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hibernate 继承映射

發布時間：2025/6/15 编程问答 24 豆豆

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在域模型中，類與類之間除了關聯關系和聚集關系，還可以存在繼承關系，在下圖所示的域模型中，Deparment類和Employee類之間為一對多的雙向關聯關系，Employee類有兩個子類：Skiller類和Sales類。由于Java只允許一個類最多有一個直接的父類，因此Employee類、 Skiller類和Sales類構成了一棵繼承關系樹。

?在面向對象的范疇中，還存在多態的概念，多態建立在繼承關系的基礎上。簡單地理解，多態是指當一個Java應用變量被聲明為Employee類時，這個變量實際上既可以引用Employee類自己的實例，Skiller類的實例，也可以引用Sales類的實例。Department類的getEmps()方法通過HibernateAPI從數據庫中檢索出所有Employee對象。getEmps()方法返回的集合既可以包含Employee類自己的實例，Skiller類的實例，也可以引用Sales類的實例。，這種查詢被稱為多態查詢。數據庫表之間并不存在繼承關系，那么如何把域模型的繼承關系映射到關系數據模型中呢？hibernate有以下三種映射方式：

繼承關系樹的根類對應一個表：對關系數據模型進行非常規設計，在數據庫表中加入額外的區分子類型的字段。通過這種方式，可以使關系數據模型支持繼承關系和多態。

繼承關系樹的每個類對應一個表(子類與父類通過外鍵關聯)：在關系數據模型中用外鍵參照關系來表示繼承關系。

繼承關系樹的每個具體類對應一個表：關系數據模型完全不支持域模型中的繼承關系和多態。

1.繼承關系樹的根類對應一個表employee（整個繼承樹一張表）：

employee的表結構如下所示：

mysql> desc employee;
+------------+--------------+------+-----+---------+----------------+
| Field????? | Type???????? | Null | Key | Default | Extra????????? |
+------------+--------------+------+-----+---------+----------------+
| id???????? | int(11)????? | NO?? | PRI | NULL??? | auto_increment |
| type?????? | int(11)????? | NO?? |???? | NULL??? |??????????????? |
| name?????? | varchar(255) | YES? | UNI | NULL??? |??????????????? |
| depart_id? | int(11)????? | YES? | MUL | NULL??? |??????????????? |
| skill????? | varchar(255) | YES? |???? | NULL??? |??????????????? |
| saleAmount | int(11)????? | YES? |???? | NULL??? |??????????????? |
+------------+--------------+------+-----+---------+----------------+

實體類Department和Employee請參看我前面的文章，Skiller和Sales分別如下所示：

Java代碼 ?

package?com.reiyen.hibernate.domain;??

public?class?Skiller?extends?Employee?{??

????private?String?skill;??

//setter和getter方法??

}??

? Java代碼 ?

package?com.reiyen.hibernate.domain;??

public?class?Sales?extends?Employee?{??

????private?int?saleAmount;??

//setter和getter方法??

}??

?Employee.hbm.xml映射文件如下：

Xml代碼 ?

<?xml?version="1.0"?>??

<!DOCTYPE?hibernate-mapping?PUBLIC???

????"-//Hibernate/Hibernate?Mapping?DTD?3.0//EN"??

????"http://hibernate.sourceforge.net/hibernate-mapping-3.0.dtd">??

<hibernate-mapping?package="com.reiyen.hibernate.domain">??

????<class?name="Employee"?discriminator-value="0">??

????????<id?name="id">??

????????????<generator?class="native"?/>??

????????</id>??

??????????

????????<discriminator?column="type"?type="int"></discriminator>??

????????<property?name="name"?unique="true"/>??

??????????

????????<many-to-one?name="department"?column="depart_id"?/>??

????????<subclass?name="Skiller"?discriminator-value="1">??

????????????<property?name="skill"?/>??

????????</subclass>??

????????<subclass?name="Sales"?discriminator-value="2">??

????????????<property?name="saleAmount"?/>??

????????</subclass>??

????</class>??

</hibernate-mapping>??

?測試類如下：

Java代碼 ?

public?class?Many2One?{??

????public?static?void?main(String[]?args)?{??

????????add();??

?????????query(1);//1??

????}??

????static?void?query(int?empId)?{??

????????Session?s?=?null;??

????????Transaction?tx?=?null;??

????????try?{??

????????????s?=?HibernateUtil.getSession();??

????????????tx?=?s.beginTransaction();??

????????????Employee?emp?=?(Employee)?s.get(Employee.class,?empId);//2??

????????????System.out.println(emp.getClass());??

????????????tx.commit();??

????????}?finally?{??

????????????if?(s?!=?null)??

????????????????s.close();??

????????}??

????}??

????static?void?add()?{??

????????Session?s?=?null;??

????????Transaction?tx?=?null;??

