我有以下JAXB解组代码,我试图概括.这是它的样子:
private Object getResponseObject(String stubbedXmlFile,
Class jaxbInterfaceClass,
AbstractRepository repository) {
Object responseObject = null;
try {
JAXBContext jc = JAXBContext.newInstance(jaxbInterfaceClass);
Unmarshaller u = jc.createUnmarshaller();
InputStream resourceAsStream = TestProvidePensionValuationRepository.class.getClassLoader()
.getResourceAsStream(stubbedXmlFile);
BufferedReader br = new BufferedReader(new InputStreamReader(resourceAsStream));
StringBuilder sb = new StringBuilder();
String line;
while ((line = br.readLine()) != null) {
sb.append(line);
}
br.close();
ByteArrayInputStream byteStream =
new ByteArrayInputStream(sb.toString().getBytes());
responseObject = u.unmarshal(byteStream);
} catch (JAXBException e) {
e.printStackTrace();
} catch (IOException e) {
e.printStackTrace();
}
return responseObject;
}
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我的目标不是返回一个Object,而是返回作为参数传入的同一个类(例如jaxbInterfaceClass).我认为这可能与Generics有关,但作为仿制药的相对新手,我很难过.任何帮助赞赏!
这根本不是一个JAXB问题,只需对所需的泛型进行一些调整:
private <T> T getResponseObject(String stubbedXmlFile,
Class<T> jaxbInterfaceClass,
AbstractRepository repository) {
T responseObject = null;
// code as before
responseObject = (T) u.unmarshal(byteStream);
return responseObject;
}
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这可能会引发编译器警告,但这种情况并非罕见,因为代码将泛型代码与非泛型代码(如JAXB)相连接.
然后你可以用以下方法调用它:
MyType x = getResponseObject(file, MyType.class, repo);
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