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在SQL和Hibernate映射中根据另一个表中被引用的行的count()设置列的值到我的表

[英]Setting the value of a column according to the count() of the referred rows in another table to my table in SQL and hibernate mapping

我的数据库中有以下表格。 我想列booked_itemstestuser表包含与该用户相关联的行数testbooking表。 当然可以在程序内部完成,但是如果可以在休眠映射和表DDL中完成则很脏。

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>

PS。 我想问两个独立的问题。 一次用于Hibernate,一次用于SQL,但我担心它将被视为重复项。 对于Hibernate需要了解的SQL我很感兴趣。 并且它取决于SQL的方言吗? 我目前正在询问mysql方言。

您可以在用户类上使用与SQL公式关联的属性。

<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: 好像重复

使用JPA / Hibernate计算属性

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