mybatis一对多一对一

版权声明：本文为博主原创文章，遵循 CC 4.0 BY-SA 版权协议，转载请附上原文出处链接和本声明。

本文链接： https://blog.csdn.net/caoxuecheng001/article/details/102585083

一对多：一个用户对应多个银行账户

用户表：user字段：id username birthday sex address

账户表：account字段：id uid money

实体类：

public class User implements Serializable {

    private Integer id;
    private String username;
    private String address;
    private String sex;
    private Date birthday;

    //一对多关系映射：主表实体应该包含从表实体的集合引用
    private List<Account> accounts;
    //省略getset方法
}

public class Account implements Serializable {

    private Integer id;
    private Integer uid;
    private Double money;

    //从表实体应该包含一个主表实体的对象引用
    private User user;
    //get set...
}

IUserDao.xml代码

    <!-- 定义User的resultMap-->
    <resultMap id="userAccountMap" type="user">
        <id property="id" column="id"></id>
        <result property="username" column="username"></result>
        <result property="address" column="address"></result>
        <result property="sex" column="sex"></result>
        <result property="birthday" column="birthday"></result>
        <!-- 配置user对象中accounts集合的映射 -->
        <collection property="accounts" ofType="account">
            <id column="aid" property="id"></id>
            <result column="uid" property="uid"></result>
            <result column="money" property="money"></result>
        </collection>
    </resultMap>

    <!-- 查询所有 -->
    <select id="findAll" resultMap="userAccountMap">
      select u.*,a.id as aid,a.uid,a.money from account a,user u where a.uid =u.id;
    </select>

分析：

IUserDao:

    List<User> findAll();

这样一对多就可以了。

一对一：一个银行账户对应一个用户。

IAccountDao.xml

<!-- 定义封装account和user的resultMap -->
    <resultMap id="accountUserMap" type="account">
        <id property="id" column="aid"></id>
        <result property="uid" column="uid"></result>
        <result property="money" column="money"></result>
        <!-- 一对一的关系映射：配置封装user的内容-->
        <association property="user" column="uid" javaType="user">
            <id property="id" column="id"></id>
            <result column="username" property="username"></result>
            <result column="address" property="address"></result>
            <result column="sex" property="sex"></result>
            <result column="birthday" property="birthday"></result>
        </association>
    </resultMap>

    <!-- 查询所有 -->
    <select id="findAll" resultMap="accountUserMap">
        select u.*,a.id as aid,a.uid,a.money from account a , user u where u.id = a.uid;
    </select>

IAccountDao

   List<Account> findAll();

这样就OK了