Hibernate association mapping-CodePudding

I am working on a bike rental system project. I am able to do CRUD operations via Rest API on Bike, and Customer and create a Contract object. For this, I have exposed CRUD endpoints(s) for

/bike
/customer
/contract

The POST /contract/final endpoint should return a complete contract overview with contractid, bike details, and customer

The association I am using is as below

class Contract {

@Id
@GeneratedValue(strategy= GenerationType.SEQUENCE)
private id;
@ManyToOne(fetch=FetchType.EAGER, cascade = CascadeType.ALL)
@JoinColumn(name="customerId")
private Customer customer;
@ManyToOne(fetch=FetchType.EAGER, cascade = CascadeType.ALL)
@JoinColumn(name="bikeId")
private Bike bike; 
}

class Customer{
@Id
@GeneratedValue(strategy = GenerationType.IDENTITY)
@Column(name = "customer_id")
private Long customerId;
private String firstName;
}

class Bike{
@Id
@GeneratedValue(strategy = GenerationType.IDENTITY)
@Column(name = "bike_id")
private long bikeId;
private String model;
}

So I just want to be able to create/modify the bike|customer records and create a contract with the already created customer(s) and bike(s).

As of now, the contract request will create the customer and bike records apart from records created separately using /customer, /bike endpoints - which I do not want.

Is the association mapping I have used correct for this requirement? What should ideally be the relationship between these entities given the below contraints?

One customer can have many contracts but one contract will have only one customer
One bike can have many contracts but one contract will have only one bike

How can I make use of the existing customer/bike records while creating a contract?

CodePudding user response：

Do you want to have the ability to change a Bike model when you create a new Contract? Obviously no! cascade = CascadeType.ALL is about such changes. So throw it away.
Do you want to load all the Contract data, like a Bike, when you load a Contract? It is not so obvious, but my advice don't do this! Let's change fetch=FetchType.EAGER to fetch=FetchType.LAZY.
You use two strategies to generate an id — strategy= GenerationType.SEQUENCE, strategy = GenerationType.IDENTITY. Use only one.
Don't use a primitive type for an id. Change long bikeId to Long bikeId.
@JoinColumn specifies a database column name, not a class property. Change @JoinColumn(name="customerId") to @JoinColumn(name="CUSTOMER_ID"). Use uppercase for column names.

class Contract {

    @Id
    @GeneratedValue(strategy = GenerationType.IDENTITY)
    @Column(name = "ID")
    private Long id;
    
    @ManyToOne(fetch = FetchType.LAZY)
    @JoinColumn(name = "CUSTOMER_ID")
    private Customer customer;
    
    @ManyToOne(fetch = FetchType.LAZY)
    @JoinColumn(name = "BIKE_ID")
    private Bike bike;
    
}

Assign existing Bike and Customer to the Contract, using getReferenceById() method. It doesn't hit a database at all.

Customer customer = customerRepository.getReferenceById(customerId);
contract.setCustomer(customer);

Another way is to use a transient customer.

Customer customer = new Customer();
customer.setId(customerId);
contract.setCustomer(customer);

But standard repository save() method uses merge(). So it will be an additional SQL request to load a customer. You can use custom repository with pure save() method to avoid this https://vladmihalcea.com/best-spring-data-jparepository/