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我是 Spring Security 的新手，当我尝试访问作为 jsp 页面的登录页面时，我得到了一个 http 404。我在用

  1. 雄猫 7
  2. Spring框架3.1.1

这是tomcat Web.xml

<?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>
<context-param>
    <param-name>log4jConfigLocation</param-name>
    <param-value>classpath:log4j.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>
<!-- Initializes log4j -->
<listener>
    <listener-class>org.springframework.web.util.Log4jConfigListener</listener-class>
</listener>

<filter>
    <filter-name>springSecurityFilterChain</filter-name>
    <filter-class>org.springframework.web.filter.DelegatingFilterProxy</filter-class>
</filter>

<filter-mapping>
    <filter-name>springSecurityFilterChain</filter-name>
    <url-pattern>/*</url-pattern>
</filter-mapping>
<!-- Processes application requests -->
<servlet>
    <servlet-name>formVilleServlet</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>formVilleServlet</servlet-name>
    <url-pattern>/</url-pattern>
</servlet-mapping>

这是spring root-context.xml

<?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"
xmlns:mvc="http://www.springframework.org/schema/mvc"
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">

<mvc:resources location="/resources/**" mapping="/resources/"/>

<!-- 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.ResourceBundleViewResolver">
   <beans:property name="basename" value="views"></beans:property>
</beans:bean>
<context:component-scan base-package="com.xtremesoftwaresolutions.formville" />

4

1 回答 1

3

当您执行以下操作时，您映射 Spring MVC Servlet 以处理所有请求：

<url-pattern>/</url-pattern>

因此，当您尝试访问您的 JSP 页面时，例如/login.jspSpring MVC 捕获它并尝试找到合适的控制器和操作来管理它。因此，可能最简单的方法是创建只返回 JSP 页面的控制器。我建议尝试这样的事情：

<mvc:view-controller path="/login.jsp" view-name="/login.jsp" />
于 2012-08-19T21:38:27.520 回答