相关疑难解决方法(0)

我已经开始学习Spring MVC阅读本教程:http://viralpatel.net/blogs/spring-3-mvc-create-hello-world-application-spring-3-mvc/

好的,这对我来说非常清楚.

在这个例子中,我有web.xml文件来配置我的Web应用程序:

<?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_2_5.xsd"
id="WebApp_ID" version="2.5">

<display-name>Spring3MVC</display-name>
<welcome-file-list>
    <welcome-file>index.jsp</welcome-file>
</welcome-file-list>

<servlet>
    <servlet-name>spring</servlet-name>
    <servlet-class>
        org.springframework.web.servlet.DispatcherServlet
    </servlet-class>
    <load-on-startup>1</load-on-startup>
</servlet>
<servlet-mapping>
    <servlet-name>spring</servlet-name>
    <url-pattern>*.html</url-pattern>
</servlet-mapping>

以及用于配置mu DispatcherServlet 的spring-servlet.xml文件:

<?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: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-3.0.xsd
http://www.springframework.org/schema/context
http://www.springframework.org/schema/context/spring-context-3.0.xsd">

<context:component-scan
    base-package="net.viralpatel.spring3.controller" />

<bean id="viewResolver"
    class="org.springframework.web.servlet.view.UrlBasedViewResolver">
    <property name="viewClass"
        value="org.springframework.web.servlet.view.JstlView" />
    <property name="prefix" value="/WEB-INF/jsp/" />
    <property name="suffix" value=".jsp" />
</bean>

而且,正如你在previus链接中看到的那样,我只有一个控制器类来处理"/ hello"URL的HTTP请求......好吧......这对我来说很明显......

在这个例子旁边,我通过STS\Eclipse中的相​​关模板项目创建了一个新的Spring MVC项目.

此示例与te previus非常相似:我总是使用web.xml文件来配置我的Web应用程序,配置DispatcherServlet的文件和处理HTTP请求的控制器类.

但是我有些不同,我无法理解.

这是我的 …

如何在方法之上将请求映射到没有显式注释的方法？例如,以下请求:

http://somedomain:8080/SampleSpring/access/loginFailed

应该映射到

"AccessController"的"loginFailed"方法,不需要对方法的显式注释,如:

@RequestMapping("/access/loginFailed")

这是我的弹簧配置:

<context:component-scan base-package="com.robikcodes.samplespring"/>
<mvc:annotation-driven/>

<bean class="org.springframework.web.servlet.mvc.support.ControllerClassNameHandlerMapping">
        <property name="basePackage" value="com.robikcodes.samplespring.controller"/>
        <property name="caseSensitive" value="true"/>
        <property name="defaultHandler">
            <bean class="org.springframework.web.servlet.mvc.UrlFilenameViewController"/>
        </property>
</bean>   

<bean class="org.springframework.web.servlet.view.InternalResourceViewResolver">
            <property name="prefix" value="/WEB-INF/views/"/>
            <property name="suffix" value=".jsp"/>
</bean>

这是我的控制器:

@Controller
public class AccessController{

    @RequestMapping(method = RequestMethod.GET)
    public void login(ModelMap m) {}

    @RequestMapping(method = RequestMethod.GET)
    public String loginFailed(ModelMap m) {
        m.addAttribute("error", "true");
        return "access/login";
    }

    @RequestMapping(method = RequestMethod.GET)
    public String logout(ModelMap m) {
        m.addAttribute("logoutStatus","true");
        return "access/login";
    }
}

我收到以下错误(似乎只有登录方法被正确映射):

org.apache.catalina.LifecycleException: org.springframework.beans.factory.BeanCreationException: Error creating bean with name 'org.springframework.web.servlet.mvc.method.annotation.RequestMappingHandlerMapping#0': …