“No match Found” when using matcher's group method

问题

I'm using Pattern/Matcher to get the response code in an HTTP response. groupCount returns 1, but I get an exception when trying to get it! Any idea why?

Here's the code:

//get response code
String firstHeader = reader.readLine();
Pattern responseCodePattern = Pattern.compile("^HTTP/1\\.1 (\\d+) OK$");
System.out.println(firstHeader);
System.out.println(responseCodePattern.matcher(firstHeader).matches());
System.out.println(responseCodePattern.matcher(firstHeader).groupCount());
System.out.println(responseCodePattern.matcher(firstHeader).group(0));
System.out.println(responseCodePattern.matcher(firstHeader).group(1));
responseCode = Integer.parseInt(responseCodePattern.matcher(firstHeader).group(1));

And here's the output:

HTTP/1.1 200 OK
true
1
Exception in thread "Thread-0" java.lang.IllegalStateException: No match found
 at java.util.regex.Matcher.group(Unknown Source)
 at cs236369.proxy.Response.<init>(Response.java:27)
 at cs236369.proxy.ProxyServer.start(ProxyServer.java:71)
 at tests.Hw3Tests$1.run(Hw3Tests.java:29)
 at java.lang.Thread.run(Unknown Source)

回答1:

pattern.matcher(input) always creates a new matcher, so you'd need to call matches() again.

Try:

Matcher m = responseCodePattern.matcher(firstHeader);
m.matches();
m.groupCount();
m.group(0); //must call matches() first
...

回答2:

You are constantly overwriting the matches you got by using

System.out.println(responseCodePattern.matcher(firstHeader).matches());
System.out.println(responseCodePattern.matcher(firstHeader).groupCount());

Each line creates a new Matcher object.

You should go

Matcher matcher = responseCodePattern.matcher(firstHeader);
System.out.println(matcher.matches());
System.out.println(matcher.groupCount());

来源：https://stackoverflow.com/questions/5674268/no-match-found-when-using-matchers-group-method

标签

java

regex

http-headers