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Spring security用户权限数据库配置

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  关于安全性服务,有两个重要的概念需要理解,一是认证,即判断用户是否在应用程序中存在,二是授权,即判断用户是否有权限访问资源。  

 

  Spring security为我们提供的HTTP Basic,Digest等认证服务,因为要在配置文件中对用户和权限进行配置,所以局限性很强。除此之外,它还允许我们在数据库中对用户和权限进行配置。那么如何实现呢?

 

  在系统中定义用户,角色,权限这三种实体,一个用户可以拥有多个角色,一个角色可以被多个用户拥有,所以用户与角色之间是多对多的关系,为了易于理解,这里加入第三种实体权限,作为用户和角色的中间关联实体,把用户与角色间的多对多关系拆为两个一对多的关联关系。这样一个用户就对应着多个权限,一个权限对应着一个用户,而一个角色对应着多个权限,一个权限对应着一个角色。

 

  一。User.java

 

package com.template.security;

import java.util.List;

/**
 * Created by IntelliJ IDEA.
 * User: Zhong Gang
 * Date: 11-7-28
 * Time: 下午9:46
 * To change this template use File | Settings | File Templates.
 */
public class User {
    private Integer id;
    private String username;
    private String password;
    private Boolean enabled;

    private List<Authority> authorities;

    public User() {
    }
}

 

 

  二。User.hbm.xml

 

<?xml version="1.0"?>

<!DOCTYPE hibernate-mapping PUBLIC "-//Hibernate/Hibernate Mapping DTD 3.0//EN"
        "http://hibernate.sourceforge.net/hibernate-mapping-3.0.dtd">

<hibernate-mapping default-access="field">

    <class name="com.template.security.User" table="user" dynamic-insert="true" dynamic-update="true">
        <id name="id" column="id" type="java.lang.Integer">
            <generator class="native"/>
        </id>
        <property name="username" column="username" type="java.lang.String"/>
        <property name="password" column="password" type="java.lang.String"/>
        <property name="enabled" column="enabled" type="java.lang.Boolean"/>

        <bag name="authorities">
            <key column="userid"/>
            <one-to-many class="com.template.security.Authority"/>
        </bag>
    </class>

</hibernate-mapping>

 

 

  三。Role.java

 

package com.template.security;

/**
 * Created by IntelliJ IDEA.
 * User: Zhong Gang
 * Date: 11-7-28
 * Time: 下午9:47
 * To change this template use File | Settings | File Templates.
 */
public class Role {
    private Integer id;
    private String name;

    public Role() {
    }
}

 

 

  四。Role.hbm.xml

 

<?xml version="1.0"?>

<!DOCTYPE hibernate-mapping PUBLIC "-//Hibernate/Hibernate Mapping DTD 3.0//EN"
        "http://hibernate.sourceforge.net/hibernate-mapping-3.0.dtd">

<hibernate-mapping default-access="field">

    <class name="com.template.security.Role" table="role" dynamic-insert="true" dynamic-update="true">
        <id name="id" column="id" type="java.lang.Integer">
            <generator class="native"/>
        </id>
        <property name="name" column="name" type="java.lang.String"/>
    </class>

</hibernate-mapping>

 

 

  五。Authority.java

 

package com.template.security;

/**
 * Created by IntelliJ IDEA.
 * User: Zhong Gang
 * Date: 11-7-28
 * Time: 下午9:47
 * To change this template use File | Settings | File Templates.
 */
public class Authority {
    private Integer id;
    private User user;
    private Role role;

    public Authority() {
    }
}

 

 

  六。Authority.hbm.xml

 

<?xml version="1.0"?>

<!DOCTYPE hibernate-mapping PUBLIC "-//Hibernate/Hibernate Mapping DTD 3.0//EN"
        "http://hibernate.sourceforge.net/hibernate-mapping-3.0.dtd">

<hibernate-mapping default-access="field">

    <class name="com.template.security.Authority" table="authority" dynamic-insert="true" dynamic-update="true">
        <id name="id" column="id" type="java.lang.Integer">
            <generator class="native"/>
        </id>
        <many-to-one name="user" column="userid" class="com.template.security.User"/>
        <many-to-one name="role" column="roleid" class="com.template.security.Role"/>
    </class>

</hibernate-mapping>

 

 

  七。security.xml

 

<?xml version="1.0" encoding="UTF-8"?>  
  
<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.xsd">  

    <beans:import resource="datasource.xml"/>

    <http use-expressions="true">  
        <intercept-url pattern="/**" access="hasRole('ROLE_ADMIN')" requires-channel="http"/>
        
        <form-login/>
        <remember-me/>
    </http>  
    
    <authentication-manager>  
        <authentication-provider>  
            <jdbc-user-service data-source-ref="dataSource"
              users-by-username-query="select username,password,enabled as status from user where username=?"
              authorities-by-username-query="select u.username,r.name as authority from user u join authority a on a.userid=u.id join role r on r.id=a.roleid where u.username=?"/>
        </authentication-provider>  
    </authentication-manager>  
  
</beans:beans>

  这里需要重点了解这样几个元素。第一个就是jdbc-user-service,这是用来配置利用数据库进行用户权限认证服务的,data-source-ref则是用来说明使用的数据库连接等相关属性,users-by-username-query则是认证用户使用的,

authorities-by-username-query则是获取用户授权的,intercept-url可以用来定义要保护的访问路径,当然如果配置文件中有多个intercept-url,其中有些是重复的,那么Spring会以位置最靠上的作为标准来判断,access则定义了哪些角色可以访问,还有其它一些属性以后再做进一步理解,现在这些已经够用了。只要做到认证和授权,那么我们的安全框架也就基本搭建完成了。

 

  八。data.ddl

 

INSERT INTO user(id,username,password,enabled) VALUES (1,"ZhongGang","ZhongGang",true);
INSERT INTO user(id,username,password,enabled) VALUES (2,"GangGang","GangGang",true);
INSERT INTO user(id,username,password,enabled) VALUES (3,"ZhongZhong","ZhongGang",false);

INSERT INTO role(id,name) VALUES (1,"ROLE_ADMIN");
INSERT INTO role(id,name) VALUES (2,"ROLE_SUPERUSER");

INSERT INTO authority(id,userid,roleid) VALUES (1,1,1);
INSERT INTO authority(id,userid,roleid) VALUES (2,2,2);
INSERT INTO authority(id,userid,roleid) VALUES (3,3,1);

  这是我使用的模拟数据。   

 

   当然这里需要在web.xml里面配置springSecurityFilterChain,可以参见我的另一篇博文Spring security HTTP Basic。

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2 楼 Crazy_Robin 2014-01-19  
楼主的文章几乎每篇都是精华啊!真的值得好好研究一下!
1 楼 _s45 2012-03-07  
按照楼主的方法做了,实验成功,非常感谢

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