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Spring 无法访问home.jsp_Spring_Jsp - Fatal编程技术网

Spring 无法访问home.jsp

Spring 无法访问home.jsp,spring,jsp,Spring,Jsp,我有一个应用程序。由于一些问题,在我的web.xml中,我不得不将url模式从“/”更改为“*.action”,并添加了一个欢迎文件列表。这是我的web.xml: <?xml version="1.0" encoding="UTF-8"?> <web-app version="2.5" xmlns="http://java.sun.com/xml/ns/javaee" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instanc

我有一个应用程序。由于一些问题,在我的web.xml中,我不得不将url模式从“/”更改为“*.action”,并添加了一个欢迎文件列表。这是我的web.xml:

<?xml version="1.0" encoding="UTF-8"?>
<web-app version="2.5" 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 
    http://java.sun.com/xml/ns/javaee/web-app_2_5.xsd">

    <!-- The definition of the Root Spring Container shared by all Servlets and Filters -->
    <context-param>
        <param-name>contextConfigLocation</param-name>
        <param-value>/WEB-INF/spring/root-context.xml</param-value>
    </context-param>

    <!-- 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>
         <welcome-file-list>
        <welcome-file>home.jsp</welcome-file>
    </welcome-file-list>
    <servlet-mapping>
        <servlet-name>appServlet</servlet-name>
        <url-pattern>*.action</url-pattern>
    </servlet-mapping>

</web-app>

您的请求应以“.action”结尾,并尝试发送localhost:8080/myapp/some.action
并将控制器中的映射更改为“/some.action”

我尝试通过键入localhost:8080/myapp/home.action来访问它,但它不起作用将控制器中的映射更改为“/some.action”是的,我忘了更改控制器映射。现在开始工作了。谢谢。您的视图名称和请求url无关
@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 = "/")
    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 "home";
    }

}