如何使用sed替换第n次出现的字符串

问题描述

有没有办法用sed替换文件中第n次出现的字符串?

Is there any way to replace the nth occurrence of a string in a file using sed?

我使用 sed -i '0,/jack.*/s//jill/' 来替换第一次出现.

I'm using sed -i '0,/jack.*/ s//jill/' to replace the first occurrence.

如何更改它以替换第 n 次出现?

How can i change it so that it replaces the nth occurrence?

我的文件包含以下几行:

My file contents the following lines:

first line
second line
third line
jack=1
fifth line
jack=
seventh line

我不知道 jack= 之后的值，它可以是任何东西，也可以什么都不是.

I don't know the value after jack=, it can be anything or nothing.

我想用 jill 替换第二次出现的 jack= 及其后的任何内容.

I want to replace the 2nd occurrence of jack= and anything that follows it with jill.

推荐答案

首先使用 将所有换行符替换为文件中其他任何地方都没有出现的唯一字符(例如 ^)tr.您需要这样做才能为 sed 创建单个字符串.然后将它传递给 sed 并告诉它替换第 n 次出现的字符串.最后，通过 tr 将输出传回以重新创建换行符.

First replace all the newlines with a unique character that does not occur anywhere else in your file (e.g. ^) using tr. You need to do this in order to create a single string for sed. Then pass it to sed and tell it to replace the nth occurrence of your string. Finally, pass the output back through tr to recreate the newlines.

对于 n=2，命令是:

For n=2, the command is:

$ tr '\n' '^' < file | sed 's/jack/jill/2' | tr '^' '\n'
first line
second line
third line
jack
fifth line
jill
seventh line

更新:

也可以使用 sed 完成，无需先更改换行符，使用以下命令:

It can also be done with sed, WITHOUT changing the newlines first, using the following command:

$ sed ':a;N;$!ba;s/jack/jill/2' file

或者，使用 awk:

$ awk '/jack/{c+=1}{if(c==2){sub("jack","jill",$0)};print}' file

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