Spring 3安全性j_spring_security_check

Question

我正在努力学习Spring安全性如何工作，所以我下载了一些示例项目，然后我尝试将该解决方案应用到我的项目中。但是当我尝试登录时，我收到404错误，而在地址栏中我有http://localhost:8080/fit/j_spring_security_check。我试着在这里查看类似的问题，但我无法意识到，如何将它应用到我的项目中。如果有经验丰富的人可以帮助我，我会非常感激。

我的应用结构如下所示：

的applicationContext.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"
   xmlns:context="http://www.springframework.org/schema/context"
   xmlns:security="http://www.springframework.org/schema/security"
   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
    http://www.springframework.org/schema/security http://www.springframework.org/schema/security/spring-security-3.1.xsd">

<context:annotation-config/>

<context:component-scan base-package="cz.cvut.fit"/>

<import resource="classpath:applicationContext-security.xml"/>

</beans>

的applicationContext-web.xml中：

<?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:context="http://www.springframework.org/schema/context"
   xmlns:security="http://www.springframework.org/schema/security"
   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
    http://www.springframework.org/schema/security http://www.springframework.org/schema/security/spring-security-3.1.xsd">

<context:annotation-config/>

<context:component-scan base-package="cz.cvut.fit" />

<mvc:annotation-driven />

<security:global-method-security jsr250-annotations="enabled"
                                 proxy-target-class="true"/>
</beans>

的applicationContext-security.xml文件：

<beans xmlns:security="http://www.springframework.org/schema/security"
   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-3.0.xsd
             http://www.springframework.org/schema/security
             http://www.springframework.org/schema/security/spring-security-3.1.xsd">

<security:http pattern="/css/**" security="none"/>
<security:http pattern="/views/login.jsp*" security="none"/>
<security:http pattern="/views/denied.jsp" security="none"/>

<security:http auto-config="true" access-denied-page="/denied.jsp" servlet-api-provision="false">
    <security:intercept-url pattern="/views/login.jsp*" access="IS_AUTHENTICATED_ANONYMOUSLY"/>
    <security:intercept-url pattern="/views/edit/**" access="ROLE_EDIT"/>
    <security:intercept-url pattern="/views/admin/**" access="ROLE_ADMIN"/>
    <security:intercept-url pattern="/**" access="ROLE_USER"/>
    <security:form-login login-page="/views/login.jsp" authentication-failure-url="/denied.jsp"
                         default-target-url="/home.jsp"/>
    <security:logout/>
</security:http>

<security:authentication-manager>
    <security:authentication-provider>
        <security:user-service>
            <security:user name="adam" password="adampassword" authorities="ROLE_USER"/>
            <security:user name="jane" password="janepassword" authorities="ROLE_USER, ROLE_ADMIN"/>
            <security:user name="sue" password="suepassword" authorities="ROLE_USER, ROLE_EDIT"/>
        </security:user-service>
    </security:authentication-provider>
</security:authentication-manager>

</beans>

Answer 1

您正在尝试根据网页的当前上下文路径验证uri。 JSTL标记库可用于确保您可以根据应用程序的上下文轻松生成正确的URL。如果要快速实现，可以使用标记库来完成此操作。为此，您可以将jstl标记库添加到jsp的顶部：

<%@ taglib prefix="c" uri="http://java.sun.com/jsp/jstl/core" %>

然后，您可以使用以下内容发布到登录servlet。

<form action="<c:url value="/j_spring_security_check"></c:url>" method="post" role="form">

这可确保您始终发布到＆lt; your_application_context＆gt; / j_spring_security_check。

jstl的参考： http://docs.oracle.com/javaee/5/jstl/1.1/docs/tlddocs/c/url.html

Answer 2

<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>

添加你的web.xml文件。它创建你的springSecurityFilterChain的Bean。然后你得到了响应

Answer 3

路径/j_spring_security_check已在春季4更改为/login，并且已在UsernamePasswordAuthenticationFilter中处理

您可以在此处找到其来源：https://github.com/spring-projects/spring-security/blob/master/web/src/main/java/org/springframework/security/web/authentication/UsernamePasswordAuthenticationFilter.java