QueryDSL@OneToone Join-FetchMode with Hibernate

Question

@Entity
public class Customer extends EntityBase {
    @Column(name = "name", nullable = true)
    private String name;

    @OneToOne(mappedBy = "customer")
    private Address address;

    // getter, setter, ...
}

@Entity
public class Address extends EntityBase {
    @OneToOne(optional = false)
    private Customer customer;

    @Column(nullable = true)
    private String street;
    @Column(nullable = true)
    private String zip;
    @Column(nullable = true)
    private String city;

    // getter, setter, ...
}

session.createCriteria(Customer.class).list();

select
    this_.id as id1_1_1_,
    this_.name as name2_1_1_,
    address2_.id as id1_0_0_,
    address2_.city as city2_0_0_,
    address2_.customer_id as customer5_0_0_,
    address2_.street as street3_0_0_,
    address2_.zip as zip4_0_0_ 
from
    Customer this_ 
left outer join
    Address address2_ 
        on this_.id=address2_.customer_id

new HibernateQuery<Customer>(session).from(QCustomer.customer).fetchResults();

使用QueryDSLHibernate的功能：

select
    count(customer0_.id) as col_0_0_
from
    Customer customer0_

select
    customer0_.id as id1_1_,
    customer0_.name as name2_1_
from
    Customer customer0_

select
    address0_.id as id1_0_1_,
    address0_.city as city2_0_1_,
    address0_.customer_id as customer5_0_1_,
    address0_.street as street3_0_1_,
    address0_.zip as zip4_0_1_,
    customer1_.id as id1_1_0_,
    customer1_.name as name2_1_0_
from
    Address address0_
inner join
    Customer customer1_
        on address0_.customer_id=customer1_.id
where
    address0_.customer_id=?

问题:
是否可以为QueryDSL查询设置全局FetchMode。在Hibernate中，您可以用@fetch(FetchMode.join)指定这一点，但不幸的是，QueryDSL忽略了这一点。

因此，我的目标是用QueryDSL加载1000个客户，并且只运行2个查询(count+select)。

new HibernateQuery<Customer>(session)
    .from(QCustomer.customer)
    .leftJoin(QCustomer.customer.address).fetchJoin()
    .fetchResults();

new HibernateQuery<Customer>(session)
    .from(QCustomer.customer)
    .where(QCustomer.customer.address.street.in("Musterstraße 12"))
    .fetchResults();

Answer 1

一旦我有同样的问题。但是我使用的是Hibernate JPA条件查询。如果您想在一个查询中获得结果，那么一种方法是使用

@OneToOne(fetch=FetchType.EAGER)
@JoinColumn(name = "address_id", insertable = false, updatable = false, referencedColumnName = "id")
private Address address;

或者我有一个条件查询的解决方案。可能会帮助你把它转换成DSL。

创建Customer类的根。

Root<Customer> root = . . .
Join<Customer, Address> join = (Join<Customer, Address>)root.fetch(Customer_.address);