????????try?{??

????????????Department?depart?=?new?Department();??

????????????depart.setName("department?name");??

??????????????

????????????Employee?employee1?=?new?Employee();??

????????????employee1.setDepartment(depart);?//1?對象模型：建立兩個對象的關聯???

????????????employee1.setName("employee1?name1");??

??????????????

????????????Skiller?employee2?=?new?Skiller();??

????????????employee2.setDepartment(depart);?//2?對象模型：建立兩個對象的關聯???

????????????employee2.setName("employee2?name2");??

????????????employee2.setSkill("j2se");??

??????????????

????????????Sales?employee3?=?new?Sales();??

????????????employee3.setDepartment(depart);?//2?對象模型：建立兩個對象的關聯???

????????????employee3.setName("employee3?name3");??

????????????employee3.setSaleAmount(1000);??

??????????????

????????????s?=?HibernateUtil.getSession();??

????????????tx?=?s.beginTransaction();??

????????????s.save(depart);??

????????????s.save(employee1);??

????????????s.save(employee2);??

????????????s.save(employee3);??

????????????tx.commit();??

????????}?finally?{??

????????????if?(s?!=?null)??

????????????????s.close();??

????????}??

????}??

}??

?程序運行后，控制臺打印信息如下所示：

Hibernate: insert into Department (name) values (?)
Hibernate: insert into Employee (name, depart_id, type) values (?, ?, 0)
Hibernate: insert into Employee (name, depart_id, skill, type) values (?, ?, ?, 1)
Hibernate: insert into Employee (name, depart_id, saleAmount, type) values (?, ?, ?, 2)
Hibernate: select employee0_.id as id1_0_, employee0_.name as name1_0_,employee0_.depart_id as depart4_1_0_, employee0_.skill as skill1_0_,employee0_.saleAmount as saleAmount1_0_, employee0_.type as type1_0_from Employee employee0_ where employee0_.id=?
class com.reiyen.hibernate.domain.Employee

employee表中記錄如下所示：

mysql> select * from employee;
+----+------+-----------------+-----------+-------+------------+
| id | type | name??????????? | depart_id | skill | saleAmount |
+----+------+-----------------+-----------+-------+------------+
|? 1 |??? 0 | employee1 name1 |???????? 1 | NULL? |?????? NULL |
|? 2 |??? 1 | employee2 name2 |???????? 1 | j2se? |?????? NULL |
|? 3 |??? 2 | employee3 name3 |???? ? ? 1 | NULL? |?????? 1000 |
+----+------+-----------------+-----------+-------+------------+
3 rows in set (0.00 sec)

將測試代碼中注釋為1的語句改成:

Java代碼 ?

query(2);??

再運行，控制臺打印的class如下所示(因為hibernate支持多態查詢)： class com.reiyen.hibernate.domain.Skiller

打印的查詢語句還是如上面所示的沒有改變。

在上面修改的基礎上，再將測試代碼中注釋為2的語句改成:

Java代碼 ?

Employee?emp?=?(Employee)?s.get(Skiller.class,?empId);??

?再運行，則控制臺打印的查詢語句為：

Hibernate: select skiller0_.id as id1_0_,skiller0_.name as name1_0_, skiller0_.depart_id as depart4_1_0_,skiller0_.skill as skill1_0_ from Employee skiller0_ where skiller0_.id=? and skiller0_.type=1
class com.reiyen.hibernate.domain.Skiller

優點：操作效率高

缺點：如果說給employee增加子類的話,必須修改表結構,給表結構增加一個字段；同時表中對應子類的字段不能有非空約束.

2.繼承關系樹的每子類對應一個表(joined-subclass),表結構如下所示：

?修改Employee.hbm.xml映射文件如下所示：

Xml代碼 ?

<?xml?version="1.0"?>??

<!DOCTYPE?hibernate-mapping?PUBLIC???

????"-//Hibernate/Hibernate?Mapping?DTD?3.0//EN"??

????"http://hibernate.sourceforge.net/hibernate-mapping-3.0.dtd">??

<hibernate-mapping?package="com.reiyen.hibernate.domain">??

????<class?name="Employee">??

????????<id?name="id">??

????????????<generator?class="native"?/>??

????????</id>??

????????<property?name="name"?unique="true"/>??

????????<many-to-one?name="department"?column="depart_id"?/>??

????????<joined-subclass?name="Skiller"?table="skiller">??

?????????<key?column="employee_id"?/>??

?????????<property?name="skill"?/>??

????????</joined-subclass>??

????????<joined-subclass?name="Sales"?table="sales">??

?????????<key?column="employee_id"?/>??

?????????<property?name="saleAmount"?column="sale_amount"?/>??

????????</joined-subclass>??

????</class>??

</hibernate-mapping>??

