一對(duì)多（one-to-many）雙向關(guān)聯(lián)實(shí)例（Department- Employee）

這里的一對(duì)多雙向關(guān)聯(lián)是在域模型(實(shí)體對(duì)象模型)上的概念,在關(guān)系數(shù)據(jù)庫(kù)中,只存在外鍵參照關(guān)系,而且總是由"many"方參照"one"方,因?yàn)檫@樣才能消除冗余數(shù)據(jù),因上關(guān)系數(shù)據(jù)庫(kù)實(shí)際上只支持多對(duì)一或一對(duì)一的單向關(guān)聯(lián).在實(shí)體(類與類之間)各種各樣的關(guān)系中,數(shù)多對(duì)一的的單向關(guān)聯(lián)關(guān)系與數(shù)據(jù)庫(kù)中的外鍵參照關(guān)系最匹配了.

在我的前一篇文章多對(duì)一的基礎(chǔ)上進(jìn)行修改。

１.修改Department實(shí)體類：

Java代碼 ?

package?com.reiyen.hibernate.domain??

public?class?Department?{??

??

????private?int?id;??

????private?String?name;??

????private?Set<Employee>?emps;???

??

　　　//setter和getter方法??

｝??

?２.修改Department.hbm.xml文件：

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="Department"?>??

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

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

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

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

????????<set?name="emps">??

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

????????<one-to-many?class="Employee"/>??

????????</set>??

????</class>??

</hibernate-mapping>??

<set>元素的屬性name:設(shè)定待映射的持久化類的屬性名,這里為Depatment類的emps屬性.

<set>元素的兩個(gè)子元素:

<key>:設(shè)定與所關(guān)聯(lián)的持久化類對(duì)應(yīng)的表的外鍵,此處為Employee表的depart_id字段.

<one-to-many>:設(shè)定所關(guān)聯(lián)的持久化類,此處為Employee.

Hibernate根椐以上配置信息獲得如下信息:

<set>元素表明Department類的emps屬性為java.util.Set集合類型.

<class>子元素表明emps集合中存放的是一組Employee對(duì)象

<key>子元素表明employee表是通過(guò)外鍵depart_id參照department表的.

?

?３.　其余的保持原樣，測(cè)試類如下：

Java代碼 ?

public?class?Many2One?{??

??

????/**?

?????*?@param?args?

?????*/??

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

????????Department?depart?=?add();??

????????Department?department?=?queryDepart(depart.getId());??

????????????????//department.getEmps();3??

//?System.out.println("emp?size:"?+?department.getEmps().size());4????

????}??

??

????static?Department?queryDepart(int?departId)?{??

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

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

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

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

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

????????????Department?depart?=?(Department)?s.get(Department.class,?departId);??

?????????????System.out.println("emp?size:"?+?depart.getEmps().size());//1????????

?????????????????????????//?System.out.println("emps?class:"?+?depart.getEmps().getClass());???//2???

?????????????????????????//?Hibernate.initialize(depart.getEmps());??

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

????????????return?depart;??

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

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

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

????????}??

????}??

??

????static?Department?add()?{??

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

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

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

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

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

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

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

????????????employee1.setDepartment(depart);?//?對(duì)象模型：建立兩個(gè)對(duì)象的關(guān)聯(lián)???

????????????employee1.setName("employee1?name2");??

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

????????????Employee?employee2?=?new?Employee();??

????????????employee2.setDepartment(depart);?//?對(duì)象模型：建立兩個(gè)對(duì)象的關(guān)聯(lián)???

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

??

　　　　　　　//　Set<Employee>?set?=?new?HashSet<Employee>();??

????????????//????set.add(employee1);??

????????????//??set.add(employee2);??

????????????//??depart.setEmps(set);??

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

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

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

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

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

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

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

????????????return?depart;??

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

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

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

????????}??

????}??

}??

?控制臺(tái)打印信息如下：

Hibernate: insert into Department (name) values (?)
Hibernate: insert into Employee (name, depart_id) values (?, ?)
Hibernate: insert into Employee (name, depart_id) values (?, ?)
Hibernate: select department0_.id as id1_0_, department0_.name as name1_0_ from Department department0_ where department0_.id=?
Hibernate: select emps0_.depart_id as depart3_1_, emps0_.id as id1_, emps0_.id as id2_0_, emps0_.name as name2_0_, emps0_.depart_id as depart3_2_0_ from Employee emps0_ where emps0_.depart_id=?

emp size:2

數(shù)據(jù)庫(kù)中記錄如下所示：

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

mysql> select * from department;
+----+-----------------+
| id | name??????????? |
+----+-----------------+
|? 1 | department name |
+----+-----------------+
1 row in set (0.00 sec)

?

