重新学习 Hibernate fetch lazy cascade inverse 关键字

自：http://www.blogjava.net/pear/archive/2006/11/05/79251.html

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=?