基于权限管理时的用户角色权限处理。
实体类:
1.User
public class User{
private Integer uid;
private String username;
private String password;
private Set roles=new HashSet<>();
2.Role
public class Role {
private Integer rid;
private String rname;
private Set modules=new HashSet<>();
3.Module
Mapper 文件。
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
<resultMap id="userMap2" type="com.example.mapper.entity.User">
<id property="uid" column="uid"/>
<result property="username" column="username"/>
<result property="password" column="password"/>
<collection property="roles" ofType="roleMap" column="uid" select="findRole"></collection>
</resultMap>
<resultMap id="roleMap" type="com.example.mapper.entity.Role">
<id property="rid" column="rid"/>
<result property="rname" column="rname"/>
<collection property="modules" ofType="com.example.mapper.entity.Module" column="rid" select="findModule">
</collection>
</resultMap>
<select id="findUserByUserName" parameterType="java.lang.String" resultMap="userMap2">
SELECT * from USER WHERE username=#{username}
</select>
<select id="findRole" parameterType="java.lang.Integer" resultMap="roleMap">
SELECT r.* from Role r LEFT JOIN user_role ur on ur.rid =r.rid where ur.uid=#{uid}
</select>
<select id="findModule" parameterType="java.lang.Integer" resultType="com.example.mapper.entity.Module">
SELECT m.* from module m LEFT JOIN module_role mr on mr.mid =m.mid where mr.rid=#{rid}
</select>
测试结果。
User{uid=1, username='hlhdidi', password='123', roles=[Role{rid=1, rname='admin',modules=[Module{mid=2, mname='delete'}, Module{mid=3, mname='query'}, Module{mid=4, mname='update'}, Module{mid=1, mname='add'}]}]}
网友评论