?測試類不變，只是將測試代碼中注釋為1的語句改成:

Java代碼 ?

query(2);??

則控制臺打印的信息如下所示：

Hibernate: insert into Department (name) values (?)
Hibernate: insert into Employee (name, depart_id) values (?, ?)
Hibernate: insert into Employee (name, depart_id) values (?, ?)
Hibernate: insert into skiller (skill, employee_id) values (?, ?)
Hibernate: insert into Employee (name, depart_id) values (?, ?)
Hibernate: insert into sales (sale_amount, employee_id) values (?, ?)
Hibernate: select employee0_.id as id1_0_, employee0_.name as name1_0_, employee0_.depart_id as depart3_1_0_, employee0_1_.skill as skill2_0_, employee0_2_.sale_amount as sale2_3_0_, case when employee0_1_.employee_id is not null then 1 when employee0_2_.employee_id is not null then 2 when employee0_.id is not null then 0 end as clazz_0_ from Employee employee0_ left outer join skiller employee0_1_ on employee0_.id=employee0_1_.employee_id left outer join sales employee0_2_ on employee0_.id=employee0_2_.employee_id where employee0_.id=?
class com.reiyen.hibernate.domain.Skiller
?從打印的SQL語句可以看出，此時，如果保存的是Employee對象的子類的實例的話，則要在兩張表中保存記錄；如果查詢的是子類對象的話，是三張表關聯在一起進行查詢。

在上面修改的基礎上，再將測試代碼中注釋為2的語句改成:

Java代碼 ?

Employee?emp?=?(Employee)?s.get(Skiller.class,?empId);??

?再運行，則控制臺打印的查詢語句為：

Hibernate: select skiller0_.employee_id as id1_0_, skiller0_1_.name as name1_0_, skiller0_1_.depart_id as depart3_1_0_, skiller0_.skill as skill2_0_ from skiller skiller0_ inner join Employee skiller0_1_ on skiller0_.employee_id=skiller0_1_.id where skiller0_.employee_id=?
此時只關聯兩張表查詢。

數據庫中表記錄如下所示：

mysql> select * from employee;
+----+-----------------+-----------+
| id | name??????????? | depart_id |
+----+-----------------+-----------+
|? 1 | employee1 name1 |???????? 1 |
|? 2 | employee2 name2 |???????? 1 |
|? 3 | employee3 name3 |???????? 1 |
+----+-----------------+-----------+
3 rows in set (0.00 sec)

mysql> select * from skiller;
+-------------+-------+
| employee_id | skill |
+-------------+-------+
|?????????? 2 | j2se? |
+-------------+-------+
1 row in set (0.00 sec)

mysql> select * from sales;
+-------------+-------------+
| employee_id | sale_amount |
+-------------+-------------+
|?????????? 3 |??????? 1000 |
+-------------+-------------+
1 row in set (0.00 sec)

3.混合使用，假設如果Sales的屬性很多，而Skiller的屬性很少，這時可以混使用“一個類繼承體系一張表”和“每個子類一張表”，表結構如下所示：

?Employee.hbm.xml映射文件如下所示：

Xml代碼 ?

<?xml?version="1.0"?>??

<!DOCTYPE?hibernate-mapping?PUBLIC???

????"-//Hibernate/Hibernate?Mapping?DTD?3.0//EN"??

????"http://hibernate.sourceforge.net/hibernate-mapping-3.0.dtd">??

<hibernate-mapping?package="com.reiyen.hibernate.domain">??

????<class?name="Employee"?discriminator-value="0">??

????????<id?name="id">??

????????????<generator?class="native"?/>??

????????</id>??

????????<discriminator?column="type"?type="int"?/>??

????????<property?name="name"?unique="true"?/>??

????????<many-to-one?name="department"?column="depart_id"?/>??

????????

<subclass?name="Skiller"?discriminator-value="1">??

????????????<property?name="skill"?/>??

????????</subclass>??

????????<subclass?name="Sales"?discriminator-value="2">??

????????????<join?table="sales">??

????????????????<key?column="employee_id"?/>??

????????????????<property?name="saleAmount"?column="sale_amount"?/>??

????????????</join>??

????????</subclass>??

????</class>??

</hibernate-mapping>??

?此時測試類不變，只是將測試代碼中注釋為1的語句改成:

Java代碼 ?

query(2);??

然后在上面的基礎上運行原程序，則控制臺會打印出如下異常信息：

Hibernate: insert into Department (name) values (?)
Hibernate: insert into Employee (name, depart_id, type) values (?, ?, 0)
Exception in thread "main" org.hibernate.exception.SQLGrammarException: could not insert: [com.reiyen.hibernate.domain.Employee]

Caused by: com.mysql.jdbc.exceptions.MySQLSyntaxErrorException: Unknown column 'type' in 'field list'

這是因為我在hibernate.cfg.xml配置文件中配置了此項：

Xml代碼 ?

