我有这些课程：
订单：
@Table(name = "ORDERS")
public class Order {
@Id
@Column(name = "order_id")
@GeneratedValue(strategy = GenerationType.IDENTITY)
private Long id;
@Column(name = "user_id", nullable = false)
private Long userId;
@JsonManagedReference
@OneToMany(mappedBy = "order", fetch = FetchType.EAGER)
private Set<OrderDetail> orderDetails;
@JsonManagedReference(value = "order-note")
@OneToMany(mappedBy = "order", fetch = FetchType.EAGER)
private Set<Note> notes;
}
详情
@Table(name = "ORDER_DETAILS")
public class OrderDetail {
@Id
@Column(name = "order_detail_id")
@GeneratedValue(strategy = GenerationType.IDENTITY)
private Long id;
@JsonBackReference
@ManyToOne
@JoinColumn(name = "order_id")
private Order order;
@JsonManagedReference
@OneToMany(mappedBy = "orderDetail", fetch = FetchType.EAGER)
private Set<OrderSize> orderSizes;
@JsonManagedReference(value="order-detail-note")
@OneToMany(mappedBy = "orderDetail", fetch = FetchType.EAGER)
private Set<Note> notes;
}
尺寸：
@Table(name = "ORDER_SIZES")
public class OrderSize {
@Id
@Column(name = "order_size_id")
@GeneratedValue(strategy = GenerationType.IDENTITY)
private Long id;
@JsonBackReference
@ManyToOne
@JoinColumn(name = "order_detail_id")
private OrderDetail orderDetail;
}
备注
@Table(name = "NOTES")
public class Note {
@Id
@Column(name = "note_id")
@GeneratedValue(strategy = GenerationType.IDENTITY)
private Long id;
@JsonBackReference(value="order-note")
@ManyToOne
@JoinColumn(name = "order_id", nullable = true)
private Order order;
@JsonBackReference(value="order-detail-note")
@ManyToOne
@JoinColumn(name = "order_detail_id", nullable = true)
private OrderDetail orderDetail;
}
我想执行查询（使用Hibernate＆＃39;标准），并获取所有订单及其相关的详细信息，大小和注释。
public List<Order> findByUser(Long userId) {
Criteria criteria = createEntityCriteria();
criteria.add(Restrictions.eq("userId", userId));
return (List<Order>) criteria.list();
}
目前Hibernate会生成如下的n个查询：
Hibernate:
select
notes0_.order_detail_id as order_de4_17_0_,
notes0_.note_id as note_id1_15_0_,
notes0_.note_id as note_id1_15_1_,
notes0_.note_text as note_tex2_15_1_,
notes0_.order_id as order_id3_15_1_,
notes0_.order_detail_id as order_de4_15_1_,
order1_.order_id as order_id1_16_2_,
order1_.client_id as client_i2_16_2_,
order1_.company_id as company_3_16_2_,
order1_.delivery_id as delivery4_16_2_,
order1_.discount as discount5_16_2_,
order1_.order_date as order_da6_16_2_,
order1_.order_name as order_na7_16_2_,
order1_.signature as signatur8_16_2_,
order1_.order_status as order_st9_16_2_,
order1_.payment_method_id as payment10_16_2_,
order1_.user_id as user_id11_16_2_
from
NOTES notes0_
left outer join
ORDERS order1_
on notes0_.order_id=order1_.order_id
where
notes0_.order_detail_id=?
我还没有能够在一次查询中找到解决方案来获取我需要的所有内容。
答案 0 :(得分：1)
插入后：
在Order实体中使用以下注释：
@OneToMany(mappedBy = "order", fetch = FetchType.EAGER)
@Fetch(FetchMode.JOIN)
private Set<OrderDetail> orderDetails = new HashSet<>();
@OneToMany(mappedBy = "order", fetch = FetchType.EAGER)
@Fetch(FetchMode.JOIN)
private Set<Note> notes;
生成以下查询：
Hibernate:
select
this_.order_id as order_id1_3_4_,
this_.user_id as user_id2_3_4_,
notes2_.order_id as order_id2_0_6_,
notes2_.note_id as note_id1_0_6_,
notes2_.note_id as note_id1_0_0_,
notes2_.order_id as order_id2_0_0_,
notes2_.order_detail_id as order_de3_0_0_,
orderdetai3_.order_detail_id as order_de1_1_1_,
orderdetai3_.order_id as order_id2_1_1_,
order4_.order_id as order_id1_3_2_,
order4_.user_id as user_id2_3_2_,
ordersizes5_.order_detail_id as order_de2_2_7_,
ordersizes5_.order_size_id as order_si1_2_7_,
ordersizes5_.order_size_id as order_si1_2_3_,
ordersizes5_.order_detail_id as order_de2_2_3_
from
orders this_
left outer join
notes notes2_
on this_.order_id=notes2_.order_id
left outer join
order_details orderdetai3_
on notes2_.order_detail_id=orderdetai3_.order_detail_id
left outer join
orders order4_
on orderdetai3_.order_id=order4_.order_id
left outer join
order_sizes ordersizes5_
on orderdetai3_.order_detail_id=ordersizes5_.order_detail_id
where
this_.user_id=?
Hibernate:
select
orderdetai0_.order_id as order_id2_1_0_,
orderdetai0_.order_detail_id as order_de1_1_0_,
orderdetai0_.order_detail_id as order_de1_1_1_,
orderdetai0_.order_id as order_id2_1_1_
from
order_details orderdetai0_
where
orderdetai0_.order_id=?
如果您仍然面临将查询数据作为一个查询的困难，我建议使用QueryDSL [1]，其中包含示例和maven设置详细信息[2]。
<强> LE
在QueryDSL中创建查询的示例如下：
public List<Order> findByUserQueryDsl(Long userId) {
System.out.println("Finding by User ID query dsl");
EntityManager em = this.entityManagerFactory.createEntityManager();
return new JPAQuery<>(em, HQLTemplates.DEFAULT)
.select(QOrder.order)
.from(QOrder.order)
.leftJoin(QOrder.order.notes).fetchJoin()
.leftJoin(QOrder.order.orderDetails, QOrderDetail.orderDetail).fetchJoin()
.leftJoin(QOrderDetail.orderDetail.notes).fetchJoin()
.where(QOrder.order.userId.eq(userId))
.fetch();
}
哪会生成以下查询：
select
order0_.order_id as order_id1_3_0_,
notes1_.note_id as note_id1_0_1_,
orderdetai2_.order_detail_id as order_de1_1_2_,
notes3_.note_id as note_id1_0_3_,
order0_.user_id as user_id2_3_0_,
notes1_.order_id as order_id2_0_1_,
notes1_.order_detail_id as order_de3_0_1_,
notes1_.order_id as order_id2_0_0__,
notes1_.note_id as note_id1_0_0__,
orderdetai2_.order_id as order_id2_1_2_,
orderdetai2_.order_id as order_id2_1_1__,
orderdetai2_.order_detail_id as order_de1_1_1__,
notes3_.order_id as order_id2_0_3_,
notes3_.order_detail_id as order_de3_0_3_,
notes3_.order_detail_id as order_de3_0_2__,
notes3_.note_id as note_id1_0_2__
from
orders order0_
left outer join
notes notes1_
on order0_.order_id=notes1_.order_id
left outer join
order_details orderdetai2_
on order0_.order_id=orderdetai2_.order_id
left outer join
notes notes3_
on orderdetai2_.order_detail_id=notes3_.order_detail_id
where
order0_.user_id=?
[2] http://www.querydsl.com/static/querydsl/4.1.4/reference/html/ch02s02.html