Question

我在我的项目中使用spring数据来实体：

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（strategy = GenerationType.IDENTITY） private Long id;

@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;

我想写一个带有弹簧数据的查询，可以用他们自己的狗和汽车取得所有人。可以请任何身体帮助

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<>();
}

犬：

@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
}

然后只是查询人，你将获得所有关系。

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

另请注意，我是如何创建JoinTable的。如果您不自行管理，则不需要手动执行此操作（如果您在某些条件下填充它或者如果您有其他列，则需要进行管理）。

您的模型中还有一些奇怪的事情，为什么first_name，last_name和date_of_birth是唯一的？许多人可能拥有这些专栏的相似数据。

Answer 2

像这样更新您的实体：

<强>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<>();

    ...

}

<强>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;

    ...

}

<强>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，然后用狗和汽车取出所有人

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热切地抓狗和车。否则，您可能会遇到n + 1问题或Lazy fetch问题。

使用spring数据加入三个实体

2 个答案: