Java regex: newline + white space

问题:

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

回答1:

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

回答2:

String res = orig.replaceAll("[\\r\\n]+\\s", "");

回答3:

Perhaps to make it cross-platform:

String pattern = System.getProperty("line.separator") + " ";
string.replaceAll(pattern, "");

回答4:

"\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.