Question

如何使用此URL http://localhost:8080/myApp/打开index.jsp，如何使用这样的超链接 <a href="/">HOME</a>并转到index.jsp（http://localhost:8080/myApp/）？

这是我的web.xml：

<display-name>myApp</display-name>

<context-param>
    <param-name>contextConfigLocation</param-name>
    <param-value>classpath:spring/application-config.xml</param-value>
</context-param>

<listener>
    <listener-class>org.springframework.web.context.ContextLoaderListener</listener-class>
</listener>

<servlet>
    <servlet-name>myApp</servlet-name>
    <servlet-class>
       org.springframework.web.servlet.DispatcherServlet
    </servlet-class>
    <load-on-startup>1</load-on-startup>
</servlet>

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

这是我的myApp-servlet.xml：

<context:component-scan base-package="org.myApp.com" />

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

提前致谢！

Answer 1

只需使用适当的处理程序方法添加@Controller

@Controller
public class RootController {
    @RequestMapping(value = "/", method = RequestMethod.GET)
    public String root() {
        return "index";
    }
}

假设index.jsp位于/WEB-INF/view。

Spring MVC打开index.jsp on＆＃34; /＆＃34;

1 个答案: