我正在尝试实现spring security登录,我尝试过类似的方法:
spring-security.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-3.1.xsd
http://www.springframework.org/schema/security
http://www.springframework.org/schema/security/spring-security-3.1.xsd">
<http auto-config="true" use-expressions="true">
<intercept-url pattern="/login" access="permitAll" />
<intercept-url pattern="/logout" access="permitAll" />
<intercept-url pattern="/accessdenied" access="permitAll" />
<intercept-url pattern="/**" access="hasRole('ROLE_USER')" />
<form-login login-page="/login" default-target-url="/list" authentication-failure-url="/accessdenied" />
<logout logout-success-url="/logout" />
</http>
<authentication-manager alias="authenticationManager">
<authentication-provider>
<user-service>
<user name="lokesh" password="password" authorities="ROLE_USER" />
</user-service>
</authentication-provider>
</authentication-manager>
</beans:beans>
web.xml中:
<servlet>
<servlet-name>spring</servlet-name>
<servlet-class>
org.springframework.web.servlet.DispatcherServlet
</servlet-class>
<load-on-startup>1</load-on-startup>
</servlet>
<servlet-mapping>
<servlet-name>spring</servlet-name>
<url-pattern>*.do</url-pattern>
</servlet-mapping>
<welcome-file-list>
<welcome-file>index.jsp</welcome-file>
</welcome-file-list>
<context-param>
<param-name>contextConfigLocation</param-name>
<param-value>
/WEB-INF/spring-servlet.xml,
/WEB-INF/spring-security.xml
</param-value>
</context-param>
<listener>
<listener-class>org.springframework.web.context.ContextLoaderListener</listener-class>
</listener>
<!-- Spring Security …