使用Spring 3.2.RC2使控制器方法异步的问题

时间:2012-11-30 13:41:46

标签: spring-mvc asynchronous controller

我正在尝试测试Spring MVC控制器的最新异步功能,但我无法让它工作。

以下是我的异步方法的代码:

@RequestMapping(value = "/hello")
    public Callable<String> async(final Model model) {
        System.out.println("entered async controller method");
        return new Callable<String>() {
            public String call() throws Exception {
                Thread.sleep(2000L);
                model.addAttribute("message", "asyncRequest dealt with");
                System.out.println("about to return from call()");
                return "hello";
            }
        };
}

以下是web.xml的相关部分:

<servlet>
    <servlet-name>SpringMVC</servlet-name>
    <servlet-class>org.springframework.web.servlet.DispatcherServlet</servlet-class>
    <init-param>
        <param-name>contextConfigLocation</param-name>
        <param-value>/WEB-INF/spring/webmvc-config.xml</param-value>
    </init-param>
    <load-on-startup>1</load-on-startup>
    <async-supported>true</async-supported>
</servlet>

然而“即将从call()返回”永远不会在控制台中打印出来,我永远不会看到这样的日志: 08:25:17 [MvcAsync1] WebAsyncManager < / strong> - ... 在控制台......

仅供参考,我使用Spring 3.2.RC2

3 个答案:

答案 0 :(得分:6)

进一步配置async-support实施AsyncTaskExecutor

  

默认情况下,Spring MVC使用SimpleAsyncTaskExecutor来执行控制器方法返回的Callable实例。对于生产,您必须将其替换为适合您的环境的AsyncTaskExecutor实现。

创建AsyncTaskExecutor

<beans:bean id="asyncTaskExecutor" class="org.springframework.scheduling.concurrent.ThreadPoolTaskExecutor">
    <beans:property name="corePoolSize" value="5" />
    <beans:property name="maxPoolSize" value="10" />
    <beans:property name="queueCapacity" value="25" />
</beans:bean>

然后分配task-executor

<mvc:annotation-driven >
    <mvc:async-support default-timeout="30000" task-executor="asyncTaskExecutor" />
</mvc:annotation-driven>

http://blog.springsource.org/2012/05/10/spring-mvc-3-2-preview-making-a-controller-method-asynchronous/

答案 1 :(得分:2)

我终于找到了为什么我的示例无效:必须将以下内容添加到web-mvc配置文件中:

<mvc:annotation-driven>
    <mvc:async-support default-timeout="3000"/>
</mvc:annotation-driven>

答案 2 :(得分:0)

您使用以下代码执行类似的应用程序:

public String doSlowWork() {

     System.out.println("Start  slow work");
    try {
      Thread.sleep(3000);
    } catch (InterruptedException e) {
      e.printStackTrace();
    }
    System.out.println("finish slow work");

    return "index";       // return view's name
  }

返回索引但不提供方法的日志信息