我在使用JPA访问字段时出错。我怎么解决这个？

Question

在尝试向表中插入值时，我会遇到以下错误

org.hibernate.property.access.spi.PropertyAccessException: Error accessing 
field [private int com.app.demo.model.Customer.id] by reflection for 
persistent property [com.app.demo.model.Customer#id] : 
com.app.demo.model.Customer@6d1c6e1e
javax.persistence.PersistenceException: 
org.hibernate.property.access.spi.PropertyAccessException: Error accessing 
field [private int com.app.demo.model.Customer.id] by reflection for 
persistent property [com.app.demo.model.Customer#id] : com.app.demo.model.Customer@6d1c6e1e

这是我的顾客席。我正在使用MySql工作台，我正在尝试将我的值插入到这里。

我正在使用这个类将值插入到表中

@Entity
@Table(name="customer")
public class Customer {

@Id
@GeneratedValue
@Column(name = "customer_id", unique = true, nullable = false)
private int id;
@Column(name="first_name")
private String firstName;
@Column(name="last_name")
private String lastName;
@Column(name="street_address")
private String address;
@Column(name="city")
private String city;
@Column(name="state")
private String state;
@Column(name="zip_code")
private String zipcode;
@Column(name="email")
private String email;
@Column(name="paypal_email")
private String paypalEmail;


// getters and setters

// Set customer values
Customer valuedCustomer = new Customer();

valuedCustomer.setFirstName(firstName);
valuedCustomer.setLastName(lastName);
valuedCustomer.setAddress(address);
valuedCustomer.setCity(city);
valuedCustomer.setState(state);
valuedCustomer.setZipcode(zip);
valuedCustomer.setEmail(email);

// insert customer info into the customer table
EntityManagerFactory emf = Persistence.createEntityManagerFactory("pu");

EntityManager em = emf.createEntityManager();

em.getTransaction().begin();
em.persist(valuedCustomer);
em.getTransaction().commit();

我的用户表（我在这个表上进行了单元测试）

Answer 1

请尝试使用自动或标识策略。如：

@GeneratedValue(strategy = GenerationType.IDENTITY)