如何使用 Jersey JSON POJO 支持?
我有一个对象,我想在 JSON 中作为 RESTful 资源提供服务.我像这样打开了 Jersey 的 JSON POJO 支持(在 web.xml 中):
I have an object that I'd like to serve in JSON as a RESTful resource. I have Jersey's JSON POJO support turned on like so (in web.xml):
<servlet>
<servlet-name>Jersey Web Application</servlet-name>
<servlet-class>com.sun.jersey.spi.container.servlet.ServletContainer</servlet-class>
<init-param>
<param-name>com.sun.jersey.api.json.POJOMappingFeature</param-name>
<param-value>true</param-value>
</init-param>
<load-on-startup>1</load-on-startup>
</servlet>
但是当我尝试访问资源时,我得到了这个异常:
But when I try to access the resource, I get this exception:
SEVERE: A message body writer for Java type, class com.example.MyDto, and MIME media type, application/json, was not found
SEVERE: Mapped exception to response: 500 (Internal Server Error)
javax.ws.rs.WebApplicationException
...
我要服务的类并不复杂,它所拥有的只是一些公共的 final 字段和一个设置所有这些字段的构造函数.这些字段都是字符串、原语、与此类似的类或其列表(我尝试使用普通列表而不是通用列表<T>s,但无济于事).有谁知道给了什么?谢谢!
The class that I'm trying to serve isn't complicated, all it's got are some public final fields and a constructor that sets all of them. The fields are all strings, primitives, classes similar to this one, or Lists thereof (I've tried using plain Lists instead of generic List<T>s, to no avail). Does anyone know what gives? Thanks!
Java EE 6
球衣 1.1.5
玻璃鱼 3.0.1
推荐答案
Jersey-json 有一个 JAXB 实现.您收到该异常的原因是因为您没有 Provider 已注册,或者更具体地说是一个 MessageBodyWriter.你需要在你的提供者中注册一个合适的上下文:
Jersey-json has a JAXB implementation. The reason you're getting that exception is because you don't have a Provider registered, or more specifically a MessageBodyWriter. You need to register a proper context within your provider:
@Provider
public class JAXBContextResolver implements ContextResolver<JAXBContext> {
private final static String ENTITY_PACKAGE = "package.goes.here";
private final static JAXBContext context;
static {
try {
context = new JAXBContextAdapter(new JSONJAXBContext(JSONConfiguration.mapped().rootUnwrapping(false).build(), ENTITY_PACKAGE));
} catch (final JAXBException ex) {
throw new IllegalStateException("Could not resolve JAXBContext.", ex);
}
}
public JAXBContext getContext(final Class<?> type) {
try {
if (type.getPackage().getName().contains(ENTITY_PACKAGE)) {
return context;
}
} catch (final Exception ex) {
// trap, just return null
}
return null;
}
public static final class JAXBContextAdapter extends JAXBContext {
private final JAXBContext context;
public JAXBContextAdapter(final JAXBContext context) {
this.context = context;
}
@Override
public Marshaller createMarshaller() {
Marshaller marshaller = null;
try {
marshaller = context.createMarshaller();
marshaller.setProperty(Marshaller.JAXB_FORMATTED_OUTPUT, true);
} catch (final PropertyException pe) {
return marshaller;
} catch (final JAXBException jbe) {
return null;
}
return marshaller;
}
@Override
public Unmarshaller createUnmarshaller() throws JAXBException {
final Unmarshaller unmarshaller = context.createUnmarshaller();
unmarshaller.setEventHandler(new DefaultValidationEventHandler());
return unmarshaller;
}
@Override
public Validator createValidator() throws JAXBException {
return context.createValidator();
}
}
}
这会在提供的包名称中查找 @XmlRegistry
,这是一个包含 @XmlRootElement
注释 POJO 的包.
This looks up for an @XmlRegistry
within the provided package name, which is a package that contains @XmlRootElement
annotated POJOs.
@XmlRootElement
public class Person {
private String firstName;
//getters and setters, etc.
}
然后在同一个包中创建一个ObjectFactory:
then create an ObjectFactory in the same package:
@XmlRegistry
public class ObjectFactory {
public Person createNewPerson() {
return new Person();
}
}
注册 @Provider
后,Jersey 应该会为您的资源中的编组提供便利:
With the @Provider
registered, Jersey should facilitate the marshalling for you in your resource:
@GET
@Consumes(MediaType.APPLICATION_JSON)
public Response doWork(Person person) {
// do work
return Response.ok().build();
}
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