使用 spring 数据加入三个实体

Question

我在我的项目中使用 spring 数据，以下是实体：

rr:

@Id
@GeneratedValue( strategy = GenerationType.IDENTITY )
private Long   id;
private String model;

@ManyToOne
@JoinColumn( name = "owner" )
private ffperson;
private String constructor;
//getters and setters

zaa:

@Id
@GeneratedValue( strategy = GenerationType.IDENTITY )
private Long   id;

@Column( name = "name" )
private String Name;
//getters and setters

gff @ID @GeneratedValue( 策略 = GenerationType.IDENTITY ) 私人长号;

@Column( name = "lastname", unique = true )
private String lastName;

@Column( name = "firstname", unique = true )
private String firstName;

@Column( name = "birth_date", unique = true )
private Date   dateOfBirth;

gfff:

@Id
@GeneratedValue( strategy = GenerationType.IDENTITY )
private Long   id;

@ManyToOne
@JoinColumn( name = "dog_id" )
private ffdog;

@ManyToOne
@JoinColumn( name = "person_id" )
private ffperson;

我想用 spring 数据编写一个查询，它可以获取所有拥有自己的狗和汽车的人..任何人都可以帮忙

Answer 1

获取它们的最简单方法 - 更改当前数据模型。

人：

@Entity
public class Person {

    @Id
    @GeneratedValue( strategy = GenerationType.IDENTITY )
    private Long id;

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

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

    @Column(name = "birth_date")
    private Date  dateOfBirth;

    @MayToMany(fetch = FetchType.EAGER)
    @JoinTable(name = "PERSON_DOG",
               joinColumns = {@JoinColumn(name = "PERSON_ID")},
               inverseJoinColumns = {@JoinColumn(name = "DOG_ID")})
    private List<Dog> dogs = new ArrayList<>();

    @OneToMany(mappedBy = "person")
    private List<Car> cars = new ArrayList<>();
}

狗：

@Entity
public class Dog {

    @Id
    @GeneratedValue(strategy = GenerationType.IDENTITY )
    private Long  id;

    @Column(name = "name" )
    private String Name;
    //getters and setters
}

汽车：

@Entity
public class Car {

    @Id
    @GeneratedValue(strategy = GenerationType.IDENTITY )
    private Long  id;
    private String model;

    @ManyToOne
    @JoinColumn(name = "owner")
    @JsonIgnore // this will prevent the loop on response
    private Person person;

    private String constructor;
    //getters and setters
}

然后只需查询 Person 即可获得所有关系。

public interface PersonRepository extends CrudRepository<Person, Long> {
    List<Person> findAll();
}

另请注意，我是如何创建 JoinTable 的。如果您不会自己管理它，则不需要手动执行（如果您要根据特定条件填充它或者如果您有额外的列，有时需要手动操作）。

你的模型也有一些奇怪的地方，为什么first_name、last_name和date_of_birth是唯一的？许多人可能对这些列有相似的数据。

Answer 2

像这样更新您的实体：

人

@Entity
@Table(name = "person")
public class Person {

    @Id
    @GeneratedValue(strategy = GenerationType.IDENTITY)
    private Long id;

    @OneToMany(mappedBy = "person", cascade = CascadeType.ALL)
    Set<Dog> dogs = new HashSet<>();

    @OneToMany(mappedBy = "person", cascade = CascadeType.ALL)
    Set<Car> cars = new HashSet<>();

    ...

}

狗

@Entity
@Table(name="dog")
public class Dog {

    @Id
    @GeneratedValue( strategy = GenerationType.IDENTITY )
    private Long   id;

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

    @ManyToOne
    private Person person;

    ...

}

汽车

@Entity
@Table(name = "car")
public class Car {

    @Id
    @GeneratedValue(strategy = GenerationType.IDENTITY)
    private Long id;

    @ManyToOne
    private Person person;

    ...

}

现在创建一个 PersonRepository 并获取所有有狗和车的人

public interface PersonRepository extends CrudRepository<Person, Long> {

    @Query("select p from Person p join fetch p.dogs join fetch p.cars")
    List<Person> findAll();

}

注意：使用Join Fetch to fetch dogs and cars eagerly. Otherwise you may get into n + 1问题或延迟获取问题。

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