<property?name="hbm2ddl.auto">create</property>??

?所以在此次程序運行時，會刪除數據庫中的employee表，sales表，而employee表中有skiller表的外鍵關聯，所以不能刪除employee數據表了，所以拋出了上面的異常。此時你再查看數據庫表，如下所示：

所以得先手動刪除skiller數據表，然后再來運行程序：

如果Employee.hbm.xml配置文件中

Xml代碼 ?

<subclass?name="Skiller"?>??

不配置discriminator-value="1"，則會拋出如下異常：

java.lang.ExceptionInInitializerError

Caused by: org.hibernate.MappingException: Could not format discriminator value to SQL string

因為如果discriminator-value沒有顯式的給定值的話，則與name屬性的值保持一致，即為Skiller ,所以會拋出如上異常！

4.繼承關系樹的每個具體類對應一個表（union-subclass）

表結構如下所示：

Employee.hbm.xml映射文件如下：

Xml代碼 ?

<?xml?version="1.0"?>??

<!DOCTYPE?hibernate-mapping?PUBLIC???

????"-//Hibernate/Hibernate?Mapping?DTD?3.0//EN"??

????"http://hibernate.sourceforge.net/hibernate-mapping-3.0.dtd">??

<hibernate-mapping?package="com.reiyen.hibernate.domain">??

????<class?name="Employee">??

????????<id?name="id">??

????????????<generator?class="hilo"?/>??

????????</id>??

????????<property?name="name"?unique="true"?/>??

????????<many-to-one?name="department"?column="depart_id"?/>??

????????<union-subclass?name="Skiller"?table="skiller">??

?????????<property?name="skill"?/>??

????????</union-subclass>??

????????<union-subclass?name="Sales"?table="sales">??

?????????<property?name="saleAmount"?column="sale_amount"?/>??

????????</union-subclass>??

????</class>??

</hibernate-mapping>??

?此時主鍵增長不能再是:

Xml代碼 ?

<generator?class="native"?/>??

因為如果使用native的話三張表會產生相同的id值，這樣當根據id查詢Employee時就會出錯了。所以如果你配置成native時會拋出如下異常(因為Employee實體類中id對應的是int了，所以在此使用hilo（高低位）主鍵生成方式)：

org.hibernate.MappingException: Cannot use identity column key generation with <union-subclass> mapping for: com.reiyen.hibernate.domain.Skiller

運行測試程序后，此時控制臺打印信息如下所示：

Hibernate: insert into Department (name) values (?)
Hibernate: insert into Employee (name, depart_id, id) values (?, ?, ?)
Hibernate: insert into skiller (name, depart_id, skill, id) values (?, ?, ?, ?)
Hibernate: insert into sales (name, depart_id, sale_amount, id) values (?, ?, ?, ?)
Hibernate: select employee0_.id as id1_0_, employee0_.name as name1_0_, employee0_.depart_id as depart3_1_0_, employee0_.skill as skill2_0_, employee0_.sale_amount as sale1_3_0_, employee0_.clazz_ as clazz_0_ from ( select id, null as sale_amount, depart_id, null as skill, name, 0 as clazz_ from Employee union select id, null as sale_amount, depart_id, skill, name, 1 as clazz_ from skiller union select id, sale_amount, depart_id, null as skill, name, 2 as clazz_ from sales ) employee0_ where employee0_.id=?
class com.reiyen.hibernate.domain.Skiller

執行查詢時，首先使用子查詢，在子查詢中使用union將三張表的結果全成一張表，然后再在合成的結果集中進行查詢。

如果Employee是一個抽象類，你不想在數據表中對應相應的數據表，則可以設置abstract="true".如下所示：

Xml代碼 ?

<class?name="Employee"?abstract="true"?>??

? 此外，如果繼承關系中有接口，可以把它當作抽象類對待。

三種映射方式的比較和選擇
為了方便說明為三種方式按順序標號為[1]整個繼承樹一張表;[2]每子類對應一個表(joined-subclass);[4]每個具體類對應一個表（union-subclass）。
1、復雜度：

??? [1]簡單；
??? [2]表較多且之間有外鍵約束；

??? [4]包含重復字段；
2、查詢性能：

??? [1]效率高；
??? [2]需要表內連接或左外連接；

??? [4]若查詢父類需查所有子類表；
3、可維護性：

??? [1]只需修改一個表；
??? [2]若某個類屬性變化只修改這個類對應的表；

??? [4]若父類屬性變化需要修改所有子類對應的表；
綜上，選擇時，可以參考以下原則：
1、子類屬性不是非常多時，優先考慮[1]，因為其性能最佳。
2、子類屬性非常多，且對性能要求不是很嚴格時，優先考慮[2]

總結

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