CODEWITHSUNDEEP
javaregexhttp-headers
Amir Rachum
Amir Rachum

Reputation: 79695

"No match Found" when using matcher's group method

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)

Upvotes: 83

Views: 74823

Answers (4)

Marian Kl&#252;hspies
Marian Kl&#252;hspies

Reputation: 17687

I was experiencing the same, but I'm writing this answer because I noticed something else:

As others stated, you have to call either

matcher.matches()

or

matcher.find();

Before you can call

matcher.group(1);

However, if you are using the results() call, you won't explicitly have to call those methods above and the results are immediately available.

pattern
        .matcher(pathname)
        .results()
        .collect(Collectors.toList())
        .get(0)
        .group(1);

I think this is intentionally the case, but I'm just adding it here in addition

No need to call matcher.find() or matcher.matches() when using matcher.results().

Upvotes: 10

hatanooh
hatanooh

Reputation: 4287

Or you can call matcher.find(); before matcher.group(1)

Upvotes: 3

Thomas
Thomas

Reputation: 88747

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
...

Upvotes: 130

Heiko Rupp
Heiko Rupp

Reputation: 30994

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());

Upvotes: 16

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