HTTP 状态 404 - Spring MVC

Question

我刚刚在 spring tool suit 中创建了一个示例 spring mvc 项目（没有任何修改），当我执行它时，我得到了 index.jsp 的输出。但在那之后，我只在视图文件夹中添加了一个 index.html 页面，然后我做了以下两处更改

1. 在控制器类中我添加了：return 'index';

2.  <beans:property name="prefix" value="/WEB-INF/views/" />
    <beans:property name="suffix" value=".html" />
   

在 Pivotal Server Developer Edition v3.1 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。 在此先感谢

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-context.xml

<!-- 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" /> 

web.xml

上下文配置位置 /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>

Answer 1

在 web.xml 中更改 url 模式来自

<url-pattern>/</url-pattern>

到

<url-pattern>/*.html</url-pattern>

Answer 2

由于.html 文件是静态 并且不需要servlet 处理，因此使用<mvc:resources/> mapping 更有效、更简单。这需要 Spring 3.0.4+。

例如：

<mvc:resources mapping="/static/**" location="/static/" />

这会将所有以/static/ 开头的请求传递到webapp/static/ 目录。

因此，通过将index.html 放入webapp/static/ 并在您的方法中使用return "static/index.html";，Spring 应该会找到视图。

参考这个问题 - https://stackoverflow.com/a/15480301/4531387

Answer 3

你是否配置了 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>