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java - JPA - JpaRepository 子记录具有父 ID 而不是实体记录,如果父记录已获取

转载 作者:行者123 更新时间:2023-12-01 13:45:38 25 4
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这个问题在这里已经有了答案:





Hibernate bidirectional @ManyToOne, updating the not owning side not working

(3 个回答)


去年关闭。




我仍在掌握 JPA 概念,似乎无法在任何地方找到我的问题的答案!

认为

这两个类都用@GeneratedValue(strategy = GenerationType.IDENTITY) 进行了注解,都有getter 和setter。

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 并设置了我的 Controller

这是问题
当我查询所有子记录时,只有映射到特定父级的第一个子记录才会在其中包含父实体对象。其余的将有一个引用父实体的 id。

下面是一个例子:
从 POSTMAN 获取所有子项返回:
[
{
"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只有 "parent": 1child1首先映射到该父级!
同样 child4只有 "parent": 2child3首先映射到该父级!

任何人都可以解释这种行为吗?我试过 fetch = FetchType.EAGER在 parent 身上,但它没有帮助!
我希望所有 child 都有一个全面的父对象,以防止再次 DB 旅行。

提前致谢!

使用实际类(class)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() {

}
}


child
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;
}
}

最佳答案

添加 JsonIdentityInfo给 parent 和 child ,你可以添加 fetch = FetchType.EAGER在父级和 JsonIgnore 上忽略得到循环的 child 和 parent

@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;
...
}

关于java - JPA - JpaRepository 子记录具有父 ID 而不是实体记录,如果父记录已获取,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/61962472/

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