Spring 3 安全 j_spring_security_check

Question

我正在嘗試了解 spring 安全性是如何工作的，所以我下載了一些示例項目，然后我嘗試在我的項目中實施該解決方案。 但是當我嘗試登錄時，我收到404錯誤，並且在地址欄中我有http://localhost:8080/fit/j_spring_security_check 。 我試圖在這里查看類似的問題，但我無法意識到如何將其應用於我的項目。 我真的很感激，如果有人更有經驗，可以幫助我。

我的應用程序結構如下所示：

應用上下文.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>

applicationContext-web.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: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>

applicationContext-security.xml：

<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

您正在嘗試根據網頁的當前上下文路徑驗證 uri。 JSTL 標記庫可用於確保您根據應用程序的上下文輕松生成正確的 url。 如果你想快速實現它，你可以通過使用標簽庫來做到這一點。 為此，您可以將 jstl 標記庫添加到 jsp 的頂部：

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

然后您可以使用以下內容發布到登錄 servlet。

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

這確保您始終發布到 <your_application_context>/j_spring_security_check。

jstl 參考： 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>

添加您的 web.xml 文件。它是您的 springSecurityFilterChain 的創建 Bean。然后您得到了響應

Answer 3

路徑/j_spring_security_check在 spring 4 已更改為/login並在UsernamePasswordAuthenticationFilter處理

你可以在這里找到它的來源： https : //github.com/spring-projects/spring-security/blob/master/web/src/main/java/org/springframework/security/web/authentication/UsernamePasswordAuthenticationFilter.java