从XMI文件加载EMF模型实例
[英]Load EMF model instance from XMI file
问题描述
I know there are several QnA about this subject. 
我知道有几个有关此主题的QnA。 I have tried lots of solution, I am always getting same error. 我尝试了很多解决方案，但总是遇到相同的错误。
My code structure is like: 我的代码结构如下：
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);
Error: 错误：
Exception in thread "main" org.eclipse.emf.ecore.resource.impl.ResourceSetImpl$1DiagnosticWrappedException: org.eclipse.emf.ecore.xmi.PackageNotFoundException: Package with uri 'List' not found.
What does 'package with uri "" not found' mean ? “未使用uri打包”是什么意思？ Can I read xmi file directly or do I need to parse it as xml file ? 我可以直接读取xmi文件还是需要将其解析为xml文件？
I have also tried this -> https://stackoverflow.com/a/4615965/1604503 我也尝试过-> https://stackoverflow.com/a/4615965/1604503
XMIResource resource = new XMIResourceImpl(URI.createURI("model/List.xmi"));
resource.load(null);
System.out.println( resource.getContents().get(0) );
PackageNotFoundEx. PackageNotFoundEx。 and Reource$IOWrappedEx. 和Reource $ IOWrappedEx。 again :( 再次:(
Please help 请帮忙
kind regards 亲切的问候
2 个解决方案
解决方案1
1 2014-11-17 12:20:25
The reason was, I did not generate the model codes. 原因是，我没有生成模型代码。 Because of that, 'List' package could not be found. 因此，找不到“列表”包。 I did add it to my packageRegistry and that's all. 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);
解决方案2
0 2019-09-25 18:36:29
You should use URI.createFileURI(absolutePathUri) instead of URI.createURI(uri) . 您应该使用URI.createFileURI（absolutePathUri）而不是URI.createURI（uri） 。 Also, your model instance (in your case it is List.xmi file) must have xsi:schemaLocation attribute with relative path to your metamodel to avoid pre-registration of your metamodel package. 另外，您的模型实例（在您的情况下，它是List.xmi文件）必须具有xsi：schemaLocation属性，该属性具有指向元模型的相对路径，以避免预先注册元模型包。 Very important note: your uri parameter must be constructed from absolute path of your model 非常重要的注意事项：您的uri参数必须从模型的绝对路径构造
Kotlin language example: Kotlin语言示例：
Resource.Factory.Registry.INSTANCE.extensionToFactoryMap.apply {
put("ecore", EcoreResourceFactoryImpl())
put("xmi", XMIResourceFactoryImpl())
}
val resourceSet = ResourceSetImpl()
// createFileURI method is able to locate metamodel by xsi:schemaLocation
// absolute path is important here !!!
val resource = resourceSet.getResource(URI.createFileURI(File(modelPath).absolutePath), true)
resource.allContents.forEach { eObj ->
//do smth
}
To make pre-registration of dynamic EPackage refer to eclipse wiki 要预先注册动态EPackage，请参阅Eclipse Wiki。
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