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关于hiberante FetchMode

发表于: 2012-07-05   作者:为了明天   来源:转载   浏览:
摘要: 以Person, Address 为例, 其关系为一对多,数据库中有3条person, 每个分别对应2条address。查询语句如下: Java代码  1.Criteria c = session.createCriteria(Person.class); 2.c.setResultTransformer(Criteria.DISTINCT_ROOT_ENT
以Person, Address 为例, 其关系为一对多,数据库中有3条person, 每个分别对应2条address。查询语句如下:




Java代码 
1.Criteria c = session.createCriteria(Person.class);  
2.c.setResultTransformer(Criteria.DISTINCT_ROOT_ENTITY);  
3.List<Person> list = (List<Person>)c.list();  
4.for (Person p : list) {  
5.        System.out.println(p.getName());  
6.              
7.    Set<Address> addressSet = p.getAddressSet();  
8.    for (Address a : addressSet) {  
9.        System.out.println(a);  
10.    }             
11.}  



1 默认不设置FetchMode



Java代码 
1.@OneToMany(cascade = CascadeType.ALL, fetch = FetchType.LAZY)  
2.@JoinColumn(name = "P_ID")  
3.public Set<Address> getAddressSet() {  
4.    return addressSet;  
5.}  



结果生成如下



Java代码 
1.Hibernate: select this_.PERSON_ID as PERSON1_0_0_, this_.PERSON_NAME as PERSON2_0_0_   
2.from PERSON this_  
3.P1  
4.Hibernate: select addressset0_.P_ID as P3_1_, addressset0_.ADDRESS_ID as ADDRESS1_1_, addressset0_.ADDRESS_ID as ADDRESS1_1_0_, addressset0_.ADDRESS_NAME as ADDRESS2_1_0_   
5.from ADDRESS addressset0_   
6.where addressset0_.P_ID=?  
7.2   P1 A2  
8.1   P1 A1  
9.P2  
10.Hibernate: select addressset0_.P_ID as P3_1_, addressset0_.ADDRESS_ID as ADDRESS1_1_, addressset0_.ADDRESS_ID as ADDRESS1_1_0_, addressset0_.ADDRESS_NAME as ADDRESS2_1_0_   
11.from ADDRESS addressset0_   
12.where addressset0_.P_ID=?  
13.4   P2 A2  
14.3   P2 A1  
15.P3  
16.Hibernate: select addressset0_.P_ID as P3_1_, addressset0_.ADDRESS_ID as ADDRESS1_1_, addressset0_.ADDRESS_ID as ADDRESS1_1_0_, addressset0_.ADDRESS_NAME as ADDRESS2_1_0_   
17.from ADDRESS addressset0_   
18.where addressset0_.P_ID=?  
19.5   P3 A1  
20.6   P3 A2  


即:先查所有Person的id, 然后根据id查对应的Address。产生N+1问题

2. FetchMode.SELECT




Java代码 
1.@OneToMany(cascade = CascadeType.ALL, fetch = FetchType.LAZY)  
2.@Fetch(FetchMode.SELECT)  
3.@JoinColumn(name = "P_ID")  
4.public Set<Address> getAddressSet() {  
5.    return addressSet;  
6.}  


效果同默认,即hibernate默认的FetchMode是SELECT

3. FetchMode.JOIN



Java代码 
1.@OneToMany(cascade = CascadeType.ALL, fetch = FetchType.LAZY)  
2.@Fetch(FetchMode.JOIN)  
3.@JoinColumn(name = "P_ID")  
4.public Set<Address> getAddressSet() {  
5.    return addressSet;  
6.}  


效果如下



Java代码 
1.Hibernate: select this_.PERSON_ID as PERSON1_0_1_, this_.PERSON_NAME as PERSON2_0_1_, addressset2_.P_ID as P3_3_, addressset2_.ADDRESS_ID as ADDRESS1_3_, addressset2_.ADDRESS_ID as ADDRESS1_1_0_, addressset2_.ADDRESS_NAME as ADDRESS2_1_0_   
2.from PERSON this_   
3.   left outer join ADDRESS addressset2_   
4.        on this_.PERSON_ID=addressset2_.P_ID  
5.  
6.P1  
7.2   P1 A2  
8.1   P1 A1  
9.P2  
10.4   P2 A2  
11.3   P2 A1  
12.P3  
13.5   P3 A1  
14.6   P3 A2  



采用外联,用一条sql取出person及其address

4. FetchMode.SUBSELECT



Java代码 
1.@OneToMany(cascade = CascadeType.ALL, fetch = FetchType.LAZY)  
2.@Fetch(FetchMode.SUBSELECT)  
3.@JoinColumn(name = "P_ID")  
4.public Set<Address> getAddressSet() {  
5.    return addressSet;  
6.}  



效果



Java代码 
1.Hibernate: select this_.PERSON_ID as PERSON1_0_0_, this_.PERSON_NAME as PERSON2_0_0_   
2.from PERSON this_  
3.P1  
4.Hibernate: select addressset0_.P_ID as P3_1_, addressset0_.ADDRESS_ID as ADDRESS1_1_, addressset0_.ADDRESS_ID as ADDRESS1_1_0_, addressset0_.ADDRESS_NAME as ADDRESS2_1_0_   
5.from ADDRESS addressset0_   
6.where addressset0_.P_ID in (select this_.PERSON_ID from PERSON this_)  
7.2   P1 A2  
8.1   P1 A1  
9.P2  
10.4   P2 A2  
11.3   P2 A1  
12.P3  
13.6   P3 A2  
14.5   P3 A1 


生成2条sql, 第二句用in 查关联Address数据


5. batchSize



Java代码 
1.@OneToMany(cascade = CascadeType.ALL, fetch = FetchType.LAZY)  
2.@BatchSize(size=4)  
3.@JoinColumn(name = "P_ID")  
4.public Set<Address> getAddressSet() {  
5.    return addressSet;  
6.}  


效果



Java代码 
1.Hibernate: select this_.PERSON_ID as PERSON1_0_0_, this_.PERSON_NAME as PERSON2_0_0_   
2.from PERSON this_  
3.P1  
4.Hibernate: select addressset0_.P_ID as P3_1_, addressset0_.ADDRESS_ID as ADDRESS1_1_, addressset0_.ADDRESS_ID as ADDRESS1_1_0_, addressset0_.ADDRESS_NAME as ADDRESS2_1_0_   
5.from ADDRESS addressset0_   
6.where addressset0_.P_ID in (?, ?, ?)  
7.1   P1 A1  
8.2   P1 A2  
9.P2  
10.4   P2 A2  
11.3   P2 A1  
12.P3  
13.5   P3 A1  
14.6   P3 A2  


生成2条sql,第二条用in。 这里数据库中共有3条person。
如果设size=0或1, 则效果同select一样,产生N+1问题。
如size=2, 生成3条sql,第一条相同,第二条为



Java代码 
1.select 。。。。  
2.from ADDRESS addressset0_   
3.where addressset0_.P_ID in (?, ?)  


第三条为



Java代码 
1.select。。  
2.from ADDRESS addressset0_   
3.where addressset0_.P_ID=?  



如果设size=3则生成2条sql,第一条相同,第一条为



Java代码 
1.select 。。。。  
2.from ADDRESS addressset0_   
3.where addressset0_.P_ID in (?, ?, ?)  


由上,size即为in中数据个数。

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