如何用sed中的单行替换两个模式之间的行? [英] How do I replace lines between two patterns with a single line in sed?

问题描述

这是我的输入文件:

one
two
three
four
five
six
seven
eight
nine
ten

我想把文件变成

one
two
three
NEW LINE
eight
nine
ten

与sed.也就是说，我想用单行NEW LINE替换从/four/(包括)到/seven/(包括)的行.

我可以做到

sed '/four/aNEW LINE
/four/,/seven/d' file.txt

但是我想知道是否有一种更简单的方法，尤其是一种无需重复模式的方法(就像我对/four/所需要的那样).

编辑根据 fedorquis 的评论问题，这也可能是awk(尽管出于学术"目的，我会对sed解决方案感兴趣.)

编辑2 不幸的是，输入文件表明输入文件中的单词具有逻辑顺序(一个后跟两个通过三个等).但是，在我的现实世界"问题中，情况并非如此.我不知道文件有多少行，四和七行之前或之后是什么.我所知道的唯一的事情是，有一行 four (不一定立即)紧跟着一行 7 .我很抱歉在提出问题时没有明确说明这一点，尤其是因为 fedorqui 在他的回答上花了很多心血.

解决方案

这可能对您有用(GNU sed& bash):

sed $'/^four/{:a;N;/^seven/McNEWLINE\nba}' file

This is my input file:

one
two
three
four
five
six
seven
eight
nine
ten

I want to turn the file into

one
two
three
NEW LINE
eight
nine
ten

with sed. That is, I want to replace the lines from /four/ (including) to /seven/ (including) with the single line NEW LINE.

I can do that with

sed '/four/aNEW LINE
/four/,/seven/d' file.txt

But I am wondering if there is a simpler way, notably one without having to repeat a pattern (as I needed to with /four/).

Edit As per fedorquis comment-question, this can also be in awk (although for "academic" purposes I'd be interested in sed solutions.)

Edit 2 Unfortunately, the input file suggests that there is a logical order of words in the input file (one followed by two followed by three etc). In my "real world" problem, this is not the case, however. I have no idea how many lines the file has, nor what is preceeded or followed by the lines four and seven. The onl thing I know is that there is a line four which is (not necessarily immediately) followed by a line seven. I am sorry for not stating this clearly when I asked the question, especially because fedorqui has put so much effort in his answer.

解决方案

This might work for you (GNU sed & bash):

sed $'/^four/{:a;N;/^seven/McNEWLINE\nba}' file

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