java - 在/上找不到资源上的 SpringMVC Controller

Question

我正在跟进来自这个 url 的教程

https://aishwaryavaishno.wordpress.com/2013/07/12/spring-mvchibernate-annotation-based-mysql-maven-json-simple-example/

根据教程，我在我的 Controller 中有这个

@Autowired
        private IMusicStoreService musicStoreService;
        @RequestMapping(value="/")
        public ModelAndView test(HttpServletResponse response) throws IOException{

            return new ModelAndView("home","musicDetForm",new MusicDetails());
        }
        @RequestMapping(value="AddSong",method = RequestMethod.POST)
        @ResponseBody
        public String addSong(@ModelAttribute("musicDetForm") MusicDetails musicDetails){
           return musicStoreService.addSong(musicDetails);
        }
        @RequestMapping(value = "SongList/",method = RequestMethod.GET)
        @ResponseBody
        public List<MusicDetails> getSongList(){
            return musicStoreService.getSongList();
        }

我已将我的 DispatcherServlet 和 contextConfigLocation 映射到我的 web.xml 文件

<context-param>
        <param-name>contextClass</param-name>
        <param-value>
            org.springframework.web.context.support.AnnotationConfigWebApplicationContext
        </param-value>
    </context-param>
    <listener>
        <listener-class>org.springframework.web.context.ContextLoaderListener</listener-class>
    </listener>

    <servlet>
        <servlet-name>SpringDispatcher</servlet-name>
        <servlet-class>org.springframework.web.servlet.DispatcherServlet</servlet-class>

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

    <context-param>
        <param-name>contextConfigLocation</param-name>
        <param-value>/WEB-INF/mvc-dispatcher-servlet.xml</param-value>
    </context-param>

    <session-config>
        <session-timeout>30</session-timeout>
    </session-config>

假设这是我来自 Controller 映射的默认 url

@RequestMapping(value="/")
        public ModelAndView test(HttpServletResponse response) throws IOException{

            return new ModelAndView("home","musicDetForm",new MusicDetails());
        }

现在当我启动我的应用程序到这个 url

http://localhost:8080/MusicStore/

我的项目结构截图

作为我的 servlet 引擎的 Tomcat 返回 HTTP 状态错误 404。

请问哪里出了问题？

Answer 1

在mvc-dispatcher-servlet.xml中加入以下代码来配置viewResolver

    <bean
        class="org.springframework.web.servlet.view.InternalResourceViewResolver">
        <property name="prefix" value="/WEB-INF/views/" />
        <property name="suffix" value=".jsp" />
   </bean>

然后在/WEB-INF/views/中添加你的jsp文件