Hi,
In fact my main problem is that I'm using JAX-WS client, following a source
code generated by wsimport.
static {
...
baseUrl =
ITFOSGETCHARACTERISTICLISTService.class.getResource("ITFOSGETCHARACTERISTICLISTService.class");
url = new URL(baseUrl, "ITF_O_S_GET_CHARACTERISTIC_LIST_V1.wsdl");
...
I tried modifying this code in order to find out the wsdl file in the resource
directory, but I don't like of my solution and it isn't working fine in all
environments. This JAX-WS serviceclient is in a jar (component) which is been
used by several applications, enterprise, web and local. Integrated tests with
JUnit is also a problem to find this files running by eclipse or maven,
apparently the class loader .getResouce( has different behaviors.
The better solution (that I realize) was put the wsdl file in the same
directory/package of the relative Xservice.class. However I don't know how to
make this jar with maven.
Att,
Cleiton Garcia.
Fone (0xx47) 3276-4167
-----Mensagem original-----
De: Jesse Farinacci [mailto:[email protected]]
Enviada em: Thursday, April 08, 2010 9:23 AM
Para: Maven Users List
Assunto: Re: Packing xml or wsdl in a jar, but in resource directory
Hi Cleiton,
On Thu, Apr 8, 2010 at 6:31 AM, Cleiton Dos Santos Garcia
<[email protected]> wrote:
>
> I need put some files in different java packages and this files are ".xml and
> .wsdl". I can't put it in the resources directory, I need put a wsdl file in
> the same package as your relative class. Is it possible?
> In eclipse it is a default behavior in "export-> jar", packing all .class and
> xml, or wsdl files. Anyone can help me how to it with maven?
>
What is stopping you from creating a mirrored directory hierarchy
within src/main/resources and dumping your files into the proper
location? They will end up at the right location in the packaged
artifact.
-Jesse
--
There are 10 types of people in this world, those
that can read binary and those that can not.
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