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我正在尝试使用 HttpInvokerServiceExporter + HttpInvokerProxyFactoryBean，但无论我做什么都会出现异常：

org.springframework.remoting.RemoteAccessException: Could not access HTTP invoker remote service at [http://localhost:9999/testcaseapp/testcaseservice]; nested exception is java.io.IOException: Did not receive successful HTTP response: status code = 404, status message = [Not Found]

为简单起见，我创建了一个测试用例。

@RunWith(SpringJUnit4ClassRunner.class)
@ContextConfiguration
public class RemoteTest {

    private static final Logger logger = LoggerFactory.getLogger("TestsLogger");

    static interface TestCaseService {
        public Integer add(Integer arg1, Integer arg2);
    }
    static class TestCaseServiceImpl implements TestCaseService {
        public Integer add(Integer arg1, Integer arg2) {
            return (arg1 != null ? arg1.intValue() : 0) + (arg2 != null ? arg2.intValue() : 0);
        }
    }

    @Configuration
    static class Config {
        @Bean
        public HttpInvokerServiceExporter httpInvokerServiceExporter() {
            HttpInvokerServiceExporter httpInvokerServiceExporter = new HttpInvokerServiceExporter();
            httpInvokerServiceExporter.setService(new TestCaseServiceImpl());
            httpInvokerServiceExporter.setServiceInterface(TestCaseService.class);
            return httpInvokerServiceExporter;
        }
        @Bean
        public HttpInvokerProxyFactoryBean httpInvokerProxyFactoryBean() {
            HttpInvokerProxyFactoryBean httpInvokerProxyFactoryBean = new HttpInvokerProxyFactoryBean();
            httpInvokerProxyFactoryBean.setServiceInterface(TestCaseService.class);
            httpInvokerProxyFactoryBean.setServiceUrl("http://localhost:9999/testcaseapp/testcaseservice");
            httpInvokerProxyFactoryBean.afterPropertiesSet();
            return httpInvokerProxyFactoryBean;
        }
    }

    @Autowired
    private TestCaseService[] testCaseServices;
    private static Server server;

    @BeforeClass
    public static void setUp() {
        try {
            server = new Server();
            SelectChannelConnector connector = new SelectChannelConnector();
            connector.setPort(9999);
            server.addConnector(connector);
            //
            WebAppContext webAppContext = new WebAppContext();
            webAppContext.setContextPath("/testcaseapp");
            webAppContext.setWar("src/test/java/" + RemotingTest.class.getPackage().getName().replace('.', '/'));
            server.setHandler(webAppContext);
            //
            server.start();
        } catch (Exception ex) {
            logger.info("Could not permorm the set up: {}", ex.toString());
        }
    }

    @AfterClass
    public static void destroy() {
        try {
            server.stop();
        } catch (Exception e) {
        }
    }

    @Test
    public void addTest() {
        for (TestCaseService testCaseService : testCaseServices) {
            Integer sum = testCaseService.add(10, 5);
            Assert.assertNotNull(sum);
            Assert.assertEquals(15, sum.intValue());
        }
    }

}

我也尝试过创建一个 TestCaseService bean

@Bean public TestCaseService testCaseService() ...

并将其作为 httpInvokerServiceExporter 参数提供

@Bean public HttpInvokerServiceExporter httpInvokerServiceExporter(TestCaseService testCaseService)
...
httpInvokerServiceExporter.setService(testCaseService);

但结果还是一样。

我究竟做错了什么？谢谢！

4

2 回答 2

2

我认为问题在于 Servlet 不可访问。

服务器端

确保您的 WEB-INF/web.xml（在公开方法 -SERVER- 的应用程序上）中有以下代码：

<web-app>
...
<servlet>
    <servlet-name>remoting</servlet-name>
    <servlet-class>org.springframework.web.servlet.DispatcherServlet</servlet-class>
    <load-on-startup>1</load-on-startup>
</servlet>

<servlet-mapping>
    <servlet-name>remoting</servlet-name>
    <url-pattern>/services/*</url-pattern>
</servlet-mapping>
...
</web-app>

在这里，远程方法在“服务”下提供服务，即调用该方法的 URL 应该是：

http://localhost:8080/sample/services/list

您必须通过创建一个 bean（在我的例子中，在 WEB-INF/remoting-servlet.xml 下）将此 Servlet 定义为可访问的：

<bean name="/list" class="org.springframework.remoting.httpinvoker.HttpInvokerServiceExporter">
    <property name="service" ref="myObjectQueryService" />
    <property name="serviceInterface" value="com.kategor.myapp.sample.service.ObjectQueryService" />
</bean>

客户端

如果您在客户端下使用 Spring（不像您的示例中那样），则必须定义一个用于访问远程资源的 bean，定义一些 bean（每个公共资源一个）：

在这种情况下，它将是：

<bean id="listService" class="org.springframework.remoting.httpinvoker.HttpInvokerProxyFactoryBean">
    <property name="serviceUrl" value="http://localhost:8080/sample/services/list" />
    <property name="serviceInterface" value="com.kategor.myapp.sample.service.ObjectQueryService" />
</bean>

在你的例子中是对的。

这样，调用服务“listService”，您将拥有类 com.kategor.myapp.sample.service.ObjectQueryService 中可用的所有方法

@Controller
public class HomeController {

    // This is the remote service definition
    @Autowired
    private ObjectQueryService<MyObject, Long> objectQueryService;

    /* .... */

    /**
     * List all Objects retrieved through Web Service from a remote Server
     */
    @RequestMapping(value = "listRemoteWS", method = RequestMethod.GET)
    public String listRemoteWS(Locale locale, Model model) {

        StringBuilder result = new StringBuilder();
        try {

            // The Remote Service is called
            List objs = objectQueryService.findAll(0, 10);
            result.append(objs.size() + " objs found");

        for (MyObject o : objs) {
            result.append("<br>* ").append(o.getId()).append(" = ").append(o.getName());
        }


        } catch (Exception e) {

            result.append("No objs have been found");
            e.printStackTrace();

        }

        model.addAttribute("result", result);

        return "index";
    }

}

所以我认为问题来自 URL：也许服务不可见或者这不是它的正确路径。

有关更多信息，请查看此链接（第一个非常有用）：

https://github.com/JamesEarlDouglas/barebones-spring-mvc/tree/master/reference/spring-remoting

http://www.ibm.com/developerworks/web/library/wa-spring3webserv/index.html

于 2012-11-13T15:41:55.490 回答
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对我来说，问题是 tomcat 获取了相同应用程序的两个版本。这在调试模式下从 STS 运行客户端时引发了上述错误。所以解决方案是清理应用程序在tomcat中所有扩展的webapp文件夹。然后重新部署应用程序。

于 2016-10-10T13:15:32.517 回答