在Mybatis中association標籤多層巢狀的問題

2022-03-10 16:00:15

association標籤多層巢狀問題

mybatis裡查詢使用巢狀association標籤時，發現內層的association查詢的結果一直為null

排查

檢查sql執行情況，發現有資料返回，排除
檢查property的值是否和pojo中的對應，值一致，排除
檢查column的值是否和資料庫的相對應，相對應，排除

那麼應該是mybatis沒有把資料對映到位了，經過排查是association中columnPrefix被不對應

<resultMap id="BaseResultMap" type="a.b.c.d.e">
    <id column="id" property="id" />
    <result property="workTime" column="work_time" />
    <result property="model" column="model" />
    <result property="status" column="status" />
    <association property="interfaceUpstream" javaType="interfaceUpstream" columnPrefix="ui_">
      <id column="id" property="id" />
      <result property="interfaceName" column="interface_name" />
      <result property="interfaceType" column="interface_type" />
      <result property="frequency" column="frequency" />
      <result property="address" column="address" />
      <result property="templateOrSql" column="template_or_sql" />
      <result property="status" column="status" />
      <association property="systemInfo" javaType="SystemInfo" columnPrefix="sys_">
        <id column="id" property="id"/>
        <result property="systemName" column="system_name"/>
        <result property="systemNameEN" column="system_name_en"/>
        <result property="belong" column="belong"/>
        <result property="status" column="status"/>
      </association>
      <association property="serverInfo" javaType="ServerInfo" columnPrefix="ser_">
        <id column="id" property="id"/>
        <result property="ftpIp" column="ftp_ip"/>
        <result property="ftpPort" column="ftp_port"/>
        <result property="ftpAccount" column="ftp_account"/>
        <result property="ftpPassword" column="ftp_password"/>
      </association>
    </association>
  </resultMap>

<sql id="base_select">
    SELECT
    ii.Id,
    ii.model,
    ii.status,
    ii.work_time,
    ui.id AS ui_id,
    ui.interface_name AS ui_interface_name,
    ui.interface_type AS ui_interface_type,
    ui.frequency AS ui_frequency,
    ui.address AS ui_address,
    ui.template_or_sql AS ui_template_or_sql,
    ui.status AS ui_status,
    sys.id AS sys_id,
    sys.system_name AS sys_system_name,
    sys.system_name_en AS sys_system_name_en,
    sys.belong AS sys_belong,
    sys.status AS sys_status,
    ser.id AS ser_id,
    ser.ftp_ip AS ser_ftp_ip,
    ser.ftp_port AS ser_ftp_port,
    ser.ftp_account AS ser_ftp_account,
    ser.ftp_password AS ser_ftp_password
  </sql>

從程式碼上看沒有什麼問題

原因是association在進行多層巢狀時，mybatis會將外層association的columnPrefix值與內層的進行併合，

如外層columnPrefix值位ui_, 內層為sys_，那麼在SQL中就不能這樣 sys.id AS sys_id 了，需要將ui_字首加上，變成 sys.id AS ui_sys_id ,這樣mybatis在匹配的時候才會將資料對映到對應association上

正常程式碼如下

SELECT
    ii.Id,
    ii.model,
    ii.status,
    ii.work_time,
    ui.id AS ui_id,
    ui.interface_name AS ui_interface_name,
    ui.interface_type AS ui_interface_type,
    ui.frequency AS ui_frequency,
    ui.address AS ui_address,
    ui.template_or_sql AS ui_template_or_sql,
    ui.status AS ui_status,
    sys.id AS ui_sys_id,
    sys.system_name AS ui_sys_system_name,
    sys.system_name_en AS ui_sys_system_name_en,
    sys.belong AS ui_sys_belong,
    sys.status AS ui_sys_status,
    ser.id AS ui_ser_id,
    ser.ftp_ip AS ui_ser_ftp_ip,
    ser.ftp_port AS ui_ser_ftp_port,
    ser.ftp_account AS ui_ser_ftp_account,
    ser.ftp_password AS ui_ser_ftp_password

