Java Malformed URL Exception

user3772689 picture user3772689 · Jun 25, 2014 · Viewed 58k times · Source

I'm trying to make an http POST request in an android app I'm building, but no matter what url I use for the request, Eclipse keeps raising a Malformed URL Exception. I've tried a line of code from one of the android tutorials:

URL url = new URL("https://wikipedia.org");

And even that triggers the error. Is there a reason Eclipse keeps raising this error for any URL I try to create?

Answer

Boann picture Boann · Jun 25, 2014

It is not raising the exception, it's complaining that you haven't handled the possibility that it might, even though it won't, because the URL in this case is not malformed. (Java's designers thought this concept, "checked exceptions", was a good idea, although in practice it hasn't worked well.)

To shut it up, add throws MalformedURLException, or its superclass throws IOException, to the method declaration. For example:

public void myMethod() throws IOException {
    URL url = new URL("https://wikipedia.org/");
    ...
}

Alternatively, catch and rethrow the annoying exception as an unchecked exception:

public void myMethod() {
    try {
        URL url = new URL("https://wikipedia.org/");
        ...
    } catch (IOException e) {
        throw new RuntimeException(e);
    }
}

Java 8 added the UncheckedIOException class for rethrowing IOExceptions when you cannot otherwise handle them. In earlier Java versions, use RuntimeException.