Question

我仍在掌握JPA概念，似乎无法在任何地方找到问题的答案！

假设

这两个类都使用@GeneratedValue（strategy = GenerationType.IDENTITY）进行注释，所有的获取器和设置器都有。

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

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

然后我实现了标准的JpaRepository并设置了控制器

这是问题所在
当我查询所有子记录时，只有第一个映射到特定父项的子记录将具有父实体对象。其余的将具有引用父实体的ID。

这里是一个示例：从POSTMAN获取所有孩子将返回：

[
    {
        "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
    }
]

如您所见，child2仅具有"parent": 1，因为child1首先映射到该父对象！同样，child4仅具有"parent": 2，因为child3首先映射到该父对象！

任何人都可以解释这种行为吗？我在父母上尝试过fetch = FetchType.EAGER，但没有帮助！我希望所有孩子都有一个全面的父对象，以防止再次发生DB行程。

谢谢！

使用实际类别更新问题：

父母

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() {

    }
}

孩子

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;
    }
}

Answer 1

将JsonIdentityInfo添加到父级和子级，您可以在父级上添加fetch = FetchType.EAGER，并在JsonIgnore上添加，以忽略获取循环子级和父级

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

，像这样：

@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;
    ...
}

JPA-JpaRepository子记录具有父ID而不是实体记录（如果已获取父记录）

1 个答案: