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Ternary m-to-m-to-m relation in springboot/postgresql

I am trying to implement a ternary relation with the entities Customer, CustomerServant and Room. The teranry relation also has the additional attributes StartDate, EndDate and also an Id. Therefore I created this class to get the relation table:
#Entity(name = "bookings")
#Table(name = "bookings")
public class Bookings {
#Id
#SequenceGenerator(
name = "bookings_sequence",
sequenceName = "bookings_sequence",
allocationSize = 1
)
#GeneratedValue(
strategy = SEQUENCE,
generator = "bookings_sequence"
)
#Column(
name = "bookings_id",
updatable = false
)
private Long Id;
#Column(
name = "Travel_Start_Date",
nullable = false
)
private String travelStartDate ;
#Column(
name = "Travel_End_Date",
nullable = false
)
private String travelEndDate;
#ManyToOne ( cascade = CascadeType.ALL, fetch = FetchType.EAGER)
#JoinColumn(name = "customer", insertable = false, updatable = false)
private Customer customer = new Customer();
#ManyToOne
#JoinColumn(name = "room", insertable = false, updatable = false)
private Room room;
#ManyToOne
#JoinColumn(name = "customerServant", insertable = false, updatable = false)
private CustomerServant customerServant;
What I am struggling with is what I should write in the other three classes which are referencing each other.
I do understand that to create a single m-to-m relation with additional relational attributes I can implement a class as I did above and simply write: e.g.
class Bookings {
int Id;
Date StartDate;
Date EndDate;
#ManyToOne
#JoinColumn(name = "customer_id")
Customer customer;
#ManyToOne
#JoinColumn(name = "Room")
Room room;
}
class Customer {
//...
#OneToMany(mappedBy = "customer")
Set<Bookings> bookings;
}
class Room {
#OneToMany(mappedBy = "room")
Set<Bookings> bookings;
but what do I do when I want a ternary relation?

This is how I solved it using the info provided in the article sent by #warmCabin (thorben-janssen.com/ternary-associations):
Using an association table class like above with the attributes startDate, endDate, booking_id and the references to the three entities Room, Customer, CustomerServant, provided that theres an according #ManyToOne annotation for the three referencing entities, just like in the question above.
e.g. like above
#ManyToOne (cascade = CascadeType.ALL, fetch = FetchType.EAGER)
#JoinColumn(name = "customer", insertable = false, updatable = false)
private Customer customer = new Customer();
Then I added this to all entities that are connected via the ternary relation:
#OneToMany(mappedBy = "<enter varName from ternary relation table>")
Set<T> var = new HashSet<>();
With this line of code, when a relation is stored e.g. a booking like in my case, this relation gets stored in every participating entity and every stored relation gets the Ids of all participating entities.
In my case it would look like this, for the three entities:
#Entity(name="Room")
#Table(name="Room")
class Room {
//attributes...
#OneToMany(mappedBy = "room")
Set<Bookings> bookings = new HashSet<>();
//...
}
#Entity(name="Customer")
#Table(name="Customer")
class Customer {
//attributes...
#OneToMany(mappedBy = "customer")
Set<Bookings> bookings = new HashSet<>();
//...
}
#Entity(name="CustomerServant")
#Table(name="CustomerServant")
class CustomerServant {
//attributes...
#OneToMany(mappedBy = "customerServant")
Set<Bookings> bookings = new HashSet<>();
//...
}
In the article cited above it is said that its also possible to create such a relation with a HashMap, but because the first appraoch, with an own entity class as a ternary relation, works for me, I didnt check if it would also work in my case. If necessary check the article thorben-janssen.com/ternary-associations.

