在Java Jersey资源中获取POST JSON时出错

Question

我无法让Jersey RESTful服务正常工作。

我收到以下错误

    The server encountered an internal error () that prevented it from fulfilling this request.

exception

java.lang.NullPointerException
    org.eclipse.persistence.internal.oxm.record.SAXUnmarshaller.unmarshal(SAXUnmarshaller.java:806)
    org.eclipse.persistence.oxm.XMLUnmarshaller.unmarshal(XMLUnmarshaller.java:602)
    org.eclipse.persistence.jaxb.JAXBUnmarshaller.unmarshal(JAXBUnmarshaller.java:399)
    com.sun.jersey.json.impl.BaseJSONUnmarshaller.unmarshalJAXBElementFromJSON(BaseJSONUnmarshaller.java:111)
    com.sun.jersey.json.impl.BaseJSONUnmarshaller.unmarshalFromJSON(BaseJSONUnmarshaller.java:100)
    com.sun.jersey.json.impl.provider.entity.JSONRootElementProvider.readFrom(JSONRootElementProvider.java:129)
    com.sun.jersey.core.provider.jaxb.AbstractRootElementProvider.readFrom(AbstractRootElementProvider.java:111)
    com.sun.jersey.spi.container.ContainerRequest.getEntity(ContainerRequest.java:488)
    com.sun.jersey.server.impl.model.method.dispatch.EntityParamDispatchProvider$EntityInjectable.getValue(EntityParamDispatchProvider.java:123)
    com.sun.jersey.server.impl.inject.InjectableValuesProvider.getInjectableValues(InjectableValuesProvider.java:46)
    com.sun.jersey.server.impl.model.method.dispatch.AbstractResourceMethodDispatchProvider$EntityParamInInvoker.getParams(AbstractResourceMethodDispatchProvider.java:153)
    com.sun.jersey.server.impl.model.method.dispatch.AbstractResourceMethodDispatchProvider$TypeOutInvoker._dispatch(AbstractResourceMethodDispatchProvider.java:183)
    com.sun.jersey.server.impl.model.method.dispatch.ResourceJavaMethodDispatcher.dispatch(ResourceJavaMethodDispatcher.java:75)
    com.sun.jersey.server.impl.uri.rules.HttpMethodRule.accept(HttpMethodRule.java:302)
    com.sun.jersey.server.impl.uri.rules.ResourceClassRule.accept(ResourceClassRule.java:108)
    com.sun.jersey.server.impl.uri.rules.RightHandPathRule.accept(RightHandPathRule.java:147)
    com.sun.jersey.server.impl.uri.rules.RootResourceClassesRule.accept(RootResourceClassesRule.java:84)
    com.sun.jersey.server.impl.application.WebApplicationImpl._handleRequest(WebApplicationImpl.java:1511)
    com.sun.jersey.server.impl.application.WebApplicationImpl._handleRequest(WebApplicationImpl.java:1442)
    com.sun.jersey.server.impl.application.WebApplicationImpl.handleRequest(WebApplicationImpl.java:1391)
    com.sun.jersey.server.impl.application.WebApplicationImpl.handleRequest(WebApplicationImpl.java:1381)
    com.sun.jersey.spi.container.servlet.WebComponent.service(WebComponent.java:416)
    com.sun.jersey.spi.container.servlet.ServletContainer.service(ServletContainer.java:538)
    com.sun.jersey.spi.container.servlet.ServletContainer.service(ServletContainer.java:716)
    javax.servlet.http.HttpServlet.service(HttpServlet.java:722)
* Closing connection #0

服务是：

@Path("/agents")
public class AgentService {

    @POST 
    @Consumes({MediaType.APPLICATION_JSON + ";charset=utf-8"})
    @Produces(MediaType.APPLICATION_JSON + ";charset=utf-8")
    public String createAgent(AgentTO input) {
        try{
//          Agent agent = input.createAgent()
//          PersistenceService.getInstance().save(agent);
            return input.getName();
        } catch (Exception se) {
            se.printStackTrace();
        }
        return null;
    }

这是我要接收的对象：

@XmlRootElement
public class AgentTO {

    private String name;
    private String phone;

    public AgentTO(){}

    public AgentTO(String phone, String name){
        this.phone = phone;
        this.name = name;               
    }

我在做这样的POST：

curl -H "Content-Type: application/json" -v -X POST -d '{"phone":"123","name":"asd2"}' http://127.0.0.1:8080/project/agents

在此之前，我已经制作了参考资料，但这似乎并不想工作：P

你能帮我吗？

问候，马丁

Answer 1

您可以使用Gson从正文请求中解组JSON。 请在这里检查。