java - JSON Jersey 自动序列化包装器问题

Question

我有一个 POJO

public class Graph {
    public int v;
    public int e;
}

和一个非常简单的服务

@Service("graph-service#default")
public class DefaultGraphService implements GraphService {
    public Response createGraph(Graph graph) {
        return Response.ok().build();
    }
}

它实现了一个非常简单的接口(interface)

@Path( "graph-service" )
public interface GraphService {

    @Path( "create-graph" )
    @POST
    @Consumes(MediaType.APPLICATION_JSON)
    public Response createGraph(Graph graph);
}

我有一个简单的 spring-context.xml 设置如下

<beans xmlns="http://www.springframework.org/schema/beans"
   xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
   xmlns:context="http://www.springframework.org/schema/context"
   xmlns:jaxrs="http://cxf.apache.org/jaxrs"
   xsi:schemaLocation="http://www.springframework.org/schema/beans
       http://www.springframework.org/schema/beans/spring-beans-3.0.xsd
       http://www.springframework.org/schema/context 
       http://www.springframework.org/schema/context/spring-context-3.0.xsd
       http://cxf.apache.org/jaxrs
       http://cxf.apache.org/schemas/jaxrs.xsd">

    <import resource="classpath:META-INF/cxf/cxf.xml"/>
    <import resource="classpath:META-INF/cxf/cxf-servlet.xml"/>

    <context:annotation-config/>
    <context:component-scan base-package="me.jamesphiliprobinson.graphs"/>

    <jaxrs:server id="testServices" address="/testServices">
        <jaxrs:serviceBeans>
            <ref bean="graph-service#default"/>
        </jaxrs:serviceBeans>
        <jaxrs:providers>
            <ref bean="jsonProvider"/>
        </jaxrs:providers>
    </jaxrs:server>
    <bean id="jsonProvider" class="org.codehaus.jackson.jaxrs.JacksonJsonProvider" />
</beans>

如果我在 tomcat 中启动该服务，它工作得很好，我可以 curl 一个对象吗

{"v":2,"e":1}

没有任何困难。但是，如果我运行这个测试

@Test
public void testCreateGraph() {
    UncertaintyGraphService service =
            JAXRSClientFactory.create( "http://localhost:" + port + "/" + getRestServicesPath() + "/testServices/",
                                       GraphService.class );
    Graph graph = new Graph();
    graph.e = 1;
    graph.v = 2;
    Response result = service.createGraph(graph);
    assertNotNull(result);
}

然后它就失败了

No message body writer has been found for class : class me.jamesphiliprobinson.graphs.Graph, ContentType : application/json

如果我添加一个

@XmlRootElement

到 POJO，然后服务序列化图形对象，但似乎发送

{"graph":{"e":1,"v",2}}

相反，我认为这是有道理的，但反序列化似乎仍然期望

{"e":1,"v":2}

当我收到错误时

WARNING: WebApplicationException has been caught : Unrecognized field "graph" (Class me.jamesphiliprobinson.graphs.Graph), not marked as ignorable

我肯定错过了一些非常简单的东西。我希望将这些项目序列化为

{"v":2,"e":1}

但是如果它们能正确反序列化，我就可以使用根元素。即

{"graph":{"v":2,"e":1}}

Answer 1

查看您在哪里注册 JacksonJsonProvider在服务器上？

<bean id="jsonProvider" class="org.codehaus.jackson.jaxrs.JacksonJsonProvider" />

这是为了在服务器端处理 POJO 之间的 JSON(反)序列化。但客户也需要同样的支持。您可以使用重载的

• JAXRSClientFactory.create(String baseAddress, Class<T> cls, List<?> providers)

在客户端注册提供程序。只需添加 JacksonJsonProvider到List的providers .

JAXRSClientFactory.create(baseUri, Service.class, Arrays.asList(new JacksonJsonProvider())