JPA Persist parent and child with one to many relationship

Question

I want to persist parent entity with 20 child entities, my code is below

Parent Class

@OneToMany(mappedBy = "parentId")
private Collection<Child> childCollection;

---

Child Class

@JoinColumn(name = "parent_id", referencedColumnName = "parent_id")
@ManyToOne(optional=false)
private Parent parent;

---

String jsonString = "json string containing parent properties and child  collection" 

ObjectMapper mapper = new ObjectMapper();
Parent parent = mapper.readValue(jsonString, Parent.class);

public void save(Parent parent) {
    Collection<Child> childCollection = new ArrayList<>() ;

    for(Child tha : parent.getChildCollection()) { 
        tha.setParent(parent);
        childCollection.add(tha);
    }

    parent.setChildCollection(childCollection);
    getEntityManager().persist(parent);
 }

So if there are 20 child tables then I have to set parent reference in each of them for that I have to write 20 for loops? Is it feasible? is there any other way or configuration where I can automatically persist parent and child?

Answer 1

Fix your Parent class:

@OneToMany(mappedBy = "parent")

mappedBy property should point to field on other side of relationship. As JavaDoc says:

The field that owns the relationship. Required unless the relationship is unidirectional.

Also you should explicitely persist Child entity in cycle:

for(Child tha : parent.getChildCollection()) { 
    ...
    getEntityManager().persist(tha);
    ...
}

As Alan Hay noticed in comment, you can use cascade facilities and let EntityManager automatically persist all your Child entities:

@OneToMany(mappedBy = "parent", cascade = CascadeType.PERSIST)

More details about cascades (and JPA itself) you can find in Vlad Mihalcea's blog.

Answer 2

Generally, @JoinColumn indicates that the entity is the owner of the relationship & mappedBy indicates that the entity is the inverse of the relationship.

So, if you are trying like following

@OneToMany(mappedBy = "parent")
private Collection<Child> childCollection;

That means it is inverse of the relationship and it will not set parent reference to its child.

To set parent reference to its child, you have to make the above entity owner of the relationship in the following way.

@OneToMany(cascade = CascadeType.ALL)
@JoinColumn
private Collection<Child> childCollection;

You need not set any child reference because above code will create a column in the child table.

Answer 3

As pointed out in the comments you must take care of the object graph consistency with child/parent relationship. This consistency won't come free when JSON is coming directly from i.e. a POST request.

You have to annotate the parent and child field with @JsonBackReference and @JsonManagedReference.

Parent class:

@OneToMany(mappedBy = "parentId")
@JsonBackReference
private Collection<Child> childCollection;

Child class:

@JoinColumn(name = "parent_id", referencedColumnName = "parent_id")
@ManyToOne(optional=false)
@JsonManagedReference
private Parent parent;

Similar question with answer is here

Furthermore, if you use @JsonBackReference/@JsonManagedReference on javax.persistence annotated classes in combination with Lombok's @ToString annotation you will incur in stackoverflow error.

Just exclude childCollection and parent field from the @ToString annotation with @ToString( exclude = ...)

The same will happen with Lombok's generated equals() method (@Data, @EqualsAndHashCode). Just implements those methods by hand or to use @Getter and @Setter annotations only.

Answer 4

I would let the parent persist it's own children

package com.greg;

import java.util.ArrayList;
import java.util.List;

import javax.persistence.CascadeType;
import javax.persistence.Column;
import javax.persistence.Entity;
import javax.persistence.FetchType;
import javax.persistence.GeneratedValue;
import javax.persistence.GenerationType;
import javax.persistence.Id;
import javax.persistence.JoinColumn;
import javax.persistence.OneToMany;

@Entity(name = "PARENT")
public class Parent {

    @Id
    @GeneratedValue(strategy = GenerationType.AUTO)
    private Long id;

    @Column(name = "NAME")
    private String name;

    @Column(name = "DESCRIPTION")
    private String description;

    @OneToMany(cascade = CascadeType.ALL, fetch=FetchType.EAGER)
    @JoinColumn(name = "parent", referencedColumnName = "id", nullable = false)
    private List<Child> children = new ArrayList<Child>();

    public Long getId() {
        return id;
    }

    public void setId(Long id) {
        this.id = id;
    }

    public String getName() {
        return name;
    }

    public void setName(String name) {
        this.name = name;
    }

    public String getDescription() {
        return description;
    }

    public void setDescription(String description) {
        this.description = description;
    }

    public List<Child> getChildren() {
        return children;
    }

    public void setChildren(List<Child> children) {
        this.children = children;
    }

}

Answer 5

I am using lombok to generate getter and setter properties on my entity classes. I was also facing issue of NULL referenceID on child entity when I was trying to save parent Entity having child. On my parent entity when I add children then I set "this" reference of parent on child. In my example, I have User table and Address table where a User can have many addresses.
I have created domain classes as below.

e.g address.setUser(this);

package com.payment.dfr.entities;

import lombok.Data;

import javax.persistence.*;
import java.math.BigInteger;
@Entity
@Data
@Table(name="User")
public class User {

    @Id
    @GeneratedValue
    private BigInteger RecordId;
    private String Name;
    private String Email;

    @Getter(AccessLevel.NONE)
    @Setter(AccessLevel.NONE)
    @OneToMany(mappedBy = "user", cascade = CascadeType.ALL)
    private List<Address> addresses = new ArrayList<>();

    public void addAddress(Address address){
        address.setUser(this);
        addresses.add(address);
    }
    
}

@Entity
@Data
@Table(name="UserAddress")
public class Address {

    @Id
    @GeneratedValue
    private BigInteger RecordId;
    private String AddressLine;
    private String City;

    @ManyToOne(cascade = CascadeType.ALL)
    @JoinColumn(name="UserId")
    private User user;

}


This is how I save user with address 

    User newUser = new User();
    newUser.setName("Papa");
    newUser.setEmail("manish@gmail.com");
    Address address1 = new Address();
    address1.setAddressLine("4401 Central Ave");
    address1.setCity("Fremont");
    newUser.addAddress(address1);
    Address address2 = new Address();
    address2.setAddressLine("4402 Central Ave");
    address2.setCity("Fremont");

    newUser.addAddress(address2);
    User user1 = userRepository.save(newUser);
    log.info(user1.getRecordId().toString());