Java JPA-如果父记录已获取，则JPA存储子记录具有父id而不是实体记录

我仍然掌握着JPA的概念，似乎在任何地方都找不到我问题的答案

假定

这两个类都用@GeneratedValue（strategy=GenerationType.IDENTITY）进行注释，它们都有getter和setter

正如您所看到的

child2

只有

“parent”：1

，因为

child1

首先映射到该父级！ 类似地，

child4

只有

父项：2

，因为

child3

首先映射到该父项

有人能解释一下这种行为吗？我尝试了在父对象上使用fetch=FetchType.EAGER，但没有效果！ 我希望所有的孩子都有一个全面的父对象，以防止另一个DB trip

提前谢谢

用实际课程更新问题：

---

家长

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() {

    }
}

---

将

JsonIdentityInfo

添加到父级和子级，您可以在父级上添加

fetch=FetchType.EAGER

，并将

JsonIgnore

忽略获取子级和父级

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

，例如：

@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;
    ...
}

@JsonIdentityInfo（generator=ObjectIdGenerators.PropertyGenerator.class，property=“id”）
母公司{
....
@OneToMany（mappedBy=“parent”，fetch=FetchType.LAZY）
@杰索尼奥雷
收集儿童；
....
}
@JsonIdentityInfo（generator=ObjectIdGenerators.PropertyGenerator.class，property=“id”）
孩子{
...
@JoinColumn（name=“parent”，referencedColumnName=“id”，nullable=true）
@ManyToOne（可选=false，fetch=FetchType.EAGER）
父母；
...
}

谢谢您的回复@sc0der我已经有了这些注释，除了

fetch=FetchType.EAGER

但是即使这样也不能解决问题。问题依然存在。我甚至想放弃我的整个数据库，因为我只是在后来添加了这种关系，但我怀疑这会有帮助（我认为JPA足够聪明）。我想我做的唯一一件事就是扔掉儿童桌，以确保FK的适当娱乐，但即使这样，我也没有得到任何运气！我之前尝试过实现

Serializable

，它没有改变任何东西，我将其删除。我刚把它放回去，结果是一样的。很明显，有什么东西阻止了对父对象的多次检索，我想知道这是什么！事实上，我甚至不认为这与多次检索有关，这只是嵌入以前检索到的实体的结果的问题！JSON问题？太令人困惑了！检查这一条：还有这一条：我真的遇到了这些文章，并尝试了回来，管理了引用，但没有成功！用实际的两个类更新我的问题可以吗？（希望合适）

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;
    }
}

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

@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;
    ...
}