一篇非常好的 Hibernate? 文章
出自： http://guxing.blog.enorth.com.cn/article/174717.shtml
Hibernate最讓人頭大的就是對(duì)集合的加載形式。
書(shū)看了N次了，還是沒(méi)有真正理解Hibernate。所以下午專門(mén)做了下測(cè)試，對(duì)配置文件的意思加深了認(rèn)識(shí)。
假設(shè)有兩個(gè)表，Photos(一）? ---? picture(多）Photo包含picture集合
結(jié)論1： HQL代碼 > fetch（配置） > lazy （配置）
結(jié)論2： 默認(rèn) lazy="true"
結(jié)論3： fetch 和 lazy 主要是用來(lái)級(jí)聯(lián)查詢的，?? 而 cascade 和 inverse 主要是用來(lái)級(jí)聯(lián)插入和修改的
結(jié)論4： 如果你是用spring來(lái)幫你管理你的session, 并且是自動(dòng)提交，延遲加載就等于沒(méi)加載~_~(當(dāng)然除非你手動(dòng)重新打開(kāi)session然后手動(dòng)Hibernate.initialize(set);然后關(guān)閉session.
結(jié)論5:? cascade主要是簡(jiǎn)化了在代碼中的級(jí)聯(lián)更新和刪除。
j結(jié)論6：老爸可以有多個(gè)孩子，一個(gè)孩子不能有多個(gè)老爸，而且老爸說(shuō)的算, 孩子圍著老爸轉(zhuǎn)。所以Photos老爸要有權(quán)力所以 cascade?這個(gè)關(guān)鍵子都是送給老爸的，?也就是級(jí)聯(lián)更新，老爸改姓了，兒子也得跟著改，呵呵?！安蝗?，就沒(méi)有零花錢(qián)咯”。而Picture兒子整體挨罵，但是還是要維護(hù)父子之間良好的關(guān)系，對(duì)老爸百依百順，所以老爸就說(shuō)，兒子，“關(guān)系，由你來(lái)維護(hù)（inverse="true")?，不然就不給零花錢(qián)。呵?！薄?br />?????????????? <set name="pictures" inverse="true" cascade="all">
??????????????????? <key>
?????????????????????? <column name="photosid" not-null="true" />
??????????????????? </key>
???????????????? <one-to-many class="girl.domain.Picture" />
???????? ??? </set>
???????????????
測(cè)試代碼：
???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>
測(cè)試過(guò)程會(huì)對(duì)配置文件不斷修改：并且從來(lái)不曾手動(dòng)重新打開(kāi)session
測(cè)試結(jié)構(gòu)：
當(dāng)配置條件為 lazy=true一句查詢 測(cè)試代碼中沒(méi)有調(diào)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 一句查詢 有g(shù)etPicture()
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 一句查詢? 有g(shù)etPicture() 并且訪問(wèn)了里面的元數(shù)Picture 且有異常拋出
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" 兩句查詢? 肯定沒(méi)問(wèn)題，因?yàn)槿繑?shù)據(jù)都個(gè)查了出來(lái) 所以怎么調(diào)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"? 一句查詢? 效果 ＝＝ lazy="false" 呵呵，哪個(gè)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"一句查詢? 沒(méi)有g(shù)etPicture() 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" 一句查詢? 有g(shù)etPicture() 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"一句查詢 有g(shù)etPicture() 并且訪問(wèn)里面的元素Picture的ID 有異常拋出
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=?
來(lái)個(gè)兩兵交戰(zhàn) fetch="join" lazy="true"? 呵呵 結(jié)果，一句查詢， 結(jié)構(gòu)正常 所以就當(dāng)lazy不存在好了。 看來(lái)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=?