从XMI文件加载EMF模型实例

Question

我知道关于此主题有几个QnA。我尝试了很多解决方案，但总是遇到相同的错误。

我的代码结构像：

Resource.Factory.Registry reg = Resource.Factory.Registry.INSTANCE;
    Map<String, Object> m = reg.getExtensionToFactoryMap();
    m.put("xmi", new XMIResourceFactoryImpl());

    ResourceSet resSet = new ResourceSetImpl();
    Resource resource = resSet.getResource(URI.createURI("model/List.xmi"), true);
    resource.load(Collections.EMPTY_MAP);
    EObject root = resource.getContents().get(0);

错误：

线程“主”中的异常org.eclipse.emf.ecore.resource.impl.ResourceSetImpl $ 1DiagnosticWrappedException：org.eclipse.emf.ecore.xmi.PackageNotFoundException：找不到带有uri“列表”的软件包。 （文件：/// C：/Users/2/My%20Repository/UNIT%20Research%20and%20Development/com.unitbilisim.research.transformation/model/List.xmi，6，40）在org.eclipse.emf.ecore.resource.impl.ResourceSetImpl.handleDemandLoadException（ResourceSetImpl.java:319）在org.eclipse.emf.ecore.resource.impl.ResourceSetImpl.demandLoadHelper（ResourceSetImpl.java:278）在org.eclipse.emf.ecore.resource.impl.ResourceSetImpl.getResource（ResourceSetImpl.java:406）在com.unitbilisim.research.transformation.ConvertEcore2Graph.main（ConvertEcore2Graph.java:61）原因：org.eclipse.emf.ecore.xmi.PackageNotFoundException：找不到带有uri“列表”的软件包。 （文件：/// C：/Users/2/My%20Repository/UNIT%20Research%20and%20Development/com.unitbilisim.research.transformation/model/List.xmi，6，40）在org.eclipse.emf.ecore.xmi.impl.XMLHandler.getPackageForURI（XMLHandler.java:2625）在org.eclipse.emf.ecore.xmi.impl.XMLHandler.getFactoryForPrefix（XMLHandler.java:2458）在org.eclipse.emf.ecore.xmi.impl.XMLHandler.createObjectByType（XMLHandler.java:1335）在org.eclipse.emf.ecore.xmi.impl.XMLHandler.createTopObject（XMLHandler.java:1504）在org.eclipse.emf.ecore.xmi.impl.XMLHandler.processElement（XMLHandler.java:1026）在org.eclipse.emf.ecore.xmi.impl.XMIHandler.processElement（XMIHandler.java:77）在org.eclipse.emf.ecore.xmi.impl.XMLHandler.startElement（XMLHandler.java:1008）在org.eclipse.emf.ecore.xmi.impl.XMLHandler.startElement（XMLHandler.java:719）在org.eclipse.emf.ecore.xmi.impl.XMIHandler.startElement（XMIHandler.java:163）在com.sun.org.apache.xerces.internal.parsers.AbstractSAXParser.startElement（未知来源）在com.sun.org.apache.xerces.internal.impl.dtd.XMLDTDValidator.startElement（未知来源）在com.sun.org.apache.xerces.internal.impl.XMLDocumentFragmentScannerImpl.scanStartElement（未知来源）在com.sun.org.apache.xerces.internal.impl.XMLDocumentScannerImpl $ ContentDriver.scanRootElementHook（未知来源）在com.sun.org.apache.xerces.internal.impl.XMLDocumentFragmentScannerImpl $ FragmentContentDriver.next（未知来源）在com.sun.org.apache.xerces.internal.impl.XMLDocumentScannerImpl $ PrologDriver.next（未知来源）在com.sun.org.apache.xerces.internal.impl.XMLDocumentScannerImpl.next（未知来源）在com.sun.org.apache.xerces.internal.impl.XMLDocumentFragmentScannerImpl.scanDocument（未知来源）在com.sun.org.apache.xerces.internal.parsers.XML11Configuration.parse（未知来源）在com.sun.org.apache.xerces.internal.parsers.XML11Configuration.parse（未知来源）在com.sun.org.apache.xerces.internal.parsers.XMLParser.parse（未知来源）在com.sun.org.apache.xerces.internal.parsers.AbstractSAXParser.parse（未知来源）在com.sun.org.apache.xerces.internal.jaxp.SAXParserImpl $ JAXPSAXParser.parse中（未知来源）在com.sun.org.apache.xerces.internal.jaxp.SAXParserImpl.parse（未知来源）在org.eclipse.emf.ecore.xmi.impl.XMLLoadImpl.load（XMLLoadImpl.java:175）在org.eclipse.emf.ecore.xmi.impl.XMLResourceImpl.doLoad（XMLResourceImpl.java:261）在org.eclipse.emf.ecore.resource.impl.ResourceImpl.load（ResourceImpl.java:1518）在org.eclipse.emf.ecore.resource.impl.ResourceImpl.load（ResourceImpl.java:1297）在org.eclipse.emf.ecore.resource.impl.ResourceSetImpl.demandLoad（ResourceSetImpl.java:259）在org.eclipse.emf.ecore.resource.impl.ResourceSetImpl.demandLoadHelper（ResourceSetImpl.java:274）...另外2个

'带有uri的软件包“”未找到是什么意思？我可以直接读取xmi文件还是需要将其解析为xml文件？

我也尝试过此-> https://stackoverflow.com/a/4615965/1604503

    XMIResource resource = new XMIResourceImpl(URI.createURI("model/List.xmi"));
    resource.load(null);
    System.out.println( resource.getContents().get(0) );

PackageNotFoundEx。和Reource $ IOWrappedEx。再次：（

请帮助

亲切的问候

Answer 1

原因是，我没有生成模型代码。因此，找不到“列表”包。我确实将它添加到我的packageRegistry中，仅此而已。

        ResourceSet resourceSet = new ResourceSetImpl();

        // register UML
        Map packageRegistry = resourceSet.getPackageRegistry();
        packageRegistry.put(list.ListPackage.eNS_URI, list.ListPackage.eINSTANCE);

        // Register XML resource as UMLResource.Factory.Instance
        Map extensionFactoryMap = Resource.Factory.Registry.INSTANCE.getExtensionToFactoryMap();
        extensionFactoryMap.put("xmi", new XMIResourceFactoryImpl());

        Resource resource = (Resource) resourceSet.createResource(uri);


        // try to load the file into resource
        resource.load(null);

Answer 2

您应使用URI.createFileURI（uri）而不是URI.createURI（uri）。另外，您的模型实例（在您的情况下是List.xmi文件）必须具有xsi：schemaLocation属性，该属性具有指向您的元模型的相对路径，以避免预先注册元模型包。

科特林语示例：

    val extensionMap = Resource.Factory.Registry.INSTANCE.extensionToFactoryMap
    extensionMap["ecore"] = XMIResourceFactoryImpl()
    extensionMap["xmi"] = XMIResourceFactoryImpl()

    val resourceSet = ResourceSetImpl()

    // createFileURI method is able to locate metamodel by xsi:schemaLocation
    val resource = resourceSet.getResource(URI.createFileURI(modelPath), true)
    resource.allContents.forEach { eObj ->
        //do smth
    }

要预先注册动态EPackage，请参考eclipse wiki