1

我有两个 emf 模型 A 和 B,其中 B 仅与 A 不同,因为它有一个额外的子节点。

现在我想使用代码中的 emf compare 来做:

1)读取模型A和B并创建模型C,它是A和B的合并模型。基本上这对应于A +来自B的额外节点。

我看过:

http://dev.eclipse.org/viewcvs/viewvc.cgi/org.eclipse.emf/org.eclipse.emf.compare/examples/org.eclipse.emf.compare.examples.standalone/src/org/eclipse/ emf/compare/examples/standalone/ExampleLauncher.java?view=co&root=Modeling_Project

但是我看不到如何使用代码计算最终的合并模型:

DiffModel diff = CompareUtils.compare(model1, model2, Collections.<String, Object> emptyMap());
CompareUtils.merge(diff);

任何显示如何计算合并模型的示例?

我现在尝试过:

      private void bob() {
        ResourceSet resourceSet = new ResourceSetImpl();
        Map extensionMap = (Map) resourceSet.getResourceFactoryRegistry()
            .getExtensionToFactoryMap().put("xmi", new XMIResourceFactoryImpl());
        try {

      Region region01 = StatemachineFactoryImpl.eINSTANCE.createRegion();
      addResourceToModel(resourceSet, region01, "st1.xmi");
      State state01 = StatemachineFactoryImpl.eINSTANCE.createState();
      state01.setName("aaaa");
      region01.getState().add(state01);
      if (state01.eResource() == null) {
        System.out.println("state01 NOT contained in resource!");
        return;
      }

      Region region02 = StatemachineFactoryImpl.eINSTANCE.createRegion();
      addResourceToModel(resourceSet, region02, "st2.xmi");
      State state02 = StatemachineFactoryImpl.eINSTANCE.createState();
      state02.setName("bbbb");
      region02.getState().add(state02);
      if (state02.eResource() == null) {
        System.out.println("state02 NOT contained in resource!");
        return;
      }

      final MatchModel match = MatchService.doMatch(region01, region02,
          Collections.<String, Object> emptyMap());
      final DiffModel diff = DiffService.doDiff(match, false);
      final List<DiffElement> differences = new ArrayList<DiffElement>(
          diff.getOwnedElements());
      MergeService.merge(differences, true);

      // Prints the results
      addResourceToModel(resourceSet, match, "match.xmi");
      addResourceToModel(resourceSet, diff, "diff.xmi");

      if (match.eResource() != null)
        System.out.println(ModelUtils.serialize(match)); // Throws an
                                                         // exception!
      if (diff.eResource() != null)
        System.out.println(ModelUtils.serialize(diff));

    } catch (final InterruptedException e) {
      e.printStackTrace();
    } catch (IOException e) {
      e.printStackTrace();
    }

  }

  private void addResourceToModel(ResourceSet resourceSet, EObject obj,
      String path) {
    Resource res = resourceSet.createResource(URI.createURI(path));
    res.getContents().add(obj);
  }

但是这条线:

  if (match.eResource() != null)
    System.out.println(ModelUtils.serialize(match)); // Throws an
                                                     // exception!

即使 match.eResource() != null

我收到此错误:

org.eclipse.emf.ecore.resource.Resource$IOWrappedException: The object 'statemachine.impl.StateImpl@11ce012 (name: bbbb)' is not contained in a resource.
    at org.eclipse.emf.ecore.xmi.impl.XMLSaveImpl.endSave(XMLSaveImpl.java:306)
    at org.eclipse.emf.ecore.xmi.impl.XMLSaveImpl.save(XMLSaveImpl.java:235)
    at org.eclipse.emf.ecore.xmi.impl.XMLResourceImpl.doSave(XMLResourceImpl.java:254)
    at org.eclipse.emf.ecore.xmi.impl.XMLResourceImpl.save(XMLResourceImpl.java:229)
    at org.eclipse.emf.compare.util.ModelUtils.serialize(ModelUtils.java:429)

我已根据此处的文档将 Region 添加到资源中:

http://wiki.eclipse.org/index.php/EMF-FAQ#I_get_a_DanglingHREFException: eg.2C .22org.eclipse.emf.ecore.xmi.DanglingHREFException:_The_object_.27com.example.Foo.402f5dda_.28.29.27_is_not_contained_in_a_resource。 .22_What_do_I_need_to_do.3F

并且该州包含在该地区中,所以我不明白为什么我会得到例外……有什么想法吗?

4

3 回答 3

2

那这个呢?

Model1 targetModel = EcoreUtil.copy(model1);
addResourceToModel(targetModel) // assign the copied model to a resource
MatchModel match = MatchService.doMatch(targetModel, model2,
                    Collections.<String, Object> emptyMap());
DiffModel diff = DiffService.doDiff(match, false);
EList<DiffElement> differences = diff.getDifferences();
for (DiffElement diffElement : differences) {
    MergeService.merge(diffElement, true);
}
于 2011-04-18T15:02:11.773 回答
2

图尔,

您获得的堆栈跟踪意味着“合并”对象之一未包含在资源中:合并时,我们复制一个引用状态机(名称为“bbbb”)的对象,然后我们需要从复制的对象......以及我们引用的状态机(它是从您的其他模型复制还是直接引用?您应该调试才能看到这一点)本身并不包含在任何资源中。

State state02 = StatemachineFactoryImpl.eINSTANCE.createState();
state02.setName("bbbb");
region02.getState().add(state02);
if (state02.eResource() == null) {
    System.out.println("state02 NOT contained in resource!");
    return;
}

这应该确保“bbbb”确实包含在资源中。

在此行之后:

MergeService.merge(differences, true);

如果“state02.eResource() == null”,您可以尝试再次检查吗?如果是,那是你的问题。否则,您必须确保这不会返回:

for (State state : region01.getState()) {
    if (state.eResource() == null) {
       System.err.println(state.getName() + " is not contained in a resource);
       return;
    }
}
于 2011-04-19T08:42:25.230 回答
0

您的例外:“org.eclipse.emf.ecore.resource.Resource$IOWrappedException:对象'statemachine.impl.StateImpl@11ce012(名称:bbbb)'不包含在资源中。” . StateImpl@11ce012(名称:bbbb)在一个名为 st2.xmi 的资源中,但匹配元素在另一个名为“match.xml”的资源中。这两个资源是不同的,它们不相互引用。所以匹配元素不能引用StateImpl。为了解决这个问题,所有元素(state01、state02、match、diff)必须保存在一个资源中。代码是:

res.getContents().add(stat01);
res.getContents().add(stat02);
res.getContents().add(match);
res.getContents().add(diff);

顺便说一句,条件“ state02.eResource() == null”不是必要的。

于 2017-06-27T10:34:18.327 回答