JAXB无法创建抽象类的实例现在
@XmlSeeAlso({ Dog.class, Cat.class })
public abstract class Animal {}
@XmlRootElement(name="dog")
public class Dog extends Animal {}
@XmlRootElement(name="cat")
public class Cat extends Animal {}
@WebService(name = "WebServiceDemo", serviceName = "WebServiceDemo")
@SOAPBinding(style = SOAPBinding.Style.DOCUMENT, use = SOAPBinding.Use.LITERAL,
parameterStyle = SOAPBinding.ParameterStyle.WRAPPED)
public class WebServiceDemo {
@WebMethod
public String service(@WebParam(name = "animal") Animal animal) {
System.out.println("animal service calling.....");
return animal;
}
}
当我打电话从客户端与动物CALSS这种服务方法的参数时我收到错误 -JAXB无法创建抽象类的实例现在
产生的原因:javax.xml .bind.UnmarshalException:无法创建动物实例 - 带有链接的异常: [java.lang.InstantiationException] at com.sun.xml.bind.v2.runtime.unmarshaller.UnmarshallingContext.handleEvent(UnmarshallingContext.java: 616) at com.sun.xml.bind.v2.runtime.unmarshaller.Loader.reportError(Loader.java:244) at com.sun.xml.bind.v2.runtime.unmarshaller.UnmarshallingContext.createInstance(UnmarshallingContext.java:583) at com.sun.xml.bind.v2.runtime.unmarshaller.StructureLoader.startElement(StructureLoader.java: 181) at com.sun.xml.bind.v2.runtime.unmarshaller.XsiTypeLoader.startElement(XsiTypeLoader.java:73) at com.sun.xml.bind.v2.runtime.unmarshaller.UnmarshallingContext._startElement(UnmarshallingContext。 java:455) at com.sun.xml.bind.v2.runtime.unmarshaller.UnmarshallingContext.startElement(UnmarshallingContext.java:433) at com.sun.xml.bind.v2.runtime.unmarshaller.InterningXmlVisitor.startElement( InterningXmlVisitor.java:71) at com.sun.xml.bind.v2.runtime.unmarshaller.SAXConnector.startElement(SAXConnector.java:137) at com.sun.xml.bind.unmarshalle r.DOMScanner.visit(DOMScanner.java:240) at com.sun.xml.bind.unmarshaller.DOMScanner.visit(DOMScanner.java:277) at com.sun.xml.bind.unmarshaller.DOMScanner.visit( DOMScanner.java:246) at com.sun.xml.bind.unmarshaller.DOMScanner.scan(DOMScanner.java:123) at com.sun.xml.bind.v2.runtime.unmarshaller.UnmarshallerImpl.unmarshal0(UnmarshallerImpl。 java:314) at com.sun.xml.bind.v2.runtime.unmarshaller.UnmarshallerImpl.unmarshal(UnmarshallerImpl.java:293) at com.sun.xml.bind.v2.runtime.unmarshaller.UnmarshallerImpl.unmarshal( UnmarshallerImpl.java:244) at org.jboss.ws.core.jaxws.JAXBDeserializer.deserialize(JAXBDeserializer.java:71)
抽象类不能被实例化,这是Java中它们的基本规则。从javadocs:
抽象类是声明为abstract,它可能会或可能不会 包含抽象方法的类。抽象类不能实例化,但它们可以被分类。
Jaxb会在内部尝试将您的xml解组到java对象。但是,如果它无法创建动物的对象,它将如何工作。因此它抛出异常。您需要为JaxB提供一个非抽象类来工作。
那么,你不能实例化一个抽象类的定义... – fge
可能[this](http://stackoverflow.com/questions/9580028/jaxb-and-abstract-classes)有帮助吗? – NINCOMPOOP