我使用spring mvc 3.2.4和jquery 1.9.0进行长轮询.我的应用程序部署在Tomcat 7.0.42上.我的spring配置文件如下:
应用程序Web.xml
<web-app 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"
version="3.0">
<welcome-file-list>
<welcome-file>index.html</welcome-file>
</welcome-file-list>
<servlet>
<servlet-name>app</servlet-name>
<servlet-class>
org.springframework.web.servlet.DispatcherServlet
</servlet-class>
<load-on-startup>1</load-on-startup>
<async-supported>true</async-supported>
</servlet>
<servlet-mapping>
<servlet-name>app</servlet-name>
<url-pattern>/</url-pattern>
</servlet-mapping>
<listener>
<listener-class>
org.springframework.web.context.ContextLoaderListener
</listener-class>
</listener>
<context-param>
<param-name>contextConfigLocation</param-name>
<param-value>
/WEB-INF/app-servlet.xml
</param-value>
</context-param>
</web-app>
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Spring Configration xml为: -
<?xml version="1.0" encoding="UTF-8"?>
<beans xmlns="http://www.springframework.org/schema/beans"
xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
xmlns:p="http://www.springframework.org/schema/p"
xmlns:mvc="http://www.springframework.org/schema/mvc"
xmlns:context="http://www.springframework.org/schema/context"
xsi:schemaLocation="http://www.springframework.org/schema/beans
http://www.springframework.org/schema/beans/spring-beans-3.2.xsd
http://www.springframework.org/schema/mvc http://www.springframework.org/schema/mvc/spring-mvc-3.2.xsd
http://www.springframework.org/schema/context
http://www.springframework.org/schema/context/spring-context-3.2.xsd">
<context:annotation-config/>
<mvc:annotation-driven/>
<mvc:default-servlet-handler/>
<context:component-scan base-package="com.webchat"/>
<bean id="defferedResult" class="com.exp.DeferredResultContainer"></bean>
</beans>
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发布数据的控制器看起来像
@RequestMapping(value = "/postComment", method = RequestMethod.POST)
public @ResponseBody String postComment(HttpServletRequest request) {
deferredResultContainer.updateAllResults(request.getParameter("comment")); …Run Code Online (Sandbox Code Playgroud)