Hibernate saveOrUpdate（）保存但不更新

Question

我正在学习hibernate和Spring Boot（也是Thymeleaf）,遇到了以下问题：

环境：我有一个表单，处理创建和更新数据基于ID参数在GET方法，如果ID等于数据库中的一行，那么更新，否则保存。ID存储在隐藏的输入标记中，并与表单数据一起插入到实体对象中。然后将该对象发送给一个服务，该服务调用DAO来使用saveOrUpdate（）方法保存或更新数据。

问题总结：在我的服务中，我有一个用@Transactional注释的方法，它调用我的DAO，然后它打开一个会话并执行saveOrUpdate（）。问题是，这段代码正确地将新数据保存到我的数据库中，但它不更新现有数据，在日志中saveOrUpdate（）只选择而不插入。

我得HTML表单与Thymeleaf

<form action="#" th:action="@{saveCustomer}" th:object="${newCustomer}" method="post">
            <input type="hidden" th:field="*{id}" />
            <table>
                <tr>
                    <td><label>First Name:</label></td>
                    <td><input type="text" th:field="*{firstName}"/></td>
                </tr>
                <tr>
                    <td><label>Last Name:</label></td>
                    <td><input type="text" th:field="*{lastName}"/></td>
                </tr>
                <tr>
                    <td><label>Email:</label></td>
                    <td><input type="text" th:field="*{email}"/></td>
                </tr>
                <tr>
                    <td><label></label></td>
                    <td><input type="submit" value="Submit" class="save"/></td>
                </tr>
                <tr>
                    <td><label></label></td>
                    <td><input type="reset" value="Reset" class="save"/></td>
                </tr>
            </table>
        </form>

控制器方法

    @PostMapping("/saveCustomer")
    public String saveCustomer(@ModelAttribute("newCustomer") Customer newCustomer) {
        customerService.saveCustomer(newCustomer);
        return "redirect:/list";
    }

服务方式

import javax.transaction.Transactional;

    @Override
    @Transactional
    public void saveCustomer(Customer newCustomer) {
        customerDAO.saveCustomer(newCustomer);
    }

    @Autowired
    private EntityManagerFactory entityManagerFactory;

    @Override
    public void saveCustomer(Customer newCustomer) {
        try (Session session = entityManagerFactory.unwrap(SessionFactory.class).openSession()) {
//            If I use transaction like this everything works correctly!
//            Transaction transaction = session.beginTransaction();
            session.saveOrUpdate(newCustomer);
//            transaction.commit();
        } catch (Exception e) {
            e.printStackTrace();
        }
    }

import javax.persistence.*;

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

    @Id
    @GeneratedValue(strategy = GenerationType.IDENTITY)
    @Column(name = "id")
    private int id;

    @Column(name = "first_name")
    private String firstName;

    @Column(name = "last_name")
    private String lastName;

    @Column(name = "email")
    private String email;

    public int getId() {
        return id;
    }

    public void setId(int id) {
        this.id = id;
    }

    public String getFirstName() {
        return firstName;
    }

    public void setFirstName(String firstName) {
        this.firstName = firstName;
    }

    public String getLastName() {
        return lastName;
    }

    public void setLastName(String lastName) {
        this.lastName = lastName;
    }

    public String getEmail() {
        return email;
    }

    public void setEmail(String email) {
        this.email = email;
    }

    @Override
    public String toString() {
        return "Customer{" +
                "id=" + id +
                ", firstName='" + firstName + '\'' +
                ", lastName='" + lastName + '\'' +
                ", email='" + email + '\'' +
                '}';
    }
}

Hibernate: select customer0_.id as id1_0_, customer0_.email as email2_0_, customer0_.first_name as first_na3_0_, customer0_.last_name as last_nam4_0_ from customer customer0_ order by customer0_.last_name
Hibernate: select customer0_.id as id1_0_0_, customer0_.email as email2_0_0_, customer0_.first_name as first_na3_0_0_, customer0_.last_name as last_nam4_0_0_ from customer customer0_ where customer0_.id=?
Hibernate: select customer0_.id as id1_0_, customer0_.email as email2_0_, customer0_.first_name as first_na3_0_, customer0_.last_name as last_nam4_0_ from customer customer0_ order by customer0_.last_name

我省略了declutter代码的大部分导入，因为我只包含了对象中与问题相关的部分。

Answer 1

您导入的@transactional批注错误。它应该来自包org.springframework.transaction.annotation。这里是这个注释的API文档-https://docs.spring.io/spring-framework/docs/current/javadoc-api/org/springframework/transaction/annotation/transactional.html

此外，您还可以使用Spring Data JPA，并使用JPA存储库简化数据库操作，因为它们有许多易于使用和方便的方法。使用Spring Data JPA，只需创建存储库接口（如下所示），调用其方法来保存数据：

@Repository
public interface CustomerRepository extends JpaRepository<Customer, Integer> {}

在服务中，自动连接这个存储库，并调用它的save方法：

@Autowired
private CustomerRepository customerRepository;

@org.springframework.transaction.annotation.Transactional
public Customer saveOrUpdate(Customer customer) {
    return customerRepository.save(customer); // this will save or update (if id is there)
}