Hi Graham,

Its really difficult to help without checking out the failing command first.

Could you reply back with the "javac" command you are trying to execute, along 
with the stacktrace?

-- Abhinay
________________________________
From: openjfx-dev <openjfx-dev-boun...@openjdk.java.net> on behalf of Graham 
Norris <w...@norrisfamily.org>
Sent: Tuesday, October 9, 2018 9:39 PM
To: openjfx-dev@openjdk.java.net
Subject: Re: Compiling using javac

** Reply to message from Michael Paus <m...@jugs.org> on Tue, 9 Oct 2018 
15:47:29
+0200

Thank you, I'd already found this, but it isn't exactly "documentation." Also,
it is for Linux (Windows doesn't have export,) but I have put the path directly
on the --module-path option, which should be equivalent for this.

Crucially, it does not work: I get "error: module not found: javafx.controls"
I'm not entirely sure why I'd use javafx.controls as my application does not
use that, but not finding javafx.controls seems to be unrelated to lack of use,
which is also borne out by the fact that the compile fails immediately rather
than churning out dozens of errors as it tries to compile things.

Graham.

> The documentation you are looking for can be found here:
> https://openjfx.io/openjfx-docs/
>
> Am 09.10.18 um 15:32 schrieb Graham Norris:
> > I have an existing Java application which uses JavaFX. It is currently 
> > compiled
> > (on Windows) and working using Oracle's Java 10 JDK. I am now trying to get 
> > it
> > to compile using OpenJDK 11 and OpenJFX 11, and failing miserably!
> >
> > I use the JDK's command line tools to compile the application and haven't
> > managed to find anything which shows how to make the combination of an 
> > existing
> > application, OpenJDK 11 and OpenJFX 11 compile. Basically, I can't get 
> > javac to
> > find and use either OpenJFX's jmods or jars.
> >
> > I feel there should be somewhere I could read instructions on how to use
> > OpenJFX with OpenJDK 11's javac, but if there is, I just can't find it!
> >
> > Any help appreciated.
> >
> > Graham.

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