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The UML metemodel of Papyrus [message #1027432]

Wed, 27 March 2013 00:44

yesheng chai

Messages: 64
Registered: March 2013

Member

> In my ATL project ,I need use the model of Papyrus (.uml file),
>but I can't find the UML MeteModel of Papyrus .
>I've spent a long time looking for it ,but did not find it .
>Can you tell me where to find or to download the UML MetaModel of Papyrus
>for my ATL project.
> Thank you very much !
> Look forward to your reply !

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Re: The UML metemodel of Papyrus [message #1027643 is a reply to message #1027432]

Wed, 27 March 2013 08:10

Camille Letavernier

Messages: 926
Registered: February 2011

Senior Member

Hi,

Papyrus doesn't provide its own UML Metamodel. Instead, it relies on the standard Eclipse one (from the Eclipse UML2 project).

The URI is: http://www.eclipse.org/uml2/4.0.0/UML, and you can find the Ecore file in the org.eclipse.uml2.uml plug-in

Regards,
Camille

Camille Letavernier

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Re: The UML metemodel of Papyrus [message #1027649 is a reply to message #1027643]

Wed, 27 March 2013 08:18

yesheng chai

Messages: 64
Registered: March 2013

Member

Thank you for your reply!
I find a URI ,but did not kown that whether these metemodels are UML Metamodels of Papyrus. The URI : http://dev.eclipse.org/svnroot/modeling/org.eclipse.mdt.papyrus/trunk/plugins/uml/org.eclipse.papyrus.uml.standard/model/
Unfortunately, the URL that you give to me ,can not find .

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Re: The UML metemodel of Papyrus [message #1027688 is a reply to message #1027649]

Wed, 27 March 2013 09:21

Camille Letavernier

Messages: 926
Registered: February 2011

Senior Member

Hi,

The URI is an EMF URI. Although it starts with http:/, it is not a valid HTTP page. This is just an identifier for the UML Metamodel.

To open the UML.ecore file, you should import the uml plug-in into your workspace (File -> Import -> Plug-ins and Fragments -> choose org.eclipse.uml2.uml), then look into the model/ folder.

Regards,
Camille

Camille Letavernier

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Re: The UML metemodel of Papyrus [message #1027763 is a reply to message #1027688]

Wed, 27 March 2013 11:40

yesheng chai

Messages: 64
Registered: March 2013

Member

Thank you for your reply!
I am sorry that I don't know how to use URI to get the metamodel of UML .Can you give me detailed description of the use of URI to find metamodel of UML ?
Look forward to your reply !

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Re: The UML metemodel of Papyrus [message #1027796 is a reply to message #1027763]

Wed, 27 March 2013 12:25

Camille Letavernier

Messages: 926
Registered: February 2011

Senior Member

Hi,

The metamodel URI cannot be used to retrieve the metamodel. It is an identifier.

To retrieve the metamodel:

Quote:

To open the UML.ecore file, you should import the uml plug-in into your workspace (File -> Import -> Plug-ins and Fragments -> choose org.eclipse.uml2.uml), then look into the model/ folder.

Camille

Camille Letavernier

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Re: The UML metemodel of Papyrus [message #1027799 is a reply to message #1027763]

Wed, 27 March 2013 12:29

Ed Willink

Messages: 6768
Registered: July 2009

Senior Member

http://git.eclipse.org/c/emf/org.eclipse.emf.git/tree/plugins/org.eclipse.emf.ecore/model/Ecore.ecore

On 27/03/2013 11:40, yesheng chai wrote:
> Thank you for your reply!
> I am sorry that I don't know how to use URI to get the metamodel of
> UML .Can you give me detailed description of the use of URI to find
> metamodel of UML ?
> Look forward to your reply !

Report message to a moderator

Re: The UML metemodel of Papyrus [message #1027874 is a reply to message #1027796]

Wed, 27 March 2013 14:36

yesheng chai

Messages: 64
Registered: March 2013

Member

Thank you very much !
I have found the metamodel of UML by your method .I compare the model file(.uml file) generated by Papyrus with the metamodel ,finding that the generated file is not based on UML metamodel . Can I use UML metamodel as the metamodel of the model file(.uml file) generated by Papyrus ?
Look forward to your reply !

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Re: The UML metemodel of Papyrus [message #1027880 is a reply to message #1027874]

Wed, 27 March 2013 14:42

Camille Letavernier

Messages: 926
Registered: February 2011

Senior Member

Quote:

finding that the generated file is not based on UML metamodel

What do you mean?

