我必须在我的 REST 网络服务中使用一个接口(interface)。这是接口(interface) Specs.java :
@XmlJavaTypeAdapter(MyAdapter.class)
public interface Specs {
public BaseProperties getBaseProps();
public void setBaseProps(BaseProperties baseProps);
}
我的适配器.java :
public class MyAdapter extends XmlAdapter<Object,Object>
{
public Object unmarshal(Object v)
{
return v;
}
public Object marshal(Object v)
{
return v;
}
}
RegSpecs.java
@XmlType
public class RegSpecs implements Specs{
private BaseProperties baseProps;
public BaseProperties getBaseProps()
{
return baseProps;
}
public void setBaseProps(BaseProperties baseProps)
{
this.baseProps = baseProps;
}
}
map 规范.java
@XmlType
public class MagSpecs implements Specs {
private BaseProperties baseProps;
private Features features;
public BaseProperties getBaseProps()
{
return baseProps;
}
public void setBaseProps(BaseProperties baseProps)
{
this.baseProps = baseProps;
}
public Features getFeatures() {
return features;
}
public void setFeatures(Features features) {
this.features = features;
}
}
访问此服务会引发以下错误:
javax.xml.bind.MarshalException - with linked exception: [javax.xml.bind.JAXBException: class entities.MagSpecs nor any of its super class is known to this context.]
如何修改我的上下文?我正在使用与 Jersey 1.5 捆绑在一起的 JAXB
谢谢!
编辑:为了更新我的上下文,我将这段代码添加到我的客户端(资源)类中:
public class BookService implements ContextResolver<JAXBContext>
{
private JAXBContext jaxbContext;
public BookService() {
try {
// Bootstrap your JAXBContext will all necessary classes
jaxbContext = JAXBContext.newInstance(Specs.class,MagSpecs.class, RegSpecs.class);
} catch(Exception e) {
throw new RuntimeException(e);
}
}
public JAXBContext getContext(Class<?> clazz) {
if(BookService.class == clazz) {
return jaxbContext;
}
return null;
}
在这种情况下我得到错误:
entities.Specs is an interface, and JAXB can't handle interfaces. this problem is related to the following location: at entities.Specs entities.Specs does not have a no-arg default constructor. this problem is related to the following location: at entities.Specs
最佳答案
Specs
接口(interface)的客户端需要知道 MagSpecs
可以是它的一个实例,以便它知道为了工具目的而查看它。最简单的方法是在 Specs
接口(interface)上添加一个 @XmlSeeAlso
注释:
@XmlSeeAlso({ MagSpecs.class, RegSpecs.class })
@XmlJavaTypeAdapter(MyAdapter.class) // Never needed this annotation myself...
public interface Specs {
public BaseProperties getBaseProps();
public void setBaseProps(BaseProperties baseProps);
}
一般来说,每当我使用 JAXB 注释时,我都会确保编写大量测试来检查是否可以从相关类生成 XML 模式,检查是否从每个(正常的)入口点进入 web类和接口(interface)我可以毫无异常(exception)地生成一个合理的模式。例如(对于这段时间有点长,我深表歉意):
private SchemaOutputResolver sink;
StringWriter schema;
@Before
public void init() {
schema = new StringWriter();
sink = new SchemaOutputResolver() {
@Override
public Result createOutput(String namespaceUri,
String suggestedFileName) throws IOException {
StreamResult sr = new StreamResult(schema);
sr.setSystemId("/dev/null");
return sr;
}
};
Assert.assertTrue(schema.toString().isEmpty());
}
private void testJAXB(Class<?>... classes) throws Exception {
JAXBContext.newInstance(classes).generateSchema(sink);
Assert.assertTrue(schema.toString().length() > 0);
}
@Test
public void testJAXBForSpecs() throws Exception {
testJAXB(Specs.class);
}
[编辑]:您还需要将Specs
接口(interface)更改为一个类,并让当前的实现从它继承。如果需要,它可以是一个完全抽象的类。只要您没有在类中添加重要的功能,它就应该可以工作。
关于java - Jersey REST/JAXB 错误,映射接口(interface),我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/4931960/