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, Jun 08 '11 at 12:16
Yes, if I comment out the first statement and use that one instead, it works. — Ariod, Jun 08 '11 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, Aug 14 '18 at 12:33

score 14 · Accepted Answer · edited May 23 '17 at 12:34

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 '17 at 12:34

Community

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answered Jun 08 '11 at 16:21

Ariod

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don't be sorry, they will also know how to solve this kind of question. – Nandkumar Tekale Aug 17 '12 at 13:29

score 0 · Answer 2 · edited Jul 16 '14 at 15:28

0

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

edited Jul 16 '14 at 15:28

Matt

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answered Mar 07 '12 at 01:51

Kiran

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Whilst this may theoretically answer the question, [it would be preferable](http://meta.stackexchange.com/q/8259) to include the essential parts of the answer here, and provide the link for reference. – oers Oct 26 '12 at 10:50
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The link seems no longer valid (as of 11/03/2014) – Oliver Mason Mar 11 '14 at 15:01

Java - Registering custom URL protocol handlers

2 Answers2

Linked