学习 Hibernate fetch lazy cascade inverse 关键字

大家都知道，在hibernate里为了性能考虑，引进了lazy的概念，这里我们以Parent和Child为模型来说明，

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--> public class Parent implements Serializable{

/** identifierfield */
private Longid;

/** persistentfield */
private Listchilds;

// skipallgetter/settermethod


}

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--> public class Child implements Serializable{

/** identifierfield */
private Longid;

/** persistentfield */
private net.foxlog.model.Parentparent;

//skip all getter/setter method

}

在我们查询Parent对象的时候，默认只有Parent的内容，并不包含childs的信息，如果在Parent.hbm.xml里设置lazy="false"的话才同时取出关联的所有childs内容.

问题是我既想要hibernate默认的性能又想要临时的灵活性该怎么办？ 这就是fetch的功能。我们可以把fetch与lazy="true"的关系类比为事务当中的编程式事务与声明式事务,不太准确，但是大概是这个意思。

总值，fetch就是在代码这一层给你一个主动抓取得机会.

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--> Parentparent = (Parent)hibernateTemplate.execute( new HibernateCallback(){
public ObjectdoInHibernate(Sessionsession) throws HibernateException,SQLException{
Queryq = session.createQuery(
" fromParentasparent " +
" leftouterjoinfetchparent.childs " +
" whereparent.id=:id "
);
q.setParameter( " id " , new Long( 15 ));
return (Parent)q.uniqueResult();
}

});

Assert.assertTrue(parent.getChilds().size() > 0 );

你可以在lazy="true"的情况下把fetch去掉，就会报异常. 当然，如果lazy="false"就不需要fetch了


有一个问题,使用Fetch会有重复记录的现象发生,我们可以理解为Fetch实际上不是为Parent服务的,而是为Child服务的.所以直接取Parent会有不匹配的问题.



参考一下下面的这篇文章
Hibernate集合初始化

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update:以上有些结论错误，实际上在hibernate3.2.1版本下测试，可以不出现重复记录，

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--> public void testNPlusOne() throws Exception{
Listlist = (List)hibernateTemplate.execute( new HibernateCallback(){
public ObjectdoInHibernate(Sessionsession) throws HibernateException,SQLException{
Queryq = session.createQuery(
" selectdistinctpfromnet.foxlog.model.Parentpinnerjoinfetchp.childs "
);
return q.list();
}

});

// ((Parent)(list.get(0))).getChilds();
System.out.println( " listsize= " + list.size());
for ( int i = 0 ;i < list.size();i ++ ){
Parentp = (Parent)list.get(i);
System.out.println( " ===parent= " + p);
System.out.println( " ===parent'schild'slength= " + p.getChilds().size());
}

}

打印结果如下:

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--> Hibernate:selectdistinctparent0_.idasid2_0_,childs1_.idasid0_1_,childs1_.parent_idasparent2_0_1_,childs1_.parent_idasparent2_0__,childs1_.idasid0__fromparentparent0_innerjoinchildchilds1_onparent0_.id = childs1_.parent_id
listsize = 3
=== parent = net.foxlog.model.Parent@1401d28[id = 14 ]
=== parent ' schild ' slength = 1
=== parent = net.foxlog.model.Parent@14e0e90[id = 15 ]
=== parent ' schild ' slength = 2
=== parent = net.foxlog.model.Parent@62610b[id = 17 ]
=== parent ' schild ' slength = 3

另外，如果用open session in view模式的话一般不用fetch,但首先推荐fetch,如果非用的话因为有N+1的现象，所以可以结合batch模式来改善下性能.

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Hibernate最让人头大的就是对集合的加载形式。
书看了N次了，还是没有真正理解Hibernate。所以下午专门做了下测试，对配置文件的意思加深了认识。

假设有两个表，Photos(一） --- picture(多）Photo包含picture集合

结论1： HQL代码 > fetch（配置） > lazy （配置）
结论2： 默认 lazy="true"
结论3： fetch 和 lazy 主要是用来级联查询的， 而 cascade 和 inverse 主要是用来级联插入和修改的
结论4： 如果你是用spring来帮你管理你的session, 并且是自动提交，延迟加载就等于没加载~_~(当然
除非你手动重新打开session然后手动Hibernate.initialize(set);然后关闭session.
结论5:cascade主要是简化了在代码中的级联更新和删除。
j结论6：老爸可以有多个孩子，一个孩子不能有多个老爸，而且老爸说的算, 孩子围着老爸转。
所以Photos老爸要有权力所以 cascade这个关键子都是送给老爸的，也就是级联更新，
老爸改姓了，儿子也得跟着改，呵呵。“不然，就没有零花钱咯”。
而Picture儿子整体挨骂，但是还是要维护父子之间良好的关系，对老爸百依百顺，所
以老爸就说，儿子，“关系，由你来维护（inverse="true")，不然就不给零花钱。呵。”。
<set name="pictures" inverse="true" cascade="all">
<key>
<column name="photosid" not-null="true" />
</key>
<one-to-many class="girl.domain.Picture" />
</set>

测试代码：

Photos p = ps.getById(1);
Set<Picture> set = p.getPictures();
for(Picture pic : set){
System.out.println(pic.getId());
}

配置文件的一部分：
<set name="pictures" inverse="true" cascade="all" >
<key>
<column name="photosid" not-null="true" />
</key>
<one-to-many class="girl.domain.Picture" />
</set>