If you are using Papyrus 0.9.X (Juno) or 0.10 (Kepler), then we use this version of the UML Metamodel: http://www.eclipse.org/uml2/4.0.0/UML (Corresponding to UML 2.4.1)
If you are using Papyrus 0.8.X (Indigo), then we use the previous version of the UML Metamodel, i.e.: http://www.eclipse.org/uml2/3.0.0/UML (UML 2.3), which is located in the same plug-in (org.eclipse.uml2.uml/model/UML30.ecore)

Camille

Camille Letavernier

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Re: The UML metemodel of Papyrus [message #1028232 is a reply to message #1027880]

Thu, 28 March 2013 01:49

yesheng chai

Messages: 64
Registered: March 2013

Member

Thank you for your reply!
I hava used Papyrus to draw a simple activity diagram .The code of the Activity diagram in .uml file is:

<?xml version="1.0" encoding="UTF-8"?>
<uml:Model xmi:version="20110701" xmlns:xmi="http://www.omg.org/spec/XMI/20110701" xmlns:uml="http://www.eclipse.org/uml2/4.0.0/UML" xmi:id="_vNMBYJPNEeKDxonErf88qQ" name="model">
  <packagedElement xmi:type="uml:Activity" xmi:id="_vNNPgJPNEeKDxonErf88qQ" name="Activity1" node="_4gWO0JPNEeKDxonErf88qQ _71K1kJPNEeKDxonErf88qQ _-8DuAJPNEeKDxonErf88qQ _LnPAIJPOEeKDxonErf88qQ">
    <edge xmi:type="uml:ControlFlow" xmi:id="_ANKJgJPOEeKDxonErf88qQ" name="ControlFlow1" source="_4gWO0JPNEeKDxonErf88qQ" target="_71K1kJPNEeKDxonErf88qQ">
      <guard xmi:type="uml:LiteralBoolean" xmi:id="_ANLXoJPOEeKDxonErf88qQ" value="true"/>
      <weight xmi:type="uml:LiteralInteger" xmi:id="_ANLXoZPOEeKDxonErf88qQ"/>
    </edge>
    <edge xmi:type="uml:ControlFlow" xmi:id="_BQmjgJPOEeKDxonErf88qQ" name="ControlFlow2" source="_LnPAIJPOEeKDxonErf88qQ" target="_-8DuAJPNEeKDxonErf88qQ">
      <guard xmi:type="uml:LiteralBoolean" xmi:id="_BQnxoJPOEeKDxonErf88qQ" value="true"/>
      <weight xmi:type="uml:LiteralInteger" xmi:id="_BQnxoZPOEeKDxonErf88qQ"/>
    </edge>
    <edge xmi:type="uml:ControlFlow" xmi:id="_PaTpIJPOEeKDxonErf88qQ" name="ControlFlow3" source="_71K1kJPNEeKDxonErf88qQ" target="_LnPAIJPOEeKDxonErf88qQ">
      <guard xmi:type="uml:LiteralBoolean" xmi:id="_PaUQMJPOEeKDxonErf88qQ" value="true"/>
      <weight xmi:type="uml:LiteralInteger" xmi:id="_PaU3QJPOEeKDxonErf88qQ"/>
    </edge>
    <node xmi:type="uml:InitialNode" xmi:id="_4gWO0JPNEeKDxonErf88qQ" name="InitialNode1" outgoing="_ANKJgJPOEeKDxonErf88qQ"/>
    <node xmi:type="uml:OpaqueAction" xmi:id="_71K1kJPNEeKDxonErf88qQ" name="eat" outgoing="_PaTpIJPOEeKDxonErf88qQ" incoming="_ANKJgJPOEeKDxonErf88qQ"/>
    <node xmi:type="uml:ActivityFinalNode" xmi:id="_-8DuAJPNEeKDxonErf88qQ" name="ActivityFinalNode1" incoming="_BQmjgJPOEeKDxonErf88qQ"/>
    <node xmi:type="uml:OpaqueAction" xmi:id="_LnPAIJPOEeKDxonErf88qQ" name="sleep" outgoing="_BQmjgJPOEeKDxonErf88qQ" incoming="_PaTpIJPOEeKDxonErf88qQ"/>
  </packagedElement>
</uml:Model>

I know that node,edge,weight,and guard in the code are references,but I can't find them and their attributes(type ,id ,name ,outgoing and so on)in the metemodel of UML.
I want to use the model file (.uml file )in my ATL project and I also want to use the UML metamodel as the metamodel of the generated model file(.uml file).
I want to know that whether my method can work ?
Look forward to your reply .

