SQL和hibernate映射中根据另一个表中引用行的count（）设置列的值到我的表答案

【问题标题】：Setting the value of a column according to the count() of the referred rows in another table to my table in SQL and hibernate mappingSQL和hibernate映射中根据另一个表中引用行的count（）设置列的值到我的表
【发布时间】：2012-12-28 08:16:47
【问题描述】：

我的数据库中有以下表格。我希望testuser 表中的booked_items 列包含testbooking 表中与用户关联的行数。它当然可以在程序内部完成，但如果它可以在休眠映射和表 DDL 中完成，那就很脏了。

CREATE TABLE `testuser` (
  `id` bigint(20) NOT NULL,
  `username` varchar(30) DEFAULT NULL,
  `password` varchar(128) DEFAULT NULL,
  `booked_items` int(11) DEFAULT NULL,
  PRIMARY KEY (`id`),
  UNIQUE KEY `username` (`username`)
) ENGINE=InnoDB DEFAULT CHARSET=latin1;

CREATE TABLE `testbooking` (
  `id` bigint(20) NOT NULL AUTO_INCREMENT,
  `user_id` bigint(20) NOT NULL,
  `item_id` bigint(20) NOT NULL,
  `start_date` date DEFAULT NULL,
  `end_date` date DEFAULT NULL,
  PRIMARY KEY (`id`),
  KEY `FKEC747367A12BE486` (`item_id`),
  KEY `FKEC7473671463A66C` (`user_id`),
  CONSTRAINT `FKEC7473671463A66C` FOREIGN KEY (`user_id`) REFERENCES `testuser`
(`id`),
  CONSTRAINT `FKEC747367A12BE486` FOREIGN KEY (`item_id`) REFERENCES `testitem`
(`id`)
) ENGINE=InnoDB AUTO_INCREMENT=7 DEFAULT CHARSET=latin1;

CREATE TABLE `testitem` (
  `id` bigint(20) NOT NULL AUTO_INCREMENT,
  `title` varchar(30) DEFAULT NULL,
  PRIMARY KEY (`id`)
) ENGINE=InnoDB AUTO_INCREMENT=38 DEFAULT CHARSET=latin1;

testuser 表的休眠映射为：

<?xml version="1.0" encoding="utf-8"?>
<!DOCTYPE hibernate-mapping PUBLIC "-//Hibernate/Hibernate Mapping DTD 3.0//EN"
"http://www.hibernate.org/dtd/hibernate-mapping-3.0.dtd">
<hibernate-mapping>
    <class name="my.hibernate.actors.User" discriminator-value="user"
        table="testuser" catalog="efeu">
        <id name="id" type="java.lang.Long">
            <column name="id" />
            <generator class="increment" />
        </id>
        <discriminator column="user_type" type="string" length="8" />
        <property name="username" type="java.lang.String" not-null="true"
            unique="true" unique-key="true">
            <column name="username" length="30" />
        </property>
        <property name="password" type="java.lang.String" not-null="true">
            <column name="password" length="128" />
        <subclass name="my.hibernate.actors.Subscriber"
            discriminator-value="subsc">
            <property name="bookedItems" type="java.lang.Integer" column="booked_items" />
            <set name="bookings" inverse="true" cascade="all-delete-orphan">
                <key column="user_id" />
                <one-to-many class="my.hibernate.operations.Booking" />
            </set>
        </subclass>
        <subclass name="my.hibernate.actors.Clerk"
            discriminator-value="clerk" />
    </class>
</hibernate-mapping>

testbooking 表的休眠映射为：

<?xml version="1.0" encoding="utf-8"?>
<!DOCTYPE hibernate-mapping PUBLIC "-//Hibernate/Hibernate Mapping DTD 3.0//EN"
"http://www.hibernate.org/dtd/hibernate-mapping-3.0.dtd">
<hibernate-mapping>
    <class name="my.hibernate.operations.Booking" table="testBooking" catalog="efeu">
        <id name="id" type="java.lang.Long">
            <column name="id" />
            <generator class="native" />
        </id>

        <many-to-one name="userID" column="user_id" cascade="persist"
            not-null="true" class="my.hibernate.actors.Subscriber" />
        <many-to-one name="itemID" column="item_id"  cascade="persist"
            not-null="true" class="my.hibernate.items.Item" />

        <property name="startDate" type="java.sql.Date" not-null="true">
            <column name="start_date" length="10" />
        </property>

        <property name="endDate" type="java.sql.Date">
            <column name="end_date" length="10" />
        </property>
    </class>
</hibernate-mapping>

附言。我想针对这个问题提出 2 个单独的问题。一次用于 Hibernate，一次用于 SQL，但我担心它会被视为重复。对于 Hibernate 需要了解的 SQL，我很想知道。它是否取决于 SQL 的方言？我目前正在询问mysql方言。

【问题讨论】：

标签： mysql sql hibernate hibernate-mapping

【解决方案1】：

您可以使用用户类的属性和与之关联的 sql 公式。

<property name="booked_items" type="long" formula="(SELECT COUNT(tb.id) FROM testbooking AS tb WHERE tb.user_id = id)"/>

看看这个答案：好像是重复的

Calculated property with JPA / Hibernate

【讨论】：