java - JAXB 2.0 验证不起作用

Question

我正在开发 JAXB 2.0，目前正在验证部分，因为它没有按预期工作。下面是验证码

public void validateXMLToSchema(Unmarshaller ummarshaller,String xsdFileName) throws SAXException, JAXBException{
    System.out.println(getClass().getResource(DEFAULT_XSD_NAME).toString());
    Schema schema;
    SchemaFactory schemaFactory=SchemaFactory.newInstance(XMLConstants.W3C_XML_SCHEMA_NS_URI);
    if(null==xsdFileName)
        schema=schemaFactory.newSchema(getClass().getResource(DEFAULT_XSD_NAME));

    else
        schema=schemaFactory.newSchema(new File(xsdFileName));

    ummarshaller.setSchema(schema);
    ummarshaller.setEventHandler( new ValidationEventHandler() {

        @Override
        public boolean handleEvent(ValidationEvent validationevent) {
            if(validationevent.getSeverity()==ValidationEvent.FATAL_ERROR || validationevent.getSeverity()==ValidationEvent.ERROR || validationevent.getSeverity()==ValidationEvent.WARNING){
                ValidationEventLocator  locator = validationevent.getLocator();
                log.info("Line:Col[" + locator.getLineNumber()
                        + ":" + locator.getColumnNumber()
                        + "]:" + validationevent.getMessage());
            }
            return true;
        }
    });

}

这是对该方法的调用

Destination destination=new Destination();
    try {
         destination=(Destination)unmarshal(Destination.class,new FileInputStream(new File("C:/Users/Raisonne/Desktop/jaxb/jaxb-ri-20101119/bin/destination.xml")));
    } catch (FileNotFoundException e) {
        e.printStackTrace();
    } catch (JAXBException e) {
        e.printStackTrace();
    }
    System.out.println(destination.getName());

}

public static <T> T unmarshal( Class<T> docClass, InputStream inputStream )
throws JAXBException, SAXException {
String packageName = docClass.getPackage().getName();
JAXBContext jc = JAXBContext.newInstance( packageName );
Unmarshaller u = jc.createUnmarshaller();
XMLValidator xmlValidator=new XMLValidator();
xmlValidator.validateXMLToSchema(u, null);

根据 XSD，我有几个字段作为必填字段，但即使删除它们，它也会给我错误，而它没有提供任何内容并将我的 xml 文件解析为相应的对象任何人都可以指出出了什么问题吗？

这是XSD的部分

    <xs:element name="destination" type="Destination"/>
  <xs:complexType name="Destination">
    <xs:sequence>
      <xs:element name="name" type="xs:string"/>
      <xs:element name="destinationID" type="xs:string" minOccurs="0"/>
      <xs:element name="shortDescription" type="xs:string" minOccurs="0"/>
      <xs:element name="longDescription" type="xs:string" minOccurs="0"/>
      <xs:element name="stateID" type="xs:string"/>
      <xs:element name="typeCode" type="xs:int"/>
      <xs:element name="countryCode" type="xs:string"/>
      <xs:element name="categories" type="xs:string"/>
      <xs:element name="transport" type="Transport" minOccurs="0" maxOccurs="1"/>
      <xs:element name="culture" type="Culture" minOccurs="0" maxOccurs="1"/>
      <xs:element name="events" type="Events" minOccurs="0" maxOccurs="1"/>
      <xs:element name="placesToVisit" type="PlacesToVisit" minOccurs="0" maxOccurs="1"/>
      <xs:element name="contacts" type="Contact" minOccurs="0" maxOccurs="1"/>
      <xs:element name="addresses" type="address" minOccurs="0" maxOccurs="1"/>
    </xs:sequence>
  </xs:complexType>
</xs:schema>

以及生成的Java文件

@XmlElement(required = true)
protected String name;
protected String destinationID;
protected String shortDescription;
protected String longDescription;
@XmlElement(required = true)
protected String stateID;

我正在从 xml 文件中删除 stateID，但验证部分仍然没有警报

提前致谢

Answer 1

你的代码片段有点难以理解，解码实际上发生了吗？您可能需要将最后一行添加到解码方法中:

public static <T> T unmarshal( Class<T> docClass, InputStream inputStream ) throws JAXBException, SAXException {
    String packageName = docClass.getPackage().getName();
    JAXBContext jc = JAXBContext.newInstance( packageName );
    Unmarshaller u = jc.createUnmarshaller();
    XMLValidator xmlValidator=new XMLValidator();
    xmlValidator.validateXMLToSchema(u, null);
    u.unmarshal(inputStream);
}