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

@Controller

public class RootController {

@RequestMapping(value = "/", method = RequestMethod.GET)

public String root() {

return "index";

}

}

假设index.jsp/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

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