OneToMany relationship does not store entity field

When saving my entities, child entities that work through the #OneToMany relationship are not saved to their tables. I can’t understand what’s the matter.
Employee:
#Entity
#Table(name = "EMPLOYEE", schema = PUBLIC)
public class Employee {
private String name;
private String lastname;
#OneToMany(mappedBy = "employee", cascade = CascadeType.ALL, orphanRemoval = true)
List<EmployeePhoneNumber> employeePhoneNumbers = new ArrayList<>();
}
EmployeePhoneNumber:
#Entity
#Table(name = "EMPLOYEE_PHONES", schema = PUBLIC)
public class EmployeePhoneNumber {
#Id
#SequenceGenerator(allocationSize = 1, name = "SEQ_EMPLOYEE_PHONES", schema = PUBLIC,
sequenceName = "EMPLOYEE_PHONES_ID_SEQ")
#GeneratedValue(generator = "SEQ_EMPLOYEE_PHONES", strategy = GenerationType.SEQUENCE)
#Column(name = "ID", unique = true, nullable = false)
private Long id;
#ManyToOne
#JoinColumn(name = "employee_id", referencedColumnName = "id",
nullable = false, insertable = false, updatable = false)
private Employee employee;
#Column(name = "PHONE_NUMBER", unique = true, nullable = false)
private String phoneNumber;
#Enumerated(EnumType.STRING)
#Column(name = "NUMBER_TYPE", nullable = false)
private PhoneNumberType phoneNumberType;
}
How I set those fields and then save the entity:
EmployeePhoneNumber workPhone = new EmployeePhoneNumber();
workPhone.setPhoneNumber(workPhone);
workPhone.setPhoneNumberType(PhoneNumberType.WORK_PHONE);
EmployeePhoneNumber mobilePhone = new EmployeePhoneNumber();
mobilePhone.setPhoneNumber(mobilePhone);
mobilePhone.setPhoneNumberType(PhoneNumberType.MOBILE_PHONE);
EmployeePhoneNumber corporatePhone = new EmployeePhoneNumber();
corporatePhone.setPhoneNumber(corporatePhoneNumber);
corporatePhone.setPhoneNumberType(PhoneNumberType.CORPORATE_PHONE);
List<EmployeePhoneNumber> employeePhoneNumbers = employee.getEmployeePhoneNumbers();
employeePhoneNumbers.add(workPhone);
employeePhoneNumbers.add(mobilePhone);
employeePhoneNumbers.add(corporatePhone);
employee.setEmployeePhoneNumbers(employeePhoneNumbers);
employeeRepository.save(employee);
Upon completion of the method, I do not have a single error, everything works out correctly, only the tables are not filled - why?

You must also set the Employee reference in EmployeePhoneNumber because Hibernate will use this to save it.
workPhone.setEmployee(employee);
mobilePhone.setEmployee(employee);
corporatePhone.setEmployee(employee);
The best solution would be to create an addEmployeePhoneNumber method on the Employee like this:
public void addEmployeePhoneNumber(EmployeePhoneNumber phoneNumber) {
phoneNumber.setEmployee(this);
employeePhoneNumbers.add(phoneNumber);
}
That way you will not forget to set both sides of the relationship.

how to find id of foreign key in hibernate

I have two entities viz:
State
#Entity
#Table(name = "State")
public class StateEntity {
#Column(name = "id", length = 36, nullable = false, unique = true)
private String id;
#ManyToOne (fetch = FetchType.LAZY)
#JoinColumn(name = "InsurerId", nullable = false)
private InsurerEntity insurer;
#Column(name ="StateName", length = 50, nullable = false)
private String stateName;
//getters and setters
}
Insurer
#Entity
#Table(name = "Insurer")
public class InsurerEntity {
#Column(name = "InsurerId", length = 36, nullable = false, unique = true)
private String insurerId;
#Column(name = "InsurerName", length = 100, nullable = true)
private String insurerName;
#OneToMany(mappedBy = "state", fetch = FetchType.LAZY)
private List<StateEntity> stateEntityList;
//getters and setters
}
the insurer's id gets saved in state database and I want to retrieve it using hibernate query but I cant't seem to find the solution for that
How to write this query SELECT InsurerId FROM State; in Hibernate query using CriteriaBuilder, CriteriaQuery and Root..

If you want to select all Insurers's Ids for all states:
String selectionQuery = "SELECT s.insurer.insurerId FROM State s";
List<String> insurersIds = session.createQuery(selectionQuery).list();
If you want to select the Insurer's Id of a certain state:
String selectionQuery = "SELECT s.insurer.insurerId FROM State s WHERE s.id = :stateId";
String insurerId = (String) session.createQuery(selectionQuery).setParameter("stateId", stateId).getSingleResult(); //This should be placed in a try/catch block to handle org.hibernate.NonUniqueResultException
Edit:
You should update your Insurer entity as Prasad wrote in his answer.

