關於Hiberante FetchMode

以Person, Address 爲例， 其關係爲一對多，數據庫中有3條person, 每個分別對應2條address。查詢語句如下： 

Criteria c = session.createCriteria(Person.class);
c.setResultTransformer(Criteria.DISTINCT_ROOT_ENTITY);
List<Person> list = (List<Person>)c.list();
for (Person p : list) {
        System.out.println(p.getName());

    Set<Address> addressSet = p.getAddressSet();
    for (Address a : addressSet) {
        System.out.println(a);
    }
}

1 默認不設置FetchMode 

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

結果生成如下 

Hibernate: select this_.PERSON_ID as PERSON1_0_0_, this_.PERSON_NAME as PERSON2_0_0_
from PERSON this_
P1
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_
from ADDRESS addressset0_
where addressset0_.P_ID=?
2   P1 A2
1   P1 A1
P2
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_
from ADDRESS addressset0_
where addressset0_.P_ID=?
4   P2 A2
3   P2 A1
P3
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_
from ADDRESS addressset0_
where addressset0_.P_ID=?
5   P3 A1
6   P3 A2

即：先查所有Person的id， 然後根據id查對應的Address。產生N＋1問題 

2. FetchMode.SELECT 

@OneToMany(cascade = CascadeType.ALL, fetch = FetchType.LAZY)
@Fetch(FetchMode.SELECT)
@JoinColumn(name = "P_ID")
public Set<Address> getAddressSet() {
    return addressSet;
}

效果同默認，即hibernate默認的FetchMode是SELECT 

3. FetchMode.JOIN 

@OneToMany(cascade = CascadeType.ALL, fetch = FetchType.LAZY)
@Fetch(FetchMode.JOIN)
@JoinColumn(name = "P_ID")
public Set<Address> getAddressSet() {
    return addressSet;
}

效果如下 

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_
from PERSON this_
   left outer join ADDRESS addressset2_
        on this_.PERSON_ID=addressset2_.P_ID

P1
2   P1 A2
1   P1 A1
P2
4   P2 A2
3   P2 A1
P3
5   P3 A1
6   P3 A2

採用外聯，用一條sql取出person及其address 

4. FetchMode.SUBSELECT 

@OneToMany(cascade = CascadeType.ALL, fetch = FetchType.LAZY)
@Fetch(FetchMode.SUBSELECT)
@JoinColumn(name = "P_ID")
public Set<Address> getAddressSet() {
    return addressSet;
}

效果 

Hibernate: select this_.PERSON_ID as PERSON1_0_0_, this_.PERSON_NAME as PERSON2_0_0_
from PERSON this_
P1
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_
from ADDRESS addressset0_
where addressset0_.P_ID in (select this_.PERSON_ID from PERSON this_)
2   P1 A2
1   P1 A1
P2
4   P2 A2
3   P2 A1
P3
6   P3 A2
5   P3 A1

生成2條sql， 第二句用in 查關聯Address數據 


5. batchSize 

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

效果 

Hibernate: select this_.PERSON_ID as PERSON1_0_0_, this_.PERSON_NAME as PERSON2_0_0_
from PERSON this_
P1
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_
from ADDRESS addressset0_
where addressset0_.P_ID in (?, ?, ?)
1   P1 A1
2   P1 A2
P2
4   P2 A2
3   P2 A1
P3
5   P3 A1
6   P3 A2

生成2條sql，第二條用in。 這裏數據庫中共有3條person。 
如果設size＝0或1， 則效果同select一樣，產生N＋1問題。 
如size＝2， 生成3條sql，第一條相同，第二條爲 

select 。。。。
from ADDRESS addressset0_
where addressset0_.P_ID in (?, ?)

第三條爲 

select。。
from ADDRESS addressset0_
where addressset0_.P_ID=?

如果設size=3則生成2條sql，第一條相同，第一條爲 

select 。。。。
from ADDRESS addressset0_
where addressset0_.P_ID in (?, ?， ？)

由上，size即爲in中數據個數。