4

私はまだ JPA の概念を把握しており、私の質問に対する答えがどこにも見つからないようです!

推定

どちらのクラスにも @GeneratedValue(strategy = GenerationType.IDENTITY) の注釈が付けられています。すべてのゲッターとセッターがあります。

Parent{
    ....
    @OneToMany(mappedBy = "parent")
    Collection<Child> children;
    ....
}

Child{
    ...
    @JoinColumn(name = "parent", referencedColumnName = "id", nullable = true)
    @ManyToOne(optional = false)
    Parent parent;
    ...
}

次に、標準の JpaRepository を実装し、コントローラーをセットアップしました

ここに問題があり
ます すべての子レコードを照会すると、特定の親にマップされた最初の子レコードだけが親エンティティ オブジェクトを持ちます。残りは、親エンティティを参照する ID を持ちます。

次に例を示します: POSTMAN からすべての子を取得すると、次のように返されます。

[
    {
        "id": 1,
        "name": "child1",
        "parent": {
            "id": 1,
            "firstName": "..."
            ...
            }
    },
    {
        "id": 2,
        "name": "child2",
        "parent": 1
    }
    {
        "id": 3,
        "name": "child3",
        "parent": {
            "id": 2,
            "firstName": "..."
            ...
            }
    },
    {
        "id": 4,
        "name": "child4",
        "parent": 2
    }
]

ご覧のとおり、最初にその親にマップされているchild2だけです! 同様に、最初にその親にマップされただけです!"parent": 1child1child4"parent": 2child3

誰でもこの動作を説明できますか? 親を試しfetch = FetchType.EAGERてみましたが、役に立ちませんでした！私は、すべての子が包括的な親オブジェクトを持ち、別の DB トリップを防ぐことを期待しています。

前もって感謝します！

実際のクラスで質問を更新する:

親

package backend.application.payroll.models;

import com.fasterxml.jackson.annotation.*;
import lombok.Data;

import java.io.Serializable;
import java.math.BigDecimal;
import java.util.Date;
import java.util.HashSet;
import java.util.Set;

import javax.persistence.*;

@Data
@Entity
@Table(name = "employees")
@JsonIdentityInfo(
        generator = ObjectIdGenerators.PropertyGenerator.class,
        property = "id")
public class Employee implements Serializable {

    @Id
    @GeneratedValue(strategy = GenerationType.IDENTITY)
    private long id;
    @Column(name = "emp_code", nullable = false)
    private String empCode;
    @Column(name = "first_name", nullable = false)
    private String firstName;
    @Column(name = "middle_name", nullable = true)
    private String middleName;
    @Column(name = "last_name", nullable = false)
    private String lastName;
    @Column(name = "dob", nullable = false)
    @Temporal(TemporalType.DATE)
    private Date dob;
    @Column(name = "id_number", nullable = true)
    private String idNumber;
    @Column(name = "passport_number", nullable = true)
    private String passportNumber;
    @Column(name = "email_address", nullable = true)
    private String emailAddress;
    @OneToOne(cascade = CascadeType.DETACH)
    @JoinColumn(name = "pay_grade", referencedColumnName = "id", nullable = true)
    private Salary payGrade;
    @Column(name = "basic_pay", nullable = true)
    private BigDecimal basicPay;
    @OneToOne(cascade = CascadeType.DETACH)
    @JoinColumn(name = "department", referencedColumnName = "id", nullable = true)
    private Department department;
    @OneToOne(cascade = CascadeType.DETACH)
    @JoinColumn(name = "position", referencedColumnName = "id", nullable = true)
    private Position position;
    @Column(name = "tax_number", nullable = true)
    private String taxNumber;
    @Column(name = "hire_date", nullable = true)
    @Temporal(TemporalType.DATE)
    private Date hireDate;
    @Column(name = "address1", nullable = true)
    private String address1;
    @Column(name = "address2", nullable = true)
    private String address2;
    @Column(name = "postal_code", nullable = true)
    private String postalCode;
    //country
    @Column(name = "phone_number", nullable = true)
    private String phoneNumber;
    //banking details

    //HERE IT WORKS FINE SINCE IT'S ONETOONE - YOU CAN IGNORE
    @OneToOne(mappedBy = "employee")
    //@JsonManagedReference//used in conjunction with @JsonBackReference on the other end - works like @JsonIdentityInfo class annotation.
    private User user;

    //THIS IS WHAT CAUSING THE PROBLEM
    @OneToMany(mappedBy = "owner", fetch = FetchType.LAZY)
    //@JsonBackReference
    @JsonIgnore
    private Set<Costcentre> costcentres = new HashSet<>();

    public Employee() {

    }
}

子

package backend.application.payroll.models;

import com.fasterxml.jackson.annotation.JsonIdentityInfo;
import com.fasterxml.jackson.annotation.JsonManagedReference;
import com.fasterxml.jackson.annotation.ObjectIdGenerators;
import lombok.Data;

import javax.persistence.*;
import java.io.Serializable;


@Entity
@Table(name = "costcentres")
@Data
@JsonIdentityInfo(generator = ObjectIdGenerators.PropertyGenerator.class, property = "id")
public class Costcentre implements Serializable {

    @Id
    @GeneratedValue(strategy = GenerationType.IDENTITY)
    private long id;
    @Column(name = "name", nullable = false)
    private String name;
    @Column(name = "description", nullable = true)
    private String description;
    @ManyToOne(cascade = CascadeType.DETACH, fetch = FetchType.EAGER)
    @JoinColumn(name = "owner", referencedColumnName = "id", nullable = true)
    //@JsonManagedReference
    private Employee owner; //CULPRIT

    public Costcentre() {

    }
    public Costcentre(long id, String name, String description) {
        super();
        this.id = id;
        this.name = name;
        this.description = description;
    }
}
4

1 に答える 1

0

JsonIdentityInfo親と子に追加すると、fetch = FetchType.EAGER親に追加できJsonIgnore、無視して循環的な子と親を取得できます

@JsonIdentityInfo(generator = ObjectIdGenerators.PropertyGenerator.class, property = "id")

、 このような：

@JsonIdentityInfo(generator = ObjectIdGenerators.PropertyGenerator.class, property = "id")
Parent{
    ....
    @OneToMany(mappedBy = "parent", fetch = FetchType.LAZY)
    @JsonIgnore
    Collection<Child> children;
    ....
}

@JsonIdentityInfo(generator = ObjectIdGenerators.PropertyGenerator.class, property = "id")
Child{
    ...
    @JoinColumn(name = "parent", referencedColumnName = "id", nullable = true)
    @ManyToOne(optional = false, fetch = FetchType.EAGER)
    Parent parent;
    ...
}
于 2020-05-22T19:36:16.900 に答える