2

我是 Java 和 Java EE 的新手。你能告诉我如何在成功登录后从 Active Directore 中检索用户详细信息，如全名、公司、电话、部门、邮件等，所以：

我的 web.xml：

    <?xml version="1.0" encoding="UTF-8"?>
<web-app version="3.1" xmlns="http://xmlns.jcp.org/xml/ns/javaee"     xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xsi:schemaLocation="http://xmlns.jcp.org/xml/ns/javaee http://xmlns.jcp.org/xml/ns/javaee/web-app_3_1.xsd">
<context-param>
    <param-name>contextConfigLocation</param-name>
    <param-value>/WEB-INF/applicationContext.xml
                 /WEB-INF/applicationContext-security.xml
    </param-value>
</context-param>
<listener>
    <listener-class>org.springframework.web.context.ContextLoaderListener</listener-class>
</listener>
<servlet>
    <servlet-name>dispatcher</servlet-name>
    <servlet-class>org.springframework.web.servlet.DispatcherServlet</servlet-class>
    <load-on-startup>2</load-on-startup>
</servlet>
<servlet-mapping>
    <servlet-name>dispatcher</servlet-name>
    <url-pattern>/</url-pattern>
</servlet-mapping>
<session-config>
    <session-timeout>
        30
    </session-timeout>
</session-config>
<welcome-file-list>
    <welcome-file>redirect.jsp</welcome-file>
</welcome-file-list>
<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>

我的 applicationContextsecurity.xml：

<?xml version="1.0" encoding="UTF-8"?>
<beans:beans xmlns:security="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-3.2.xsd http://www.springframework.org/schema/security http://www.springframework.org/schema/security/spring-security-3.1.xsd">

<!-- LDAP server details --> 
<security:authentication-manager>
    <security:authentication-provider ref="ldapActiveDirectoryAuthProvider" />
</security:authentication-manager>

<beans:bean id="grantedAuthoritiesMapper" class="org.mops.security.ActiveDirectoryGrantedAuthoritiesMapper"/>

<beans:bean id="ldapActiveDirectoryAuthProvider" class="org.springframework.security.ldap.authentication.ad.ActiveDirectoryLdapAuthenticationProvider">
    <beans:constructor-arg value="xxx.local" />
    <beans:constructor-arg value="ldap://xxx.local:389/" /> 
    <beans:property name="authoritiesMapper" ref="grantedAuthoritiesMapper" />
    <beans:property name="useAuthenticationRequestCredentials" value="true" />
    <beans:property name="convertSubErrorCodesToExceptions" value="true" />
</beans:bean>

<security:http auto-config="true" pattern="/**">
    <!-- Login pages -->
    <security:form-login login-page="/" default-target-url="/user/" 
        login-processing-url="/j_spring_security_check" authentication-failure-url="/?error=true" />
    <security:logout logout-success-url="/"/>

    <!-- Security zones -->
    <!--<security:intercept-url pattern="/it/**" access="ROLE_ADMIN" />
    <security:intercept-url pattern="/user/**" access="ROLE_ADMINISTRATION" /> -->
</security:http>

我可以正常登录，我可以在控制器中获取用户名：

用户控制器.java：

import org.springframework.security.core.context.SecurityContextHolder;
import org.springframework.security.core.userdetails.UserDetails;
import org.springframework.stereotype.Controller;
import org.springframework.ui.Model;
import org.springframework.web.bind.annotation.RequestMapping;


@Controller
public class UserController{
private String username;
@RequestMapping("/user")
public String User(Model model) {

    Object principal = SecurityContextHolder.getContext().getAuthentication().getPrincipal();

    if (principal instanceof UserDetails) {
      this.username = ((UserDetails)principal).getUsername();
    } else {
      this.username = principal.toString();
    }
    model.addAttribute("message", username);

    return "user";
 }


}

现在我考虑创建类，它存储所有用户详细信息并在我需要使用用户详细信息时创建此类的实例。

任何人都可以一步一步告诉我如何做到这一点？

4

1 回答 1

4

两周后，我找到了解决方案。

在pring-security.xml 中：

<beans:bean id="ldapActiveDirectoryAuthProvider" class="org.springframework.security.ldap.authentication.ad.ActiveDirectoryLdapAuthenticationProvider" >
    <beans:constructor-arg value="xxx.yyy" />
    <beans:constructor-arg value="ldap://zzz.xxx.yyy:389/" /> 
    <beans:property name="authoritiesMapper" ref="grantedAuthoritiesMapper" />
    <beans:property name="useAuthenticationRequestCredentials" value="true" />
    <beans:property name="convertSubErrorCodesToExceptions" value="true" />
    <beans:property name="userDetailsContextMapper">
    <beans:bean class="org.springframework.security.ldap.userdetails.InetOrgPersonContextMapper" />
</beans:property>
</beans:bean>

在控制器中，我们将 Principal 转换为 InetOrgPerson：

import javax.naming.NamingException;
import org.springframework.security.core.context.SecurityContextHolder;
import org.springframework.security.core.userdetails.UserDetails;
import org.springframework.stereotype.Controller;
import org.springframework.ui.Model;
import org.springframework.web.bind.annotation.RequestMapping;
import org.springframework.web.bind.annotation.RequestMethod;
import org.springframework.security.ldap.userdetails.InetOrgPerson;


@Controller
public class UserController  {
private String username;


@RequestMapping(value="/user", method = RequestMethod.GET)
public String User(Model model) throws NamingException {
Object principal = SecurityContextHolder.getContext().getAuthentication().getPrincipal();


    if (principal instanceof UserDetails) {
      this.username = ((UserDetails)principal).getUsername();

    } else {
      this.username = principal.toString();

    }
        model.addAttribute("username", username);        
        model.addAttribute("roomNumber", ((InetOrgPerson) principal).getRoomNumber());
    return "user";
}

}
于 2013-11-06T13:13:12.577 回答