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Re: [mdt-papyrus.dev] Overwrite custom UML profile

Hi,

you may use a reference based on a PATHMAP and a URI mapping normally registered by a plugin, as shown below. If you apply the profile registered this way, references towards the profile in your model will be based on the pathmap. Thus, you could relocate the profile and adapt the URI mapping accordingly. However, I don't know how to use that in a headless configuration, i.e. out of Eclipse.

  <extension
         point="org.eclipse.emf.ecore.uri_mapping">
      <mapping
            source="pathmap://your ID/"
            target="your path ...">
      </mapping>
   </extension>

  <extension
         point="org.eclipse.papyrus.uml.extensionpoints.UMLProfile">
      <profile
            description="..."
            iconpath="..."
            name="Your profile name"
            path="pathmap://your ID/your profile.profile.uml"
            provider="...">
      </profile>
   </extension>

Best regards

Ansgar

PS: It would be good to pass by the Papyrus forum for this question

On 21/02/2018 09:37, Christoph Czurda wrote:
Hello,

I have a UML file, myModel.uml. The model references a profile, myProfile.uml, in various places.

For example:

xsi:schemaLocation="http:///schemas/MyNamespace/_yEh3IAjnEei6dL15TtaBqw/30 ../ myProfileProject/myProfile.uml#_yEk6cAjnEei6dL15TtaBqw" >

<references xmi:type="ecore:EPackage" href=""/>

 I am writing a standalone Java application, so I load the model and the resources on my own. As you can see from the code snippets, the profile is referenced via relative path. As long as the user has his paths accordingly, everything works fine. However, for convenience I want to add a command line parameter to the standalone app where the user can specifiy the profile location.

How can I apply the profile to the loaded model, should the user specify a different location? I tried adding the profile from the custom path to the resource set, but the model still cannot resolve the profile, due to the relative paths hardcoded into the model.uml I suppose.

Kind regards,
Christoph


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