我刚刚在spring工具套装中创建了一个示例spring mvc项目(没有任何修改),当我执行它时,我按照index.jsp获取输出。但之后我只在视图文件夹中添加了一个index.html页面并进行了两次更改
在控制器类中,我添加了:return' index';
<beans:property name="prefix" value="/WEB-INF/views/" />
<beans:property name="suffix" value=".html" />
在Pivotal Server Developer Edition v3.1上更改后执行代码时,我收到HTTP状态404(请求的资源不可用错误。)
据我所知,不需要进行其他更改。 请提供一些建议。 提前致谢
HomeController.java
@Controller
public class HomeController {
private static final Logger logger = LoggerFactory.getLogger(HomeController.class);
/**
* Simply selects the home view to render by returning its name.
*/
@RequestMapping(value = "/", method = RequestMethod.GET)
public String home(Locale locale, Model model) {
logger.info("Welcome home! The client locale is {}.", locale);
Date date = new Date();
DateFormat dateFormat = DateFormat.getDateTimeInstance(DateFormat.LONG, DateFormat.LONG, locale);
String formattedDate = dateFormat.format(date);
model.addAttribute("serverTime", formattedDate );
return "index";
}
}
servlet的context.xml中
<!-- DispatcherServlet Context: defines this servlet's request-processing infrastructure -->
<!-- Enables the Spring MVC @Controller programming model -->
<annotation-driven />
<!-- Handles HTTP GET requests for /resources/** by efficiently serving up static resources in the ${webappRoot}/resources directory -->
<resources mapping="/resources/**" location="/resources/" />
<!-- Resolves views selected for rendering by @Controllers to .jsp resources in the /WEB-INF/views directory -->
<beans:bean class="org.springframework.web.servlet.view.InternalResourceViewResolver">
<beans:property name="prefix" value="/WEB-INF/views/" />
<beans:property name="suffix" value=".html" />
</beans:bean>
<context:component-scan base-package="com.srio.tata" />
的web.xml
contextConfigLocation的 /WEB-INF/spring/root-context.xml
<!-- Creates the Spring Container shared by all Servlets and Filters -->
<listener>
<listener-class>org.springframework.web.context.ContextLoaderListener</listener-class>
</listener>
<!-- Processes application requests -->
<servlet>
<servlet-name>appServlet</servlet-name>
<servlet-class>org.springframework.web.servlet.DispatcherServlet</servlet-class>
<init-param>
<param-name>contextConfigLocation</param-name>
<param-value>/WEB-INF/spring/appServlet/servlet-context.xml</param-value>
</init-param>
<load-on-startup>1</load-on-startup>
</servlet>
<servlet-mapping>
<servlet-name>appServlet</servlet-name>
<url-pattern>/</url-pattern>
</servlet-mapping>
答案 0 :(得分:1)
您是否配置了viewResolver
<bean id="jspViewResolver" class="org.springframework.web.servlet.view.InternalResourceViewResolver">
<property name="prefix" value="/WEB-INF/pages/"/>
<property name="suffix" value=".jsp"/>
<property name="viewClass" value="org.springframework.web.servlet.view.JstlView"/>
</bean>
答案 1 :(得分:0)
由于.html
文件是静态并且不需要servlet处理,因此使用<mvc:resources/>
mapping会更有效,也更简单。这需要Spring 3.0.4 +。
例如:
<mvc:resources mapping="/static/**" location="/static/" />
将通过所有以/static/
开头的请求传递到webapp/static/
目录。
因此,将index.html
放入webapp/static/
并使用方法中的return "static/index.html";
,Spring应该会找到该视图。
答案 2 :(得分:0)
在web.xml中更改来自
的网址格式<url-pattern>/</url-pattern>
到
<url-pattern>/*.html</url-pattern>