Java 具有左连接和分组依据的JPQL查询

我正在使用简单的Hibernate 5.0.4、Spring 4.2.3和基于Oracle 11g XE的Maven 3.3.3项目学习JPQL。完整的源代码可以在我的网站上找到

我有两种型号：

import java.util.Date;
import java.util.LinkedList;
import java.util.List;

import javax.persistence.CascadeType;
import javax.persistence.Column;
import javax.persistence.Entity;
import javax.persistence.EnumType;
import javax.persistence.Enumerated;
import javax.persistence.FetchType;
import javax.persistence.GeneratedValue;
import javax.persistence.GenerationType;
import javax.persistence.Id;
import javax.persistence.OneToMany;
import javax.persistence.SequenceGenerator;
import javax.persistence.Table;
import javax.persistence.Temporal;
import javax.persistence.TemporalType;

import lombok.Data;
import lombok.NoArgsConstructor;

@Entity
@Table(name = "T_OWNER")
@NoArgsConstructor
public @Data class OwnerModel {

    public OwnerModel(String firstName, String lastName, Integer age, OwnerType type) {
        super();
        this.firstName = firstName;
        this.lastName = lastName;
        this.type = type;
        this.since = new Date(System.currentTimeMillis());
        this.age = age;
    }

    @Id
    @GeneratedValue(generator = "owner-sequence-generator", strategy = GenerationType.SEQUENCE)
    @SequenceGenerator(name = "owner-sequence-generator", sequenceName = "OWNER_SEQ", initialValue = 1, allocationSize = 20)
    private Long id;

    @Column(name = "FIRST_NAME")
    private String firstName;

    @Column(name = "LAST_NAME")
    private String lastName;

    @Column(name = "TYPE")
    @Enumerated(EnumType.STRING)
    private OwnerType type;

    @Column(name = "SINCE")
    @Temporal(TemporalType.TIME)
    private Date since;

    @Column(name = "AGE")
    private Integer age;

    @OneToMany(mappedBy = "owner", cascade = CascadeType.ALL, fetch = FetchType.EAGER, targetEntity = CarModel.class)
    private List<CarModel> cars = new LinkedList<>();

    public void addCar(CarModel car) {
        cars.add(car);
        car.setOwner(this);
    }
}

import java.sql.Blob;
import java.sql.Clob;
import java.util.Date;

import javax.persistence.CascadeType;
import javax.persistence.Column;
import javax.persistence.Entity;
import javax.persistence.FetchType;
import javax.persistence.GeneratedValue;
import javax.persistence.GenerationType;
import javax.persistence.Id;
import javax.persistence.JoinColumn;
import javax.persistence.Lob;
import javax.persistence.ManyToOne;
import javax.persistence.SequenceGenerator;
import javax.persistence.Table;
import javax.persistence.Temporal;
import javax.persistence.TemporalType;

import lombok.AllArgsConstructor;
import lombok.Data;
import lombok.NoArgsConstructor;
import lombok.ToString;

@Entity
@Table(name = "T_CAR")
@NoArgsConstructor
@AllArgsConstructor
@ToString(exclude = "owner")
public @Data class CarModel {

    public CarModel(String name, Integer wheelsNumber, Clob spec, Blob image) {
        super();
        this.name = name;
        this.wheelsNumber = wheelsNumber;
        this.spec = spec;
        this.image = image;
        this.createdIn = new Date(System.currentTimeMillis());
    }

    @Id
    @GeneratedValue(generator = "car-sequence-generator", strategy = GenerationType.SEQUENCE)
    @SequenceGenerator(name = "car-sequence-generator", sequenceName = "CAR_SEQ", initialValue = 1, allocationSize = 20)
    private Long id;

    @Column(name = "NAME")
    private String name;

    @Column(name = "CREATED_IN")
    @Temporal(TemporalType.TIMESTAMP)
    private Date createdIn;

    @Column(name = "WHEELS_NUMBER")
    private Integer wheelsNumber;

    @Lob
    @Column(name = "SPEC")
    private Clob spec;

    @Lob
    @Column(name = "IMAGE")
    private Blob image;

    @ManyToOne(cascade = CascadeType.ALL, fetch = FetchType.EAGER, targetEntity = OwnerModel.class)
    @JoinColumn(name = "ID_OWNER")
    private OwnerModel owner;
}

