问题描述
@Tenelope您不需要返回，"WEB-INF/views/home.jsp";您可以返回"home"它，因为您已经在servlet-
context中对其进行了配置：
<beans:bean class="org.springframework.web.servlet.view.InternalResourceViewResolver">
<beans:property name="prefix" value="/WEB-INF/views/" />
<beans:property name="suffix" value=".jsp" />
</beans:bean>
我对您的建议是下载spring源（类似于IDE的Eclipse）：
http://www.springsource.org/springsource-tool-suite- download
下载后，您只需转到 File > New > Spring Template Project > Spring MVC project
创建完项目后，将为您创建工作模板项目（结构化项目，示例控制器文件，示例视图，pom.xml等）。包括"/"您正在谈论的示例。您可以构建它并将其部署到tomcat或其他任何东西。
这样，您已经获得了有效的示例，您可以从那里获取它。
顺便说一句。看一下这个 ：
https://github.com/SpringSource/spring-mvc- showcase/blob/master/MasteringSpringMVC3.pdf?raw=true
这足以使您开始使用Spring MVC。
解决方法
我刚刚开始学习Spring MVC 3，开始运行简单的hello world程序时遇到了问题
我试图让这个启动和运行，但我不能因为某些原因
我不断收到这个错误
type Status report
message /
description The requested resource (/) is not available.
这是我的servlet-context.xml档案
<?xml version="1.0" encoding="UTF-8"?>
<beans:beans xmlns="http://www.springframework.org/schema/mvc"
xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
xmlns:beans="http://www.springframework.org/schema/beans"
xmlns:context="http://www.springframework.org/schema/context"
xsi:schemaLocation="http://www.springframework.org/schema/mvc http://www.springframework.org/schema/mvc/spring-mvc-3.0.xsd
http://www.springframework.org/schema/beans http://www.springframework.org/schema/beans/spring-beans-3.0.xsd
http://www.springframework.org/schema/context http://www.springframework.org/schema/context/spring-context-3.0.xsd">
<!-- DispatcherServlet Context: defines this servlet's request-processing infrastructure -->
<!-- Enables the Spring MVC @Controller programming model -->
<annotation-driven />
<!-- Handles HTTP GET requests for /resources/** by efficiently serving up static resources in the ${webappRoot}/resources directory -->
<resources mapping="/resources/**" location="/resources/" />
<!-- Resolves views selected for rendering by @Controllers to .jsp resources in the /WEB-INF/views directory -->
<beans:bean class="org.springframework.web.servlet.view.InternalResourceViewResolver">
<beans:property name="prefix" value="/WEB-INF/views/" />
<beans:property name="suffix" value=".jsp" />
</beans:bean>
<context:component-scan base-package="com.test.test" />
</beans:beans>
web.xml 文件
<?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>
<servlet-mapping>
<servlet-name>appServlet</servlet-name>
<url-pattern>/</url-pattern>
</servlet-mapping>
</web-app>
HomeController.java
打包com.test.test;
import org.springframework.stereotype.Controller;
import org.springframework.ui.Model;
import org.springframework.web.bind.annotation.RequestMapping;
import org.springframework.web.bind.annotation.RequestMethod;
@Controller
public class HomeController {
@RequestMapping(value = "/")
public String home() {
return "WEB-INF/views/home.jsp";
}
}
我正在使用Tomcat 7。
有人知道这个错误是什么吗？