Report message to a moderator

Re: The UML metemodel of Papyrus [message #1028434 is a reply to message #1028232]

Thu, 28 March 2013 08:46

Ed Willink

Messages: 6768
Registered: July 2009

Senior Member

Hi

Please use new subjects lines for new topics.

Please do not ask the same question on multiple newsgroups.

See reply on ATL newsgroup.

Regards

Ed Willink

On 28/03/2013 01:49, yesheng chai wrote:
> Thank you for your reply!
> I hava used Papyrus to draw a simple activity diagram .The code of the
> Activity diagram in .uml file is:
> <?xml version="1.0" encoding="UTF-8"?>
> <uml:Model xmi:version="20110701"
> xmlns:xmi="http://www.omg.org/spec/XMI/20110701"
> xmlns:uml="http://www.eclipse.org/uml2/4.0.0/UML"
> xmi:id="_vNMBYJPNEeKDxonErf88qQ" name="model">
> <packagedElement xmi:type="uml:Activity"
> xmi:id="_vNNPgJPNEeKDxonErf88qQ" name="Activity1"
> node="_4gWO0JPNEeKDxonErf88qQ _71K1kJPNEeKDxonErf88qQ
> _-8DuAJPNEeKDxonErf88qQ _LnPAIJPOEeKDxonErf88qQ">
> <edge xmi:type="uml:ControlFlow" xmi:id="_ANKJgJPOEeKDxonErf88qQ"
> name="ControlFlow1" source="_4gWO0JPNEeKDxonErf88qQ"
> target="_71K1kJPNEeKDxonErf88qQ">
> <guard xmi:type="uml:LiteralBoolean" xmi:id="_ANLXoJPOEeKDxonErf88qQ"
> value="true"/>
> <weight xmi:type="uml:LiteralInteger" xmi:id="_ANLXoZPOEeKDxonErf88qQ"/>
> </edge>
> <edge xmi:type="uml:ControlFlow" xmi:id="_BQmjgJPOEeKDxonErf88qQ"
> name="ControlFlow2" source="_LnPAIJPOEeKDxonErf88qQ"
> target="_-8DuAJPNEeKDxonErf88qQ">
> <guard xmi:type="uml:LiteralBoolean" xmi:id="_BQnxoJPOEeKDxonErf88qQ"
> value="true"/>
> <weight xmi:type="uml:LiteralInteger" xmi:id="_BQnxoZPOEeKDxonErf88qQ"/>
> </edge>
> <edge xmi:type="uml:ControlFlow" xmi:id="_PaTpIJPOEeKDxonErf88qQ"
> name="ControlFlow3" source="_71K1kJPNEeKDxonErf88qQ"
> target="_LnPAIJPOEeKDxonErf88qQ">
> <guard xmi:type="uml:LiteralBoolean" xmi:id="_PaUQMJPOEeKDxonErf88qQ"
> value="true"/>
> <weight xmi:type="uml:LiteralInteger" xmi:id="_PaU3QJPOEeKDxonErf88qQ"/>
> </edge>
> <node xmi:type="uml:InitialNode" xmi:id="_4gWO0JPNEeKDxonErf88qQ"
> name="InitialNode1" outgoing="_ANKJgJPOEeKDxonErf88qQ"/>
> <node xmi:type="uml:OpaqueAction" xmi:id="_71K1kJPNEeKDxonErf88qQ"
> name="eat" outgoing="_PaTpIJPOEeKDxonErf88qQ"
> incoming="_ANKJgJPOEeKDxonErf88qQ"/>
> <node xmi:type="uml:ActivityFinalNode"
> xmi:id="_-8DuAJPNEeKDxonErf88qQ" name="ActivityFinalNode1"
> incoming="_BQmjgJPOEeKDxonErf88qQ"/>
> <node xmi:type="uml:OpaqueAction" xmi:id="_LnPAIJPOEeKDxonErf88qQ"
> name="sleep" outgoing="_BQmjgJPOEeKDxonErf88qQ"
> incoming="_PaTpIJPOEeKDxonErf88qQ"/>
> </packagedElement>
> </uml:Model>
> I know that node,edge,weight,and guard in the code are references,but
> I can't find them and their attributes(type ,id ,name ,outgoing and so
> on)in the metemodel of UML.
> I want to use the model file (.uml file )in my ATL project and I also
> want to use the UML metamodel as the metamodel of the generated model
> file(.uml file).
> I want to know that whether my method can work ?
> Look forward to your reply .

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