問題解決！

association集合巢狀

學了一下mybatis的查詢返回值的集合巢狀，先查了查官網：

這個返回集合有什麼用呢

舉個例子三張表

hr_job_department

hr_job_position

第三張表裡在表示部門和職位的時候只用了上面兩張表的主鍵

但是查詢的時候，希望表示下面這樣的結果

所以返回值是不止一個物件，這樣就用到了集合巢狀

<resultMap id="userInfoMap" type="com.advancedc.hrsys.entity.UserInfo">
		<id column="id" property="id" />
		<result column="name" property="name" />
		<result column="gender" property="gender" />
		<result column="id_card" property="idCard" />
		<result column="is_married" property="isMarried" />
		<result column="phone" property="phone" />
		<result column="priority" property="priority" />
		<result column="entry_time" property="entryTime" />
		<result column="full_time" property="fullTime" />
		<result column="created_time" property="createdTime" />
		<result column="edited_time" property="editedTime" />
		<association property="jobDepartment" column="id"
			javaType="com.advancedc.hrsys.entity.JobDepartment">
			<id column="jdid" property="id" />
			<result column="jdname" property="name" />
		</association>
		<association property="jobPosition" column="id"
			javaType="com.advancedc.hrsys.entity.JobPosition">
			<id column="jpid" property="id" />
			<result column="jpname" property="name" />
		</association>
	</resultMap>

只需要知道：

（1）column表示資料庫欄位

（2）property表示Java裡的值

而且我這裡的主鍵都是id所以會出現重名的情況，在SQL語句裡，查詢時就要賦予別名才能加以區分，返回結果resultMap就如上圖所示

<select id="queryUserInfoBySomeone" resultMap="userInfoMap" resultType="com.advancedc.hrsys.entity.UserInfo">
		SELECT
		ui.id,
		ui.name,
		ui.gender,
		ui.id_card,
		ui.is_married,
		ui.department_id,
		ui.position_id,
		ui.phone,
		ui.priority,
		ui.entry_time,
		ui.full_time,
		ui.created_time,
		ui.edited_time,
		jd.id jdid,
		jd.name jdname,
		jp.id jpid,
		jp.name jpname
		FROM
		hr_user_info ui
		INNER JOIN
		hr_job_department jd
		ON
		ui.department_id=jd.id
		INNER JOIN
		hr_job_position jp
		ON
		ui.position_id=jp.id
		<where>
			<if test="someone.id>0">
			and ui.id = #{someone.id}
			</if>
			<if test="someone.gender!=null">
			and ui.gender = #{someone.gender}
			</if>
			<if test="someone.name!=null">
			and ui.name = #{someone.name}
			</if>
			<if test="someone.idCard!=null">
			and ui.id_card = #{someone.idCard}
			</if>
			<if test="someone.isMarried!=null">
			and ui.is_married = #{someone.isMarried}
			</if>
			<if test="someone.jobDepartment!=null and someone.jobDepartment.id!=null">
			and ui.department_id = #{someone.jobDepartment.id}
			</if>
			<if test="someone.jobPosition!=null and someone.jobPosition.id!=null">
			and ui.position_id = #{someone.jonPosition.id}
			</if>
			<if test="someone.phone!=null">
			and ui.phone = #{someone.phone}
			</if>
			<if test="someone.entryTime!=null">
			and ui.entry_time = #{someone.entryTime}
			</if>
			<if test="someone.fullTime!=null">
			and ui.full_time = #{someone.fullTime}
			</if>
		</where>
	</select>

上圖用了INNER JOIN來查詢看上去挺簡潔的，有一種不簡潔的寫法如下，雖然也能得到結果，但是不知道效能對比如何

SELECT
		ui.id,
		ui.name,
		ui.gender,
		ui.id_card,
		ui.is_married,
		ui.department_id,
		ui.position_id,
		ui.phone,
		ui.priority,
		ui.entry_time,
		ui.full_time,
		ui.created_time,
		ui.edited_time,
		(select id jdid from hr_job_department jd where jd.id=ui.department_id) jdid,
		(select name jdname from hr_job_department jd where jd.id=ui.department_id) jdname,
		(select id jpid from hr_job_position jp where jp.id=ui.position_id) jpid,
		(select name jpname from hr_job_position jp where jp.id=ui.position_id) jpname
		FROM
		hr_user_info ui;

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