据我所知，这是由Hibernate生成的SQL中的错误引起的：

select
    ownermodel0_.id as col_0_0_,
    count(cars1_.id) as col_1_0_,
    ownermodel0_.id as id1_1_,
    ownermodel0_.AGE as AGE2_1_,
    ownermodel0_.FIRST_NAME as FIRST_NAME3_1_,
    ownermodel0_.LAST_NAME as LAST_NAME4_1_,
    ownermodel0_.SINCE as SINCE5_1_,
    ownermodel0_.TYPE as TYPE6_1_ 
from
    hibernate.T_OWNER ownermodel0_ 
left outer join
    hibernate.T_CAR cars1_ 
        on ownermodel0_.id=cars1_.ID_OWNER 
group by
    ownermodel0_.id

其中，

groupby

子句应包括所有列（而不仅仅是id）。我认为hibernate应该生成这样的SQL：

select
    ownermodel0_.id as col_0_0_,
    count(cars1_.id) as col_1_0_,
    ownermodel0_.id as id1_1_,
    ownermodel0_.AGE as AGE2_1_,
    ownermodel0_.FIRST_NAME as FIRST_NAME3_1_,
    ownermodel0_.LAST_NAME as LAST_NAME4_1_,
    ownermodel0_.SINCE as SINCE5_1_,
    ownermodel0_.TYPE as TYPE6_1_ 
from
    hibernate.T_OWNER ownermodel0_ 
left outer join
    hibernate.T_CAR cars1_ 
        on ownermodel0_.id=cars1_.ID_OWNER 
group by
    ownermodel0_.id,
    ownermodel0_.AGE,
    ownermodel0_.FIRST_NAME,
    ownermodel0_.LAST_NAME,
    ownermodel0_.SINCE,
    ownermodel0_.TYPE; 

但是（据我所知）和显示的JPQL查询与我的完全相同

它是Hibernate中的一个bug还是我代码中的一个隐藏bug？

感谢您的帮助和建议。

让我们最好这样做：

select 
    o.column1, o.column2, COUNT(c) 

from 
    OwnerModel o 

LEFT JOIN 
    o.cars c 

GROUP BY 
    o.column1, o.column2

---

因为

count

方法会破坏所有的查询概念

所以它对hibernate不起作用。 正如你们所看到的，Jira中有一个与此相关的问题——它处于未解决状态

你们提供的第一个链接是JPA规范，第二个链接并没有按模型分组查询，只是按简单的数字字段

---

正如@GingerHead所回答的，在所有引用中修改查询将更容易，分组发生在简单字段上，聚合发生在简单字段上，因此在建议的示例中，SQL将起作用

您的代码似乎有问题，您正在尝试获取一个急切的加载的集合，作为分组方式的一部分

尽管如此，您的

列表

始终是预填充的（它被标记为

急切的

）-因此要获得计数，只需在加载

所有者模型

实体后获得列表的长度即可

我建议您修改模型和DAO层，以便从

OwnerModel

中删除

@OneToMany List cars

字段。如果您总是需要可用

CarModels

的计数，只需将此字段作为

@Formula

表达式添加到您的

OwnerModel

实体中即可

select
    ownermodel0_.id as col_0_0_,
    count(cars1_.id) as col_1_0_,
    ownermodel0_.id as id1_1_,
    ownermodel0_.AGE as AGE2_1_,
    ownermodel0_.FIRST_NAME as FIRST_NAME3_1_,
    ownermodel0_.LAST_NAME as LAST_NAME4_1_,
    ownermodel0_.SINCE as SINCE5_1_,
    ownermodel0_.TYPE as TYPE6_1_ 
from
    hibernate.T_OWNER ownermodel0_ 
left outer join
    hibernate.T_CAR cars1_ 
        on ownermodel0_.id=cars1_.ID_OWNER 
group by
    ownermodel0_.id,
    ownermodel0_.AGE,
    ownermodel0_.FIRST_NAME,
    ownermodel0_.LAST_NAME,
    ownermodel0_.SINCE,
    ownermodel0_.TYPE; 

select 
    o.column1, o.column2, COUNT(c) 

from 
    OwnerModel o 

LEFT JOIN 
    o.cars c 

GROUP BY 
    o.column1, o.column2