for this you have to map both the class as in put #oneToMany annotation in class InsurerEntity as well
#OneToMany(fetch = FetchType.LAZY,mappedBy="StateEntity", cascade = CascadeType.ALL)
private List< StateEntity > StateEntitys;
and when you fetch states you will also get object of InsurerEntity in it from where you can access it with the getter

How to join a table with one column of a view in that table's entity class

Scenario:
I have a products table with these fields: id, code, description, photo.
I have a product_stock view from other schema with fields: prod_code, stok_tot
Misson: What I need to do in Spring Boot is to join the products with their total stock.
Rules:
I can't use #Query on this task, I need to join them at the Product Entity Class, so that stock became some kind of Transient field of product.
I can't change the fact that product's ID is a Long and ProductStock's ID is a String, but I could use product's code field instead right? (how?)
So far... I tryed to use #OneToOne and #JoinColumn to do the job, but my REST gives me the stock field as NULL.
"Estoque.java"
#Entity
#Table(name = "VW_ESTOQUE", schema = "ASICAT")
public class Estoque {
#Id
#Column(name = "CD_BEM_SERVICO", unique = true, nullable = false)
private String codigo;
#Column(name = "ESTOQUE")
private Long estoque;
// getters and setters hidden by me
}
"Produto.java"
#Entity
#NamedEntityGraph(name = "Produto.detail", attributeNodes = #NamedAttributeNode("categorias"))
public class Produto implements Serializable {
#Id
#GeneratedValue(strategy=GenerationType.SEQUENCE)
private Long id;
private String codigo;
private String descricao;
// here is where I get the null values
#Transient
#OneToOne(fetch = FetchType.LAZY)
#JoinTable(name = "VW_ESTOQUE", joinColumns = #JoinColumn(name = "CODIGO", referencedColumnName = "CODIGO"), inverseJoinColumns = #JoinColumn(name = "CD_BEM_SERVICO", referencedColumnName = "CODIGO"))
private Estoque estoque;
private String hash;
#ManyToMany(mappedBy = "produtos", fetch = FetchType.EAGER)
#BatchSize(size = 10)
private List<Categoria> categorias = new ArrayList<>();
// getters and setters hidden by me
}
In my product repository I call FindAll()

You have annotated Produto.estoque as #Transient, which means that it is not part of the persistent state of the entity. Such a field will be neither written nor read when instances of that entity are managed. That's not going to serve your purpose.
There are two things I can imagine you might have been trying to achieve:
That every time an Estoque is accessed via a Produto, it should be loaded from the DB to ensure its freshness. JPA does not provide for that, though you might want to annotate Estoque with #Cacheable(value = false), and specify the lazy fetch strategy on the Produto side of the relationship.
You want to avoid the persistence provider attempting to persist any changes to an Estoque, since it is backed by a view, not an updatable table. This we can address.
My first suggestion would be to map ASICAT.VW_ESTOQUE as a secondary table instead of an entirely separate entity. That might look something like this:
#Entity
#SecondaryTable(name = "VW_ESTOQUE", schema = "ASICAT"
pkJoinColumns = {
#PrimaryKeyJoinColumn(name = "CD_BEM_SERVICO",
referencedColumnName = "CODIGO") })
public class Produto implements Serializable {
#Id
#GeneratedValue(strategy=GenerationType.SEQUENCE)
private Long id;
private String codigo;
private String descricao;
#Column(name = "ESTOQUE", table = "VW_ESTOQUE", nullable = true,
insertable = false, updatable = false)
private Long estoque;
// ...
}
You might furthermore avoid providing a setter for the estoque property.
But the SecondaryTable approach might not work well if you cannot rely on the ESTOQUE view always to provide a row for every row of PRODUTO, as there will very likely be an inner join involved in retrievals. Moreover, you don't get lazy fetches this way. The main alternative is more or less what you present in your question: to set up a separate Estoque entity.
If you do set up a separate Estoque, however, then I would approach it a bit differently. Specifically,
I would make the relationship bidirectional, so that I could
make the Estoque entity the relationship owner.
Something like this, then:
#Entity
public class Produto implements Serializable {
#Id
#GeneratedValue(strategy=GenerationType.SEQUENCE)
private Long id;
private String codigo;
private String descricao;
// must not be #Transient:
#OneToOne(fetch = FetchType.LAZY, mappedBy = "produto", cascade = {
CascadeType.REFRESH
})
private Estoque estoque;
// ...
}
#Entity
#Table(name = "VW_ESTOQUE", schema = "ASICAT")
#Cacheable(value = false)
public class Estoque {
#Id
#Column(name = "CD_BEM_SERVICO", nullable = false,
insertable = false, updatable = false)
private String codigo;
#Column(name = "ESTOQUE", insertable = false, updatable = false)
private Long estoque;
#OneToOne(fetch = FetchType.LAZY, optional = false)
#JoinColumn(name = "CD_BEM_SERVICO", referencedColumnName = "CODIGO",
nullable = false, insertable = false, updatable = false, unique = true)
Produto produto;
// getters and setters hidden by me
}
In this case, I would avoid providing setter methods for any of the properties of Estoque, and avoid providing any constructor that allows initial property values to be set. Thus, to a first approximation, the only
way an instance's properties will take non-null values is if they are set by the persistence provider.
Additionally, since you mention Oracle, if you are using TopLink as your persistence provider then you might consider applying its #ReadOnly extension attribute to the Estoque entity, in place of or even in addition to some of these protections against trying to insert into or update the view.