雖然對(duì)象模型修改了，但數(shù)據(jù)庫(kù)關(guān)系模型并沒(méi)有改變。

?

如果把測(cè)試程序中注釋部分去掉，同時(shí)將注釋為“建立對(duì)象模型”這兩句程序注釋掉，雖然也能正確運(yùn)行，但控制臺(tái)會(huì)多打印出兩條更新語(yǔ)句，說(shuō)明數(shù)據(jù)庫(kù)多執(zhí)行了兩次更新操作，效率上有影響，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: update Employee set depart_id=? where id=?
Hibernate: update Employee set depart_id=? where id=?
Hibernate: select department0_.id as id1_0_, department0_.name as name1_0_ from Department department0_ where department0_.id=?
Hibernate: select emps0_.depart_id as depart3_1_, emps0_.id as id1_, emps0_.id as id2_0_, emps0_.name as name2_0_, emps0_.depart_id as depart3_2_0_ from Employee emps0_ where emps0_.depart_id=?
emp size:2

?

4.懶加載分析：

（1）如果將程序中標(biāo)記為1的程序注釋掉，控制臺(tái)打印信息如下：

Hibernate: select department0_.id as id0_0_, department0_.name as name0_0_ from Department department0_ where department0_.id=?

此時(shí)只是將Department對(duì)象查詢出來(lái)了，而對(duì)應(yīng)的Employee并沒(méi)有進(jìn)行查詢。

(2）如果將程序中標(biāo)記為1的程序注釋掉,同時(shí)將標(biāo)記為2的前面的注釋去掉，運(yùn)行程序，控制臺(tái)打印信息如下：

Hibernate: select department0_.id as id0_0_, department0_.name as name0_0_ from Department department0_ where department0_.id=?
emps class:class org.hibernate.collection.PersistentSet

此時(shí)可以看到其實(shí)Hibernate將hashset替換成了它自己寫(xiě)的PersistentSet,所以才能實(shí)現(xiàn)懶加載功能。同時(shí)可以發(fā)現(xiàn)當(dāng)調(diào)用department.getEmps();hibernate只是獲取了集合代理對(duì)象的引用，它并沒(méi)有去查詢數(shù)據(jù)庫(kù)來(lái)填充集合對(duì)象。你還可以進(jìn)一步測(cè)試來(lái)驗(yàn)證這個(gè)問(wèn)題。如果你將程序中標(biāo)記為1的程序注釋掉，同時(shí)去掉標(biāo)記為3的語(yǔ)句前面的注釋，你會(huì)發(fā)現(xiàn)程序能正常運(yùn)行，但如果去掉標(biāo)記為4的語(yǔ)句前面的注釋，運(yùn)行程序，則會(huì)拋出如下異常：

?org.hibernate.LazyInitializationException: failed to lazily initialize a collection of role: com.reiyen.hibernate.domain.Department.emps, no session or session was closed

(3)當(dāng)進(jìn)行department對(duì)象查詢時(shí)，從department的表結(jié)構(gòu)中也不能判斷這個(gè)Department對(duì)象是否有對(duì)應(yīng)的Employee，那為什么它不像一對(duì)一查詢時(shí)，也一次性把相關(guān)聯(lián)的對(duì)象查詢出來(lái)呢？？這是因?yàn)槿绻麑㈥P(guān)聯(lián)的對(duì)象查詢出來(lái)的話，因?yàn)閿?shù)據(jù)較多，它會(huì)嚴(yán)重影響效率，而不像一對(duì)一時(shí)，反正從對(duì)象只有一個(gè)，一次性查詢出來(lái)影響也不是太大，所以hibernate就假定你有相對(duì)應(yīng)的Employee，直接創(chuàng)建了一個(gè)集合代理對(duì)象的返回給你（把對(duì)象引用先給你），等你需要數(shù)據(jù)的時(shí)候再去查詢數(shù)據(jù)庫(kù)，以提高效率！

?

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