如何在 Web 应用程序中配置 Spring Security?
我在我的库中添加了三个 JAR 文件:security-core、security-web 和 security-config。现在我需要在带有自定义登录页面的配置 XML 文件中添加什么?
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1 回答
10
参考:http ://www.mkyong.com/spring-security/spring-security-form-login-example/
web.xml
<?xml version="1.0" encoding="UTF-8"?>
<web-app id="WebApp_ID" version="2.4"
xmlns="http://java.sun.com/xml/ns/j2ee" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
xsi:schemaLocation="http://java.sun.com/xml/ns/j2ee http://java.sun.com/xml/ns/j2ee/web-app_2_4.xsd">
<display-name>Spring</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>
<init-param>
<param-name>contextConfigLocation</param-name>
<param-value>/WEB-INF/spring/app-config.xml</param-value>
</init-param>
<load-on-startup>1</load-on-startup>
</servlet>
<servlet-mapping>
<servlet-name>Spring</servlet-name>
<url-pattern>/rest/*</url-pattern>
</servlet-mapping>
<listener>
<listener-class>org.springframework.web.context.ContextLoaderListener</listener-class>
</listener>
<context-param>
<param-name>contextConfigLocation</param-name>
<param-value>
/WEB-INF/spring/spring-security.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>
</web-app>
弹簧安全.xml
<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-3.0.3.xsd">
<http auto-config="true">
<intercept-url pattern="/login.jsp" access="ROLE_ANONYMOUS" />
<intercept-url pattern="/**" access="ROLE_USER" />
<form-login login-page="/login.jsp" default-target-url="/" />
<logout logout-url="/logout" logout-success-url="/login.jsp"/>
<intercept-url pattern="/css/**" filters="none"/>
<intercept-url pattern="/js/**" filters="none"/>
</http>
<authentication-manager>
<authentication-provider>
<user-service>
<user name="admin" password="admin" authorities="ROLE_USER" />
</user-service>
</authentication-provider>
</authentication-manager>
</beans:beans>
应用程序配置.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:task="http://www.springframework.org/schema/task"
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
http://www.springframework.org/schema/task http://www.springframework.org/schema/task/spring-task-3.0.xsd
http://www.springframework.org/schema/mvc http://www.springframework.org/schema/mvc/spring-mvc-3.0.xsd">
<mvc:annotation-driven />
<task:annotation-driven/>
<bean id="viewResolver" class= "org.springframework.web.servlet.view.freemarker.FreeMarkerViewResolver">
<property name="cache" value= "true"/>
<property name="prefix" value= ""/>
<property name="suffix" value=".ftl"/>
<property name="contentType" value="text/html;charset=UTF-8"/>
<property name="exposeSpringMacroHelpers" value="true"/>
<property name="requestContextAttribute" value="rc"/>
</bean>
</beans>
于 2012-12-24T10:05:23.597 回答