末尾打印出来。
回答by elig
grep -A 1 x file | tail -n 2
-A 1tells grep to print one line after a match line
with tailyou get the last two lines.
-A 1告诉 grep 在匹配行之后打印一行，
并tail获得最后两行。
or in a reversed way:
或以相反的方式：
tac fail | grep -B 1 x -m1 | tac
Note: You should make sure your pattern is "strong" enough so it gets you the right lines. i.e. by enclosing it with ^at the start and $at the end.
注意：您应该确保您的模式足够“强大”，以便为您提供正确的线条。即通过^在开始和$结束时将其括起来。
回答by ray
not sure how to do it using sed, but you can try awk
不确定如何使用sed，但您可以尝试awk
awk '{a=a"\n"##代码##; if (##代码## == "x"){ a=##代码##}} END{print a}' file