测试过程会对配置文件不断修改：并且从来不曾手动重新打开session

测试结构：

当配置条件为 lazy=true 一句查询 测试代码中没有调用getPicture() 正常
Hibernate: select photos0_.id as id0_0_, photos0_.userid as userid0_0_, photos0_.typeid as typeid0_0_, photos0_.name as name0_0_, photos0_.createtime as createtime0_0_, photos0_.description as descript6_0_0_, photos0_.faceid as faceid0_0_, photos0_.uri as uri0_0_ from super.photos photos0_ where photos0_.id=?

lazy=true 一句查询 有getPicture()
Hibernate: select photos0_.id as id0_0_, photos0_.userid as userid0_0_, photos0_.typeid as typeid0_0_, photos0_.name as name0_0_, photos0_.createtime as createtime0_0_, photos0_.description as descript6_0_0_, photos0_.faceid as faceid0_0_, photos0_.uri as uri0_0_ from super.photos photos0_ where photos0_.id=?

lazy=true 一句查询 有getPicture() 并且访问了里面的元数Picture 且有异常抛出
Hibernate: select photos0_.id as id0_0_, photos0_.userid as userid0_0_, photos0_.typeid as typeid0_0_, photos0_.name as name0_0_, photos0_.createtime as createtime0_0_, photos0_.description as descript6_0_0_, photos0_.faceid as faceid0_0_, photos0_.uri as uri0_0_ from super.photos photos0_ where photos0_.id=?

lazy="false" 两句查询 肯定没问题，因为全部数据都个查了出来 所以怎么调用都正常
Hibernate: select photos0_.id as id0_0_, photos0_.userid as userid0_0_, photos0_.typeid as typeid0_0_, photos0_.name as name0_0_, photos0_.createtime as createtime0_0_, photos0_.description as descript6_0_0_, photos0_.faceid as faceid0_0_, photos0_.uri as uri0_0_ from super.photos photos0_ where photos0_.id=?
Hibernate: select pictures0_.photosid as photosid1_, pictures0_.id as id1_, pictures0_.id as id2_0_, pictures0_.photosid as photosid2_0_, pictures0_.name as name2_0_, pictures0_.clicked as clicked2_0_, pictures0_.uploaddate as uploaddate2_0_, pictures0_.size as size2_0_, pictures0_.description as descript7_2_0_, pictures0_.uri as uri2_0_ from super.picture pictures0_ where pictures0_.photosid=?

fetch="join" 一句查询 效果 ＝＝ lazy="false" 呵呵，哪个效率高，我就不知道了。。。。。。。。。。。
Hibernate: select photos0_.id as id0_1_, photos0_.userid as userid0_1_, photos0_.typeid as typeid0_1_, photos0_.name as name0_1_, photos0_.createtime as createtime0_1_, photos0_.description as descript6_0_1_, photos0_.faceid as faceid0_1_, photos0_.uri as uri0_1_, pictures1_.photosid as photosid3_, pictures1_.id as id3_, pictures1_.id as id2_0_, pictures1_.photosid as photosid2_0_, pictures1_.name as name2_0_, pictures1_.clicked as clicked2_0_, pictures1_.uploaddate as uploaddate2_0_, pictures1_.size as size2_0_, pictures1_.description as descript7_2_0_, pictures1_.uri as uri2_0_ from super.photos photos0_ left outer join super.picture pictures1_ on photos0_.id=pictures1_.photosid where photos0_.id=?

不加fetch＝"join" 一句查询 没有getPicture() 正常
Hibernate: select photos0_.id as id0_0_, photos0_.userid as userid0_0_, photos0_.typeid as typeid0_0_, photos0_.name as name0_0_, photos0_.createtime as createtime0_0_, photos0_.description as descript6_0_0_, photos0_.faceid as faceid0_0_, photos0_.uri as uri0_0_ from super.photos photos0_ where photos0_.id=?

不加fetch＝"join" 一句查询 有getPicture() 正常
Hibernate: select photos0_.id as id0_0_, photos0_.userid as userid0_0_, photos0_.typeid as typeid0_0_, photos0_.name as name0_0_, photos0_.createtime as createtime0_0_, photos0_.description as descript6_0_0_, photos0_.faceid as faceid0_0_, photos0_.uri as uri0_0_ from super.photos photos0_ where photos0_.id=?

不加fetch＝"join" 一句查询 有getPicture() 并且访问里面的元素Picture的ID 有异常抛出
Hibernate: select photos0_.id as id0_0_, photos0_.userid as userid0_0_, photos0_.typeid as typeid0_0_, photos0_.name as name0_0_, photos0_.createtime as createtime0_0_, photos0_.description as descript6_0_0_, photos0_.faceid as faceid0_0_, photos0_.uri as uri0_0_ from super.photos photos0_ where photos0_.id=?

来个两兵交战 fetch="join" lazy="true" 呵呵 结果，一句查询， 结构正常 所以就当lazy不存在好了。 看来fetch 是老大。、、、、、、、、、、、、、
Hibernate: select photos0_.id as id0_1_, photos0_.userid as userid0_1_, photos0_.typeid as typeid0_1_, photos0_.name as name0_1_, photos0_.createtime as createtime0_1_, photos0_.description as descript6_0_1_, photos0_.faceid as faceid0_1_, photos0_.uri as uri0_1_, pictures1_.photosid as photosid3_, pictures1_.id as id3_, pictures1_.id as id2_0_, pictures1_.photosid as photosid2_0_, pictures1_.name as name2_0_, pictures1_.clicked as clicked2_0_, pictures1_.uploaddate as uploaddate2_0_, pictures1_.size as size2_0_, pictures1_.description as descript7_2_0_, pictures1_.uri as uri2_0_ from super.photos photos0_ left outer join super.picture pictures1_ on photos0_.id=pictures1_.photosid where photos0_.id=?