How do I process following PHP regex in Java?_问答_开发者

How do I process following PHP regex in Java:

if(preg_match("/\r\n(.*?)\$/",$req,$match)){ 开发者_Python百科$data=$match[1]; }

This line is part of following function, by the way:

function getheaders($req){
  $r=$h=$o=null;
  if(preg_match("/GET (.*) HTTP/"   ,$req,$match)){ $r=$match[1]; }
  if(preg_match("/Host: (.*)\r\n/"  ,$req,$match)){ $h=$match[1]; }
  if(preg_match("/Origin: (.*)\r\n/",$req,$match)){ $o=$match[1]; }
  if(preg_match("/Sec-WebSocket-Key2: (.*)\r\n/",$req,$match)){ $key2=$match[1]; }
  if(preg_match("/Sec-WebSocket-Key1: (.*)\r\n/",$req,$match)){ $key1=$match[1]; }
  if(preg_match("/\r\n(.*?)\$/",$req,$match)){ $data=$match[1]; }
  return array($r,$h,$o,$key1,$key2,$data);
}

Thanks in advance!

So far I have:

Matcher matcher = Pattern.compile("\r\n(.*?)\\$").matcher(req);
while(matcher.find()){
    data = matcher.group(1);
}

I am sure, however, that this is wrong.

Ok guys, thanks for your answers, but they did not help yet. May I ask you to tell me, however, what this regex means:

  if(preg_match("/\r\n(.*?)\$/",$req,$match)){ $data=$match[1]; }

I know, that if it does find a match with /\r\n(.*?)\$/ in the string $req, it will save the different kinds of mathces into the array $match. BUT: what is being matched here? And what's the difference between $match[0] and $match[1]? Maybe, if I understand this, I will be able to reconstruct the way to produce equal results in Java.

Thanks Jaroslav, but:

The string I am trying to process however (the last line of the handshake sent to me by Google Chrome, is:

Cookie: 34ad04df964553fb6017b93d35dccd5f=%7C34%7C36%7C37%7C40%7C41%7C42%7C43%7C44%7C45%7C46%7C47%7C48%7C49%7C50%7C52%7C53%7C54%7C55%7C56%7C57%7C58%7C59%7C60%7C61%7C62%7C63%7C64%7C65%7C66%7C67%7C68%7C69%7C70%7C71%7C72%7C73%7C74%7C75%7C76%7C77%7C78%7C79%7C80%7C81%7C82%7C83%7C84%7C85%7C86%7C87%7C88%7C89%7C90%7C91%7C92%7C93%7C94%7C95%7C96%7C97%7C98%7C99%7C100%7C101%7C102%7C103%7C104%7C105%7C106%7C107%7C108%7C109%7C110%7C111%7C112%7C113%7C114%7C115%7C116%7C117%7C118%7C119%7C120%7C121%7C122%7C123%7C124%7C125%7C126%7C127%7C128%7C129%7C130%7C131%7C132%7C133%7C134%7C135%7C136%7C137%7C138%7C139%7C%3B%7C%3B%7C%3B%7C%3B1%3B2%3B3%3B4%3B5%3B6%3B7%3B8%3B9%3B10%3B11%3B14%3B15%3B18%3B23%3B24%3B25%3B26%3B28%3B29%3B30%3B31%3B32%3B33%3B%7C

Hey guys, I just now realize what I have been asking was irrelevant :( But one answer has been right.

Use java.util.regex.Pattern look here for instructions for this class. Here is Regular Expressions Tutorial. And here is the example:

String p = "Host: (.*)\\r\\n";
String input = "Host: example.com\r\n";
Pattern pattern = Pattern.compile(p);
Matcher matcher = pattern.matcher(input);
if(matcher.matches()) {
  String output = matcher.group(1);
    System.out.println(output);
} else {
    System.out.println("not found");
}

Note: Matcher.find matches subsequences, Matcher.matches matches entire region. IMHO in your example \\$ at the end may cause a problem when your input is multiline and you parse it at once.

In Java there are more convenient methods for accessing headers. At client side this is HttpURLConnection.getHeaderField. At the server side there is HttpServletRequest.getHeader.

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

public class SplitDemo2 {

    private static final String REGEX = "/\\r\\n(.*?)\\$/";
    private static final String INPUT = "/GET (.*) HTTP/";

    public static void main(String[] args) {
        Pattern p = Pattern.compile(REGEX);
        Matcher m = p.matcher(INPUT); // get a matcher object
        int count = 0;

        while(m.find()) {
          count++;
          System.out.println("Match number "+count);
          System.out.println("start(): "+m.start());
          System.out.println("end(): "+m.end());
   }
}

}

More info on regex http://download.oracle.com/javase/tutorial/essential/regex/matcher.html

Explanation of your regex