Question

I have 2 entities Customer and address please find the code below, I have omitted boiler plate code for simplicity.

public class Customer  implements java.io.Serializable {
  private static final long serialVersionUID = 3116894694769321104L;
     private short customerId;
     private Address address;
     private String firstName;
     private String lastName;
     private String email;
     private boolean active;
     private Date createDate;
     private Date lastUpdate;


    // Property accessors
    @Id
    @Column(name="customer_id", unique=true, nullable=false, insertable=true, updatable=true)

    public short getCustomerId() {
        return this.customerId;
    }

    public void setCustomerId(short customerId) {
        this.customerId = customerId;
    }
    @ManyToOne(cascade={CascadeType.ALL},
        fetch=FetchType.LAZY)

        @JoinColumn(name="address_id", unique=false, nullable=false, insertable=true, updatable=true)

    public Address getAddress() {
        return this.address;
    }

    public void setAddress(Address address) {
        this.address = address;
    }

and Address class is :

public class Address  implements java.io.Serializable {


    // Fields    

     private short addressId;
     private short customerId;
     private String address;
     private String address2;
     private String district;
     private String postalCode;
     private String phone;
     private Date lastUpdate;
     private Set<Customer> customers_1 = new HashSet<Customer>(0);


    // Constructors

    /** default constructor */
    public Address() {
    }

   // Property accessors
    @Id
    @Column(name="address_id", unique=true, nullable=false, insertable=true, updatable=true)

    public short getAddressId() {
        return this.addressId;
    }

    public void setAddressId(short addressId) {
        this.addressId = addressId;
    }

    /**
     * ??????what goes here
     */
    public short getCustomerId() {
        return customerId;
    }

    /**
     * @param customerId the customerId to set
     */
    public void setCustomerId(short customerId) {
        this.customerId = customerId;
    }

I need to persist the customer id as a foreign key in address table.

Answer 1

Just use @ManyToOne relationship with Customer. So, instead of customerId in Java code you will be operates with Customer object, but at database level Hibernate will use foreign key to table with customer.