如何用模板解组？

Question

我想错误在

T bobj = (T) jaxbUnmarshaller.unmarshal(file);

它总是返回null

使用非模板测试，它工作并返回一个Customer类，仅在与模板一起使用时返回null

原始代码

XMLObj<Customer> XMLtool = new XMLObj<Customer>(Customer.class);
Customer c = XMLtool.ConvertXMLToObject("c:\\file2.xml");

public class XMLObj<T> {
    final Class<T> typeParameterClass;
    public XMLObj(Class<T> typeParameterClass) {
        this.typeParameterClass = typeParameterClass;
    }

public T ConvertXMLToObject(String path)
    {
        //Convert XML to Object
        try {
            File file = new File(path);
            if(file.exists())
            {
                JAXBContext jaxbContext;
                jaxbContext = JAXBContext.newInstance(typeParameterClass.getClass());
                Unmarshaller jaxbUnmarshaller = jaxbContext.createUnmarshaller();
                T bobj = (T) jaxbUnmarshaller.unmarshal(file);
                System.out.println(bobj);
                return bobj;
            }
            else
                Logger.getInstance().process_message("File not exist in ConvertObjectToXML");
        } catch (JAXBException e) {
            // TODO Auto-generated catch block
            Logger.getInstance().process_message(e.getMessage());
        }
        return null;
    }
}

Answer 1

尝试更改此行：

jaxbContext = JAXBContext.newInstance(typeParameterClass.getClass());

对此：

jaxbContext = JAXBContext.newInstance(typeParameterClass);

typeParameterClass.getClass()会返回java.lang.Class类的类型，而typeParameterClass本身的类型为Customer，这是您要解组的类。