Spring MVC在“ /”上打开index.jsp
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/
)?
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!
回答:
只需添加@Controller
带有适当处理程序方法的
@Controllerpublic class RootController {
@RequestMapping(value = "/", method = RequestMethod.GET)
public String root() {
return "index";
}
}
假设index.jsp
在/WEB-INF/view
。
以上是 Spring MVC在“ /”上打开index.jsp 的全部内容, 来源链接: utcz.com/qa/418382.html