Java - Registering custom URL protocol handlers

Question

I tried to register a custom URL handler for a classpath protocol, as described in another thread. Here is the code:

package com.mycompany;

import org.junit.Test;
import java.net.MalformedURLException;
import java.net.URL;
import com.mycompany.protocol.classpath.Handler;

public class ParserTest {
    @Test
    public void testParsing() throws MalformedURLException {      
        System.out.println(System.getProperty("java.protocol.handler.pkgs"));

        //URL url = new URL(null, "classpath://com.mycompany/hello-world.xml", new Handler(ClassLoader.getSystemClassLoader()));
        URL url = new URL("classpath://com.mycompany/hello-world.xml");
    }
}

The test case has the following JVM arguments:

-Djava.protocol.handler.pkgs=com.mycompany.protocol

The System.out.println line properly outputs com.mycompany.protocol, so the property is being set. However, it looks like it's not being taken into effect, because the above call will throw a java.net.MalformedURLException: unknown protocol: classpath exception.

If I provide the handler explicitly as in the commented line, everything is fine. However, I would rather not provide it explicitly - it should be done automatically.

What am I doing wrong?

Does it work if you pass the handler to the URL's constructor? (This is just a debugging proposal, not the final solution). — Paŭlo Ebermann
– Paŭlo Ebermann, Commented Jun 8, 2011 at 12:16
Yes, if I comment out the first statement and use that one instead, it works. — Ariod
– Ariod, Commented Jun 8, 2011 at 12:36
@Dario can you please advise me on how to write a custom jdbc driver? If there is a good reference somewhere, please help me with its link — KCK
– KCK, Commented Aug 14, 2018 at 12:33

Community · Accepted Answer · 2017-05-23 12:34:54Z

14

I have found the issue. The original classpath handler class that I used had a non-default constructor. Of course, because it had only a non-default constructor, the handler couldn't be instantiated. I apologize to everyone who have tried to debug this issue, I failed to see this connection.

edited May 23, 2017 at 12:34

CommunityBot

11 silver badge

answered Jun 8, 2011 at 16:21

Ariod

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1 Comment

Nandkumar Tekale Over a year ago

don't be sorry, they will also know how to solve this kind of question.

Matt · Accepted Answer · 2014-07-16 15:28:29Z

0

Probably easiest way to debug such problems is to enumerate the protocol handlers registered.

edited Jul 16, 2014 at 15:28

Matt

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answered Mar 7, 2012 at 1:51

Kiran

1

2 Comments

oers Over a year ago

Whilst this may theoretically answer the question, it would be preferable to include the essential parts of the answer here, and provide the link for reference.

Oliver Mason Over a year ago

The link seems no longer valid (as of 11/03/2014)

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