1. 包含多个 List(String)属性的情况
数据库数据
mysql> select * from ai_user;
+----+-----------+--------+
| id | user_name | status |
+----+-----------+--------+
| 1 | Answer | 1 |
| 2 | Iris | 1 |
+----+-----------+--------+
mysql> select * from ai_name;
+---------+-----------+
| user_id | user_name |
+---------+-----------+
| 1 | Answer |
| 1 | AI |
| 1 | AAL |
| 2 | Iris |
| 2 | Ellis |
| 2 | Monta |
+---------+-----------+
mysql> select * from ai_role;
+---------+-----------------+
| user_id | role_name |
+---------+-----------------+
| 1 | Admin |
| 1 | Manager |
| 1 | Coder |
| 2 | CustomerService |
+---------+-----------------+
实体类
@Data
public class User {
private Long id;
private List<String> names;
private List<String> roles;
}
Mapper 层
public interface UserMapper {
List<User> queryUsers();
}
Mapper Sql 映射文件
<resultMap id="UserMap" type="User">
<result column="id" property="id" jdbcType="BIGINT" />
<collection property="names" resultMap="NamesMap" />
<collection property="roles" resultMap="RolesMap" />
</resultMap>
<resultMap id="NamesMap" type="string">
<result column="user_name" />
</resultMap>
<resultMap id="RolesMap" type="string">
<result column="role_name" />
</resultMap>
<select id="queryUsers" resultMap="UserMap">
SELECT au.id, an.user_name, ar.role_name
FROM ai_user au
LEFT JOIN ai_name an ON an.user_id = au.id
LEFT JOIN ai_role ar ON ar.user_id = au.id
</select>
结果输出示例
{"id":1,"names":["Answer","AI","AAL"],"roles":["Admin","Manager","Coder"]}
{"id":2,"names":["Iris","Ellis","Monta"],"roles":["CustomerService"]}
2. 包含多个 List(T) 和 List(String)属性的情况
实体类
@Data
public class SysUser {
private Long id;
private String loginName;
private String userName;
private String email;
private List<Long> roleIds;
private List<SysRole> roles;
}
@Data
public class SysRole {
private Long id;
private String roleName;
private String roleDesc;
}
Mapper 接口
public interface UserMapper {
List<SysUser> findUsers(int status);
}
Mapper Sql 映射文件
<resultMap id="SysUserMap" type="com.answer.ai.entity.SysUser">
<id column="id" property="id" jdbcType="BIGINT" />
<result column="login_name" property="loginName" jdbcType="VARCHAR"/>
<result column="user_name" property="userName" jdbcType="VARCHAR"/>
<result column="email" property="email" jdbcType="VARCHAR"/>
<!-- userId 和 roleStatus 为查询 findRoleById 的入参, id 和 role_status 为查询 findUsers 的结果信息 -->
<!-- 如果查询 findRoleById 需要 roleId 作为入参, column 写法 {userId=id,roleStatus=role_status,roleId=role_id} -->
<collection property="roles" ofType="com.answer.ai.entity.SysRole" select="findRoleById" column="{userId=id,roleStatus=role_status}" />
<collection property="roleIds" ofType="Long">
<constructor>
<arg column="role_id" />
</constructor>
</collection>
</resultMap>
<!-- 查询 findUsers 的入参 #{status} 放在结果集上作为查询 findRoleById 作为入参 -->
<select id="findUsers" parameterType="Integer" resultMap="SysUserMap">
SELECT su.id, su.login_name, su.user_name, su.email, sur.role_id, #{status} AS role_status
FROM ai_user su
LEFT JOIN dc_sys_user_role sur ON sur.user_id = su.id
</select>
<select id="findRoleById" resultType="com.answer.ai.entity.SysRole">
SELECT id, role_name, role_descript AS role_desc
FROM ai_user_role
WHERE user_id = #{userId} AND role_status = #{roleStatus}
</select>
接口调用结果
{
"code": 10000,
"desc": "success",
"data": [
{
"id": 1,
"loginName": "admin",
"userName": "林**",
"email": "[email protected]",
"roleIds": [
1
],
"roles": [
{
"id": 1,
"roleName": "admin",
"roleDesc": "超级管理员"
}
]
},
{
"id": 2,
"loginName": "answer",
"userName": "杨**",
"email": "[email protected]",
"roleIds": [
2,
3
],
"roles": [
{
"id": 2,
"roleName": "answer",
"roleDesc": "普通管理员"
},
{
"id": 3,
"roleName": "develop",
"roleDesc": "开发者权限"
}
]
}
]
}
Refer
https://blog.csdn.net/u010979642/article/details/100668909
https://blog.csdn.net/u010979642/article/details/94646861?depth_1-utm_source=distribute.pc_relevant.none-task-blog-BlogCommendFromBaidu-2&utm_source=distribute.pc_relevant.none-task-blog-BlogCommendFromBaidu-2