JPA many-to-one relation - need to save only Id

Question

I have 2 classes: Driver and Car. Cars table updated in separate process. What I need is to have property in Driver that allows me to read full car description and write only Id pointing to existing Car. Here is example:

@Entity(name = "DRIVER")
public class Driver {
... ID and other properties for Driver goes here .....

    @ManyToOne(fetch=FetchType.LAZY)
    @JoinColumn(name = "CAR_ID")
    private Car car;

    @JsonView({Views.Full.class})
    public Car getCar() {
      return car;
    }
    @JsonView({Views.Short.class})
    public long getCarId() {
      return car.getId();
    }
    public void setCarId(long carId) {
      this.car = new Car (carId);
    }

}

Car object is just typical JPA object with no back reference to the Driver.

So what I was trying to achieve by this is:

1. I can read full Car description using detailed JSON View
2. or I can read only Id of the Car in Short JsonView
3. and most important, when creating new Driver I just want to pass in JSON ID of the car. This way I dont need to do unnesessery reads for the Car during persist but just update Id.

Im getting following error:

object references an unsaved transient instance - save the transient instance before flushing : com.Driver.car -> com.Car

I dont want to update instance of the Car in DB but rather just reference to it from Driver. Any idea how to achieve what I want?

Thank you.

UPDATE: Forgot to mention that the ID of the Car that I pass during creation of the Driver is valid Id of the existing Car in DB.

Answer 1

You can do this via getReference call in EntityManager:

EntityManager em = ...;
Car car = em.getReference(Car.class, carId);

Driver driver = ...;
driver.setCar(car);
em.persist(driver);

This will not execute SELECT statement from the database.

Answer 2

As an answer to okutane, please see snippet:

@JoinColumn(name = "car_id", insertable = false, updatable = false)
@ManyToOne(targetEntity = Car, fetch = FetchType.EAGER)
private Car car;

@Column(name = "car_id")
private Long carId;

So what happens here is that when you want to do an insert/update, you only populate the carId field and perform the insert/update. Since the car field is non-insertable and non-updatable Hibernate will not complain about this and since in your database model you would only populate your car_id as foreign key anyway this is enough at this point (and your foreign key relationship on the database will ensure your data integrity). Now when you fetch your entity the car field will be populated by Hibernate giving you the flexibility where only your parent gets fetched when it needs to.

Answer 3

You can work only with the car ID like this:

@JoinColumn(name = "car")
@ManyToOne(targetEntity = Car.class, fetch = FetchType.LAZY)
@NotNull(message = "Car not set")
@JsonIgnore
private Car car;

@Column(name = "car", insertable = false, updatable = false)
private Long carId;

Answer 4

That error message means that you have have a transient instance in your object graph that is not explicitly persisted. Short recap of the statuses an object can have in JPA:

• Transient: A new object that has not yet been stored in the database (and is thus unknown to the entitymanager.) Does not have an id set.
• Managed: An object that the entitymanager keeps track of. Managed objects are what you work with within the scope of a transaction, and all changes done to a managed object will automatically be stored once the transaction is commited.
• Detached: A previously managed object that is still reachable after the transction commits. (A managed object outside a transaction.) Has an id set.

What the error message is telling you is that the (managed/detached) Driver-object you are working with holds a reference to a Car-object that is unknown to Hibernate (it is transient). In order to make Hibernate understand that any unsaved instances of Car being referenced from a Driver about be saved should also be saved you can call the persist-method of the EntityManager.

Alternatively, you can add a cascade on persist (I think, just from the top of my head, haven't tested it), which will execute a persist on the Car prior to persisting the Driver.

@ManyToOne(fetch=FetchType.LAZY, cascade=CascadeType.PERSIST)
@JoinColumn(name = "CAR_ID")
private Car car;

If you use the merge-method of the entitymanager to store the Driver, you should add CascadeType.MERGE instead, or both:

@ManyToOne(fetch=FetchType.LAZY, cascade={ CascadeType.PERSIST, CascadeType.MERGE })
@JoinColumn(name = "CAR_ID")
private Car car;

Answer 5

public void setCarId(long carId) {
      this.car = new Car (carId);
    }

It is actually not saved version of a car. So it is a transient object because it hasn't id. JPA demands that you should take care about relations. If entity is new (doesn't managed by context) it should be saved before it can relate with other managed/detached objects (actually the MASTER entity can maintain it's children by using cascades).

Two ways: cascades or save&retrieval from db.

Also you should avoid set entity ID by hand. If you do not want to update/persist car by it's MASTER entity, you should get the CAR from database and maintain your driver with it's instance. So, if you do that, Car will be detached from persistence context, BUT still it will have and ID and can be related with any Entity without affects.

Answer 6

Add optional field equal false like following

@ManyToOne(optional = false) // Telling hibernate trust me (As a trusted developer in this project) when building the query that the id provided to this entity is exists in database thus build the insert/update query right away without pre-checks
private Car car; 

That way you can set just car's id as

driver.setCar(new Car(1));

and then persist driver normal

driverRepo.save(driver);

You will see that car with id 1 is assigned perfectly to driver in database

Description:

So what make this tiny optional=false makes may be this would help more https://stackoverflow.com/a/17987718

Answer 7

Use cascade in manytoone annotation @manytoone(cascade=CascadeType.Remove)