我想匹配以下内容:零长度的行,匹配持续跨非零长度的行,直到在一行中匹配特定的字符串为止。例如:比赛从零长度的线开始,一直持续到到达STOP:

Some random text I don't care about

The match starts at the beginning of this line
The match continues across this line
The match stops here STOP more
text I don't care about


有什么建议么?

谢谢

最佳答案

应该这样做:

(?ms)^[ \t]*+$\s*+((?:(?!STOP).)*+)


一些演示:

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

public class Main {
    public static void main(String[] args) {
        String text = "Some random text I don't care about"      + "\n" +
                ""                                               + "\n" +
                "The match starts at the beginning of this line" + "\n" +
                "The match continues across this line"           + "\n" +
                "The match stops here STOP more"                 + "\n" +
                "don't care about"                               + "\n" +
                ""                                               + "\n" +
                ""                                               + "\n" +
                ""                                               + "\n" +
                "foo"                                            + "\n" +
                "barSTOP"                                        + "\n" +
                "text I don't care about";
        Matcher m = Pattern.compile("(?ms)^[ \t]*+$\\s*+(?:(?!STOP).)*+").matcher(text);
        while(m.find()) {
            System.out.println("match ->"+m.group()+"<-");
        }
    }
}


将输出:

match ->
The match starts at the beginning of this line
The match continues across this line
The match stops here <-
match ->


foo
bar<-


一个小解释:

(?ms)               # enable mutli-line and dot-all
^[ \t]*+$           # match and empty line
\s*+                # match the line break
(                   # start group 1
  (?:(?!STOP).)     #   if the string 'STOP' cannot be seen, match any character
  *+                #   match the previous zero or more times (possessively)
)                   # stop group 1

08-06 11:54