Hibernate 各种映射用法

1. 值集合映射：这些集合包含的元素不是Domain Object，是一些值对象

? 1.1 Set映射：
????????<set?name="Set属性名"?table="表名">
????????????<key?column="与主键关联字段名"?/>
????????????<element?type="数据类型"?column="字段名"?/>
????????</set>
? 1.2 List映射：
????????<list?name="List属性名"?table="表名">
????????????<key?column="与主键关联字段名"?/>
????????????<list-index?column="索引字段名"?/>
????????????<element?type="数据类型"?column="字段名"?/>
????????</list>
? 1.3Map映射：
????????<map?name="Map属性名"?table="表名">
????????????<key?column="与主键关联字段名"?/>
????????????<map-key?type="数据类型"?column="KEY字段名"?/>
????????????<element?type="数据类型"?column="VALUE字段名"?/>
????????</map>

??1.4 Bag映射：
????????<bag?name="bag属性名"?table="表名">
????????????<key?column="与主键关联字段名"?/>????????
????????????<element?type="数据类型"?column="字段名"?/>
????????</bag>
? 1.5 Idbag映射：
????????<idbag?name="idbag?属性名"?table="表名">?

??????????? <collection-id column="cid" type="java.lang.String">

??????????????????? <generator name="code">public class User { private List<String> emails=new ArrayList<String>(); public void addEmail(String email) { emails.add(email); } public void removeEmail(String email) { emails.remove(email); }CREATE TABLE user ( id INT(11) NOT NULL auto_increment PRIMARY KEY);CREATE TABLE emails( userId INT(11) NOT NULL, email VARCHAR(100) NOT NULL);<bag name="emails" table="emails"> <key column="userId" /> <element column="email" type="java.lang.String" /> </bag> <idbag name="emails" table="emails"> <collection-id column="cid" type="java.lang.String"> <generator /> </collection-id> <key column="email" /> <element column="email" type="java.lang.String" /> </idbag>User user1=new User();user1.addEmail("email1@139.com");user1.addEmail("email2@139.com");user1.addEmail("email2@139.com");User user2=new User();user2.addEmail("email3@139.com");//save user//user table id 1 2

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?如果采用bag，则emails表如下：

userid???????? email
? ? 1???????????? email1
?? ?1???????? ??? email2
??? 1???????????? email2
? ? 2???????????? email3

执行user1.remove(email2),根据List的remove方法，会删除第一个匹配的元素，但是Hibernate无法定位此元素，因而采取的方式为先全部清空，再重新插入数据库，效率低下，生成的sql意思如下：

delete from emails where userId=1；

insert into emails(userId,email) values(1,email1);

insert into emails(userId,email) values(1,email2);

insert into emails(userId,email) values(2,email3);

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如果采用Idbag，则emails表如下：

?cid? ?????? userid???????? email
?cid1???????? ?1???????????? email1
?cid2???????? ?1???????? ??? email2
?cid3??????????1???????????? email2
?cid4????????? 2???????????? email3

执行user1.remove(email2),生成的sql意思如下：

delete from emails where cid=cid2；

此方式效率较高。

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2. 实体关系映射

?<many-to-one name="company" column="companyid" table="subproject_programmer"? inverse="true">
?????? ?<key column="programmerid"/>
????????<many-to-many column="subprojectid"/>
??</set>

??<bag name="subscriptions" table="subscription" >
?????????<key column="companyid"/>
?????????<one-to-many discriminator -??

?????????? value="AdjustCategory">
????? </subclass>
????? <subclass name="com.ijo.domain.common.category.TransportCategory" discriminator-?

??????????? value="TransportCategory">
????? </subclass>

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3.2 <joined-subclass name="com.ijo.domain.user.CustomerUser" table="customeruser">
??????????? <key column="userid"/>
???????</joined-subclass>

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