Java regex: newline + white space

2019-02-21 13:19发布

should be simple, but I'm going crazy with it.

Given a text like:

line number 1
line number 2
 line number 2A
line number 3
 line number 3A
 line number 3B
line number 4

I need the Java regex that deletes the line terminators then the new line begin with space, so that the sample text above become:

line number 1
line number 2line number 2A
line number 3line number 3Aline number 3B
line number 4

4条回答
干净又极端
2楼-- · 2019-02-21 14:09

Perhaps to make it cross-platform:

String pattern = System.getProperty("line.separator") + " ";
string.replaceAll(pattern, "");
查看更多
趁早两清
3楼-- · 2019-02-21 14:10

yourString.replaceAll("\n ", " "); this wont help?

查看更多
聊天终结者
4楼-- · 2019-02-21 14:15
String res = orig.replaceAll("[\\r\\n]+\\s", "");
查看更多
Lonely孤独者°
5楼-- · 2019-02-21 14:15

"\n " This is should do the trick if you are in Unix LF mode. For DOS like you need to match CRLF "\r\n ". Did check with RegexBuddy looking fine.

查看更多
登录 后发表回答