Creating a Hibernate Criteria that can filter by referenced entity

I have an entity (PersonQuestionsEntity) that has a PersonEntity and QuestionEntity as its primary key. I use a composite key to reflect this relationship.
Now, I want to create a Criteria object that can do the following: Find all PersonQuestion entities for a given questionId and a person's age.
Here is my attempt at creating a Criteria for this:
Session session = getHibernateTemplate().getSessionFactory().openSession();
Criteria criteria = session.createCriteria(PersonQuestionsEntity.class);
criteria.add(Restrictions.eq("question.questionId", "87"));
criteria = criteria.createCriteria("person");
criteria.add(Restrictions.eq("age", 23));
criteria.setMaxResults(100);
List l = criteria.list();
The problem is that I get this error:
Caused by: java.sql.SQLException: ORA-00904: "PERSONENTI1_"."AGE": invalid identifier
In the generated SQL, it seems the person is referenced as PERSONENTI4, not PERSONENTI1. If I copy the SQL and run it with PERSONENTIT4 instead of PERSONENTI4, it works (sort of -- it seems to be doing a cartesian join of some sort).
Any clues as to what I might be doing wrong? I'm very new to using Hibernate.
PersonQuestionsEntity
#Entity
#IdClass(com.anonymous.model.PersonQuestionsKey.class)
#Table(name = "PERSON_QUESTIONS")
public class PersonQuestionsEntity implements Serializable
{
private static final long serialVersionUID = -8254277382097937813L;
#Id
#ManyToOne
#JoinColumn(name = "USER_NAME", nullable = false)
private PersonEntity person;
#Id
#ManyToOne
#JoinColumn(name = "QUESTION_ID", nullable = false)
private QuestionEntity question;
#Column(name = "THEIR_ANSWER")
private int theirAnswer;
}
PersonEntity
#Entity
#Table(name = "PERSON")
public class PersonEntity implements Serializable
{
private static final long serialVersionUID = -1699435979266209440L;
#Id
#Column(name = "USER_NAME", length = 20, nullable = false)
private String userName;
#Column(name = "USER_NAME_REAL", length = 20, nullable = false)
private String userNameReal;
#Column(name = "AGE", nullable = false)
private int age;
}
PersonQuestionsKey
#Embeddable
public class PersonQuestionsKey implements Serializable
{
private static final long serialVersionUID = -264160855961369405L;
#Id
#ManyToOne
#JoinColumn(name = "USER_NAME", nullable = false)
private PersonEntity person;
#Id
#ManyToOne
#JoinColumn(name = "QUESTION_ID", nullable = false)
private QuestionEntity question;
}

First of all, you don't really need the inner criteria, simply use:
Criteria criteria = session.createCriteria(PersonQuestionsEntity.class);
criteria.add(Restrictions.eq("question.questionId", "87"));
criteria.add(Restrictions.eq("person.age", 23));
criteria.setMaxResults(100);
List l = criteria.list();
Second (regarding the join type), in such cases, I usually go with HQL that produces inner joins. The HQL might look as follows:
from PersonQeustionEntity where question.questionId = :questionId
and person.age = :age
in the result Query object you can set the parameters questionId and age to your desired inputs.