应用程序应该只允许ROLE_ADMIN用户访问/秘密地址，但它不允许任何人（授权或非授权的）访问，另一个问题是，即使我输入了错误的用户名和密码，它也允许登录并且不会重定向到error.page。
Web.xml中
<?xml version="1.0" encoding="UTF-8"?>
<web-app version="3.0" 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_3_0.xsd">
<context-param>
<param-name>org.apache.tiles.impl.BasicTilesContainer.DEFINITIONS_CONFIG</param-name>
<param-value>/WEB-INF/tiles.xml</param-value>
</context-param>
<listener>
<listener-class>org.apache.struts2.tiles.StrutsTilesListener</listener-class>
</listener>
<listener>
<listener-class>org.springframework.web.context.ContextLoaderListener</listener-class>
</listener>
<context-param>
<param-name>contextConfigLocation</param-name>
<param-value>
/WEB-INF/applicationContext.xml
/WEB-INF/medics-security.xml
/WEB-INF/login-service.xml
</param-value>
</context-param>
<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>
<filter>
<filter-name>struts2</filter-name>
<filter-class>org.apache.struts2.dispatcher.ng.filter.StrutsPrepareAndExecuteFilter</filter-class>
</filter>
<filter-mapping>
<filter-name>struts2</filter-name>
<url-pattern>/*</url-pattern>
</filter-mapping>
<session-config>
<session-timeout>
30
</session-timeout>
</session-config>
<welcome-file-list>
<welcome-file>index.jsp</welcome-file>
</welcome-file-list>
</web-app>
医务人员-security.xml文件
<?xml version="1.0" encoding="UTF-8"?>
<beans:beans xmlns='http://www.springframework.org/schema/security'
xmlns:beans='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-3.0.xsd
http://www.springframework.org/schema/security
http://www.springframework.org/schema/security/spring-security-3.0.xsd'>
<beans:import resource='login-service.xml'/>
<http auto-config="true" access-denied-page="/error.jsp">
<intercept-url pattern="/register*" access="ROLE_ADMIN" />
<intercept-url pattern="/secret*" access="ROLE_ADMIN" />
<form-login login-page="/login.html" authentication-failure-url="/login?error=true"/>
<remember-me/>
<logout/>
</http>
<authentication-manager>
<authentication-provider>
<user-service>
<user name="admin" password="secret" authorities="ROLE_ADMIN"/>
<user name="user" password="secret" authorities="ROLE_USER"/>
</user-service>
</authentication-provider>
</authentication-manager>
</beans:beans>
的applicationContext.xml
<?xml version='1.0' encoding='UTF-8'?>
<beans xmlns='http://www.springframework.org/schema/beans'
xmlns:context='http://www.springframework.org/schema/context'
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-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='com.myproject'/>
<bean id='internalResourceResolver'
class='org.springframework.web.servlet.view.InternalResourceViewResolver'>
<property name='prefix' value='/Web Pages/'/>
<property name='suffix' value='.jsp'/>
</bean>
<bean
class='org.springframework.web.servlet.mvc.annotation.DefaultAnnotationHandlerMapping'/>
<bean class='org.springframework.web.servlet.mvc.annotation.AnnotationMethodHandlerAdapter'/>
<bean id='placeholderConfig'
class='org.springframework.beans.factory.config.PropertyPlaceholderConfigurer'/>
<bean id="viewResolver"
class="org.springframework.web.servlet.view.UrlBasedViewResolver">
<property name="viewClass">
<value>
org.springframework.web.servlet.view.tiles2.TilesView
</value>
</property>
</bean>
<bean id="tilesConfigurer"
class="org.springframework.web.servlet.view.tiles2.TilesConfigurer">
<property name="definitions">
<list>
<value>/WEB-INF/tiles.xml</value>
</list>
</property>
</bean>
</beans>
j_spring_security_check.java
public class j_spring_security_check {
public String execute(){
System.out.append("here in spring check.java");
return "SUCCESS";
}
}
struts.xml中
<!DOCTYPE struts PUBLIC
"-//Apache Software Foundation//DTD Struts Configuration 2.0//EN"
"http://struts.apache.org/dtds/struts-2.0.dtd">
<struts>
<!-- Configuration for the default package. -->
<constant name="struts.action.extension" value="html"/>
<constant name="struts.enable.SlashesInActionNames" value="true"/>
<package name="default" namespace="/" extends="struts-default">
<result-types>
<result-type name="tiles" class="org.apache.struts2.views.tiles.TilesResult"/>
</result-types>
<action name="*">
<result type="tiles">{1}</result>
</action>
<action name="j_spring_security_check"
class="com.myproject.struts.j_spring_security_check">
<result name="SUCCESS" type="tiles">register</result>
</action>
</package>
</struts>
register.jsp
<%@page contentType="text/html" pageEncoding="UTF-8"%>
<!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.01 Transitional//EN"
"http://www.w3.org/TR/html4/loose.dtd">
<%@ taglib prefix="s" uri="/struts-tags" %>
<html>
<head>
<meta http-equiv="Content-Type" content="text/html; charset=UTF-8">
<title>secret page</title>
</head>
<body>
<p>register</p>
<a href="secret.jsp">secret</a>
</body>
</html>
答案 0 :(得分：1)
尝试删除此代码：
<action name="j_spring_security_check"
class="com.myproject.struts.j_spring_security_check">
<result name="SUCCESS" type="tiles">register</result>
</action>
。通常，您不需要处理此URL。 Spring Security具有负责处理凭证的内置过滤器。
尝试更换
<intercept-url pattern="/secret*" access="ROLE_ADMIN" />
模式
<intercept-url pattern="/secret.jsp*" access="ROLE_ADMIN" />
如果您有一个secret.jsp页面。