如何同时保持行尾拆分字符串? [英] How to split a string while preserving line endings?
本文介绍了如何同时保持行尾拆分字符串?的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!
问题描述
我有一个文本块,我想获得它的线条的无失去\r和\\\
底。现在,我有以下的(次优代码):
的String [] =行tbIn.Text.Split('\ N')
。选择(T => t.Replace(\r,\r\\\
)。)ToArray的();
所以我不知道 - ?有没有更好的办法做到这一点。
接受的答案
的String [] =行Regex.Split (tbIn.Text,@(小于?= \r\\\
)($)!);
解决方案
下面似乎做的工作:
的String [] =行Regex.Split(tbIn.Text,@(小于?= \r\\\
)(? !$));
(小于?= \r\\\
)采用积极的后向后\\匹配\\r\\\
不消耗它。
(?!$)使用负向前查找,以防止在输入的结束匹配,从而避免了最后一行,这只是一个空字符串。
I have a block of text and I want to get its lines without losing the \r and \n at the end. Right now, I have the following (suboptimal code):
string[] lines = tbIn.Text.Split('\n')
.Select(t => t.Replace("\r", "\r\n")).ToArray();
So I'm wondering - is there a better way to do it?
Accepted answer
string[] lines = Regex.Split(tbIn.Text, @"(?<=\r\n)(?!$)");
解决方案
The following seems to do the job:
string[] lines = Regex.Split(tbIn.Text, @"(?<=\r\n)(?!$)");
(?<=\r\n) uses 'positive lookbehind' to match after \r\n without consuming it.
(?!$) uses negative lookahead to prevent matching at the end of the input and so avoids a final line that is just an empty string.
这篇关于如何同时保持行尾拆分字符串?的文章就介绍到这了,希望我们推荐的答案对大家有所帮助,也希望大家多多支持IT屋!
查看全文