如何使用该URL打开index.jsp http://localhost:8080/myApp/,如何使用此类超链接 <a href="/">HOME</a>并转到index.jsp(http://localhost:8080/myApp/)?
http://localhost:8080/myApp/
<a href="/">HOME</a>
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!
只需添加@Controller带有适当处理程序方法的
@Controller
@Controller public class RootController { @RequestMapping(value = "/", method = RequestMethod.GET) public String root() { return "index"; } }
假设index.jsp在/WEB-INF/view。
index.jsp
/WEB-INF/view