高分求java regular expression.

[possibleonline]

有以下需求,文本如下
new line
__new001_tag00_tag
#SW
this is news 001
abbbbbbbbbbbb
new line
__new001_tag00_tag
#SW
this is news 002
abbbbbbbbbbbb
new line
new line
__new001_tag00_tag
#SW
this is news 003
abbbbbbbbbbbb

我写成这样子实现抓取3条记录:\\r\\n(\\w*?_\\w*?_\\w*?_\\w*?)\\r\\n(.*?#SW.*?)\\r\\n(.*?)\\r\\n\\ Pattern.DOTALL
现在只能匹配粗体3行，不能匹配4行,怎么样才能把第4行也匹配?

[rickylin86]

import java.util.regex.Pattern;
import java.util.regex.Matcher;


public class Test{
	public static void main(String[] args){
		String content = "new line\n"+
						"__new001_tag00_tag\n"+
						"#SW\n"+
						"this is news 001\n"+
						"abbbbbbbbbbbb\n"+
						"new line\n"+
						"__new001_tag00_tag\n"+
						"#SW\n"+
						"this is news 002\n"+
						"abbbbbbbbbbbb\n"+
						"new line\n"+
						"new line\n"+
						"__new001_tag00_tag\n"+
						"#SW\n"+
						"this is news 003\n"+
						"abbbbbbbbbbbb\n";
		String regex = "(?s)(?:new line\n)(.*?)(?=new line\n)";
		Pattern pattern = Pattern.compile(regex);
		Matcher matcher = pattern.matcher(content);
		if(matcher.find()){
			System.out.println(matcher.group(1));
		}
	}
}

[rickylin86]

result: ---------- java ---------- __new001_tag00_tag #SW this is news 001 abbbbbbbbbbbb Output completed (0 sec consumed) - Normal Termination