我要评论环境：
myeclipse+tomcat8
1.在webroot下新建：web.xml:
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<?xml version="1.0" encoding="utf-8"?>
<web-app xmlns:xsi="http://www.w3.org/2001/xmlschema-instance"
xmlns="http://java.sun.com/xml/ns/javaee" xmlns:web="http://java.sun.com/xml/ns/javaee/web-app_2_5.xsd"
xsi:schemalocation="http://java.sun.com/xml/ns/javaee http://java.sun.com/xml/ns/javaee/web-app_3_0.xsd"
version="3.0">
<display-name>springmvc</display-name>
<!-- 部署 dispatcherservlet -->
<servlet>
<servlet-name>springmvc</servlet-name> //servlet名字：springmvc
<servlet-class>org.springframework.web.servlet.dispatcherservlet</servlet-class>
<init-param>
<param-name>contextconfiglocation</param-name>
<param-value>classpath:springmvc-servlet.xml</param-value>
</init-param>
<!-- 表示容器再启动时立即加载servlet -->
<load-on-startup>1</load-on-startup>
</servlet>
<servlet-mapping>
<servlet-name>springmvc</servlet-name>
<!-- 处理所有url -->
<url-pattern>/</url-pattern> url访问根目录映射到webroot下的index.jsp
</servlet-mapping>
</web-app>
2.在webroot下的index.jsp进行编辑：
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<body>
this is my jsp page. <br>
<a href="${pagecontext.request.contextpath}/register}">register</a> //url访问根目录/register 映射到具体的jsp文件 //所以这是两个方面的处理（http与具体页面文件），要进行url映射，具体实现要在具体的servlet-servlet.xml文件进行映射了
<a href="${pagecontext.request.contextpath}/login}">login</a>
</body>
3.接着上文，controller控制器对应着具体的jsp页面，下面进行编辑：
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在src下新建controller 包，在controller中新建logincontroller与registercontroller文件：
具体如下：
registercontroller:
import javax.servlet.http.httpservletrequest;
import javax.servlet.http.httpservletresponse;
import org.springframework.web.servlet.modelandview;
import org.springframework.web.servlet.mvc.controller;
public class registercontroller implements controller {
public modelandview handlerequest(httpservletrequest arg0,
httpservletresponse arg1) throws exception {
return new modelandview("/web-inf/jsp/login.jsp");
}
}
logincontroller:
import javax.servlet.http.httpservletrequest;
import javax.servlet.http.httpservletresponse;
import org.springframework.web.servlet.modelandview;
import org.springframework.web.servlet.mvc.controller;
public class logincontroller implements controller {
public modelandview handlerequest(httpservletrequest arg0,
httpservletresponse arg1) throws exception {
return new modelandview("/web-inf/jsp/register.jsp");
}
4.接下来进行url 与具体页面的映射了：
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在web-inf下新建：
springmvc-servlet.xml: //springmvc在web.xml中定义
<?xml version="1.0" encoding="utf-8"?>
<beans xmlns="http://www.springframework.org/schema/beans"
xmlns:xsi="http://www.w3.org/2001/xmlschema-instance"
xmlns:mvc="http://www.springframework.org/schema/mvc"
xmlns:p="http://www.springframework.org/schema/p"
xmlns:context="http://www.springframework.org/schema/context"
xsi:schemalocation="
http://www.springframework.org/schema/beans
http://www.springframework.org/schema/beans/spring-beans.xsd
http://www.springframework.org/schema/context
http://www.springframework.org/schema/context/spring-context.xsd
http://www.springframework.org/schema/mvc
http://www.springframework.org/schema/mvc/spring-mvc.xsd">
<!-- logincontroller控制器类，映射到"/login" -->
<bean name="/login" class="controller.logincontroller"/> //bean中的/login对应着logincontroller,即url访问根目录/login 对应着 logincontroller的具体页面
<!-- logincontroller控制器类，映射到"/register" -->
<bean name="/register" class="controller.registercontroller"/>
</beans>
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5.发布然后运行：
url访问http://localhost:8080/springmvcdemo01/
应该出现index.jsp页面：
然后点击链接进行页面的跳转了
以上为spring mvc 的初次使用
大神看了勿喷！小弟感激不尽啊！
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