Mybatis：通过自动映射解决一对一映射字段

像是下面这样的，如果真要一个个地去把 reulst 子标签写出来，有点不太现实，还好 mybatis 提供了 autoMapping，可以作用在 association 和 resultMap 标签上：

<resultMap id="mapOfQueryMyself" type="Student" autoMapping="true">
  <association property="college" columnPrefix="c_" javaType="College">
    <id property="cno" column="cno"/>
    <result property="name" column="name"/>
    <result property="intro" column="intro"/>
  </association>
  <association property="profession" columnPrefix="p_" javaType="Profession" autoMapping="true">
    <id property="pno" column="pno"/>
    <result property="name" column="name"/>
    <result property="type" column="type"/>
    <result property="eduSys" column="edu_sys"/>
    <result property="degree" column="degree"/>
    <result property="code" column="code"/>
  </association>
  <association property="grade" columnPrefix="g_" javaType="Grade">
    <id property="gno" column="gno"/>
    <result property="name" column="name"/>
    <result property="layer" column="layer"/>
    <result property="grade" column="grade"/>
  </association>
</resultMap>

用 autoMapping 之前，联合查询中，一对一关系需要给字段别名，最好用一种格式，比如 c_xxx，在 association 中添加 columnPrefix，然后再开启 autoMapping 就可以完美匹配查询过后的字段：

<resultMap id="mapOfQueryMyself" type="Student" autoMapping="true">
  <association property="college" columnPrefix="c_" javaType="College" autoMapping="true"/>
  <association property="profession" columnPrefix="p_" javaType="Profession" autoMapping="true"/>
  <association property="grade" columnPrefix="g_" javaType="Grade" autoMapping="true"/>
</resultMap>

<select id="queryMyself" resultMap="mapOfQueryMyself">
  SELECT s.*,
         c.cno     c_cno,
         c.name    c_name,
         p.pno     p_pno,
         p.name    p_name,
         p.type    p_type,
         p.edu_sys p_edu_sys,
         p.degree  p_degree,
         p.code    p_code,
         g.gno     g_gno,
         g.name    g_name,
         g.layer   g_layer,
         g.grade   g_grade
  FROM `students` as s
         JOIN `colleges` as c on c.cno = s.college_id
         JOIN `grades` as g on g.gno = s.grade_id
         JOIN `professions` as p on p.pno = s.profession_id
  WHERE s.sno = 1;
</select>