Xml 在spring中链接到其他页面

所以，我有一个主页，用Spring完成。此主页有一个链接，该链接应指向另一个带有表单的页面。但第二页没有显示。我肯定我做了很多错事，但请说实话，我正在学习SpringMVC，我有点困惑

代码如下：

root-context.xml：

<?xml version="1.0" encoding="UTF-8"?>
<beans xmlns="http://www.springframework.org/schema/beans"
xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
xsi:schemaLocation="http://www.springframework.org/schema/beans http://www.springframework.org/schema/beans/spring-beans.xsd">

<!-- Root Context: defines shared resources visible to all other web components -->

    <bean name="newuserformcontroller" class="my.package.Controllers.NewUserFormController"/>
    <bean name="simpleUrlMapping" class="org.springframework.web.servlet.handler.SimpleUrlHandlerMapping">
    <property name="mappings">
        <props>
            <prop key="/newuserform">newuserformcontroller</prop>

        </props>

    </property>
    </bean>
</beans>

<?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.xsd
    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">

<!-- 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.development.testing" />



</beans:beans>

servlet-context.xml：

<?xml version="1.0" encoding="UTF-8"?>
<beans xmlns="http://www.springframework.org/schema/beans"
xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
xsi:schemaLocation="http://www.springframework.org/schema/beans http://www.springframework.org/schema/beans/spring-beans.xsd">

<!-- Root Context: defines shared resources visible to all other web components -->

    <bean name="newuserformcontroller" class="my.package.Controllers.NewUserFormController"/>
    <bean name="simpleUrlMapping" class="org.springframework.web.servlet.handler.SimpleUrlHandlerMapping">
    <property name="mappings">
        <props>
            <prop key="/newuserform">newuserformcontroller</prop>

        </props>

    </property>
    </bean>
</beans>

<?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.xsd
    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">

<!-- 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.development.testing" />



</beans:beans>

home.jsp：

<%@ taglib uri="http://java.sun.com/jsp/jstl/core" prefix="c" %>
<%@ page session="false" %>
<html>
<head>
<title>Home</title>
</head>
<body>
<h1>
Hello world!  
</h1>

<P>  The time on the server is ${serverTime}. </P>
<p><a href="<c:url value="/newuserform" />" > New User </a> </p>

</body>
</html>

以及newuserform.jsp：

<%@ page language="java" contentType="text/html; charset=ISO-8859-1"
pageEncoding="ISO-8859-1"%>
<!DOCTYPE html PUBLIC "-//W3C//DTD HTML 4.01 Transitional//EN" "http://www.w3.org/TR/html4/loose.dtd">
<html>
<head>
<meta http-equiv="Content-Type" content="text/html; charset=ISO-8859-1">
<title>New User Form</title>
</head>
<body>

</body>
</html>

当我点击链接时，我收到一个错误404。当用户在单击链接时收到newuserform.jsp时，我如何做到这一点

---

谢谢大家!

我错误地删除了我以前的评论。我看到你投了更高的票。这个解决方案对你有效吗？我没有投票，也没有测试它，我现在就去做如果你是说在home.jsp，将.jsp添加到/newuserform，它不起作用…404错误页面。你的newuserform.jsp文件位于“/WEB-INF/views/newuserform.jsp”吗？太好了。您可以使用“控制器”或“组件”注释来指示类是控制器。如果使用@Controller和component scan，则不需要在xml中指定控制器bean。