Question

I want to match only the nth (let's say 5th) whitespace character on a line and then replace that space with a line break (so that all my lines essentially have no more than 4 words / 4 spaces). On the REPLACE line in Notepad++, I think an \r\n would generally do the trick, and it is just the FIND pattern that is giving me problems.

I know there is a relatively simple regex solution for this that I am blanking on, but in my searches of "Find Nth instance of pattern on a line" type regex answers, the regex that has been closest to what I want is also matching everything before that 5th whitespace on a line. Here is that code that is grabbing too much.

^(?:\S+\s){5}

I know the "\S+" probably should not be there, but my attempts to pair this code down to some basic variations on this below

^(\s){5}

have all failed one way or another.

So here is what the text result of the regex should be.

FROM:

This line is too long. I want to shorten it.

This line is too long. I would really like to know why my regex is not working.

TO:

This line is too long. 

I want to shorten it.

This line is too long.

I would really like to

know why my regex is

not working.

Answer 1

Use the following regex in find:

((?:\S+\s){4}\S+)\s

Demo

Then replace it with \1\r\n.

This essentially captures everything up to the 5th space and then replaces it with the first capture group (\1) followed by \r\n.

Answer 2

如果你没有保留现有的换行符，那么一个简单的解决方案就是做两个单独的替换：

首先，使用单个空格搜索\s+和全部替换。

然后做这个替换：

找到：

(\S+ \S+ \S+ \S+ \S+)\s

用“全部替换”：

$1\r\n

Notepad++ regex to find Nth instance of a space character on a line

2 个答案: