java - 集成spring mvc和Jersey，在查看jersey端点时获取空指针

Question

我的简单 Jersey 端点:

@Component
@Path("/v1/user")
public class UserService {

    @Autowired
    com.myapp.core.service.UserService userService;

    @GET
    @Path("/get")
    public Response getUser() {
        User user = userService.getUser(1);

        String result = user.getUsername();

        return Response.status(200).entity(result).build();
    }
}

(顺便说一句，既然这是 RESTful，那么这会是 xml 还是 json 输出？？)

我的 spring mvc 页面工作正常(进入 hibernate 状态)，但是当我访问 localhost:8080/api/v1/user/get 时我看到:

java.lang.NullPointerException
    com.myapp.api.UserService.getUser(UserService.java:30)
    sun.reflect.NativeMethodAccessorImpl.invoke0(Native Method)
    sun.reflect.NativeMethodAccessorImpl.invoke(NativeMethodAccessorImpl.java:39)
    sun.reflect.DelegatingMethodAccessorImpl.invoke(DelegatingMethodAccessorImpl.java:25)
    java.lang.reflect.Method.invoke(Method.java:597)
    com.sun.jersey.spi.container.JavaMethodInvokerFactory$1.invoke(JavaMethodInvokerFactory.java:60)
    com.sun.jersey.server.impl.model.method.dispatch.AbstractResourceMethodDispatchProvider$ResponseOutInvoker._dispatch(AbstractResourceMethodDispatchProvider.java:205)
    com.sun.jersey.server.impl.model.method.dispatch.ResourceJavaMethodDispatcher.dispatch(ResourceJavaMethodDispatcher.java:75)
    com.sun.jersey.server.impl.uri.rules.HttpMethodRule.accept(HttpMethodRule.java:288)
    com.sun.jersey.server.impl.uri.rules.RightHandPathRule.accept(RightHandPathRule.java:147)
    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:1469)
    com.sun.jersey.server.impl.application.WebApplicationImpl._handleRequest(WebApplicationImpl.java:1400)
    com.sun.jersey.server.impl.application.WebApplicationImpl.handleRequest(WebApplicationImpl.java:1349)
    com.sun.jersey.server.impl.application.WebApplicationImpl.handleRequest(WebApplicationImpl.java:1339)
    com.sun.jersey.spi.container.servlet.WebComponent.service(WebComponent.java:416)
    com.sun.jersey.spi.container.servlet.ServletContainer.service(ServletContainer.java:537)
    com.sun.jersey.spi.container.servlet.ServletContainer.service(ServletContainer.java:699)
    javax.servlet.http.HttpServlet.service(HttpServlet.java:717)

我的 web.xml:

<?xml version="1.0" encoding="UTF-8"?>
<web-app version="2.5" xmlns="http://java.sun.com/xml/ns/javaee"
    xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
    xsi:schemaLocation="http://java.sun.com/xml/ns/javaee http://java.sun.com/xml/ns/javaee/web-app_2_5.xsd">

    <display-name>myapp</display-name>
    <description>myapp</description>

    <context-param>
        <param-name>contextConfigLocation</param-name>
        <param-value>
        /WEB-INF/spring/applicationContext.xml
        /WEB-INF/spring/myapp-servlet.xml
        </param-value>
    </context-param>

    <servlet>
        <servlet-name>myapp</servlet-name>
        <servlet-class>org.springframework.web.servlet.DispatcherServlet</servlet-class>
        <init-param>
            <param-name>contextConfigLocation</param-name>
            <param-value>/WEB-INF/spring/myapp-servlet.xml</param-value>
        </init-param>
        <load-on-startup>1</load-on-startup>
    </servlet>

    <servlet-mapping>
        <servlet-name>myapp</servlet-name>
        <url-pattern>/</url-pattern>
    </servlet-mapping>


    <!-- jersey -->
    <listener>
        <listener-class>
            org.springframework.web.context.ContextLoaderListener
        </listener-class>
    </listener>

    <!--<listener>-->
          <!--<listener-class>org.springframework.web.context.request.RequestContextListener</listener-class>-->
   <!--</listener>-->

    <servlet>
        <servlet-name>jersey-servlet</servlet-name>
        <servlet-class>
            com.sun.jersey.spi.spring.container.servlet.SpringServlet
        </servlet-class>
        <init-param>
            <param-name>com.sun.jersey.config.property.packages</param-name>
            <param-value>com.myapp.api</param-value>
        </init-param>
        <load-on-startup>2</load-on-startup>
    </servlet>

    <servlet-mapping>
        <servlet-name>jersey-servlet</servlet-name>
        <url-pattern>/api/*</url-pattern>
    </servlet-mapping>
    <!-- /jersey -->

</web-app>

我的应用程序上下文有用于以下用途的 bean:

dataSource、sessionFactory 和 userDao。

我的 myapp-servlet.xml 有:

<context:component-scan base-package="com.myapp" />

(applicationContext 和 myapp-servlet.xml 都具有相同的 xml header 节点等，以防出现问题？)

Answer 1

确保您的 myapp-servlet.xml 包含以下内容:

<!--  enable autowire -->
<context:annotation-config />

关于请求/响应的媒体类型 - 您可以明确指定它，如下所示:

@Produces({MediaType.APPLICATION_JSON})
@Consumes({MediaType.APPLICATION_JSON})
@Path("/v1/user")
public class UserService {...

默认值为“*/*” - 任何类型。有关 Jersey 注释的有用信息可以找到 here