在RESTAPI中将客户详细信息添加到用户

Question

我正在使用Spring Boot hibernate制作一个书店rest api，我不知道该如何进行。因此，首先创建用户（使用此处未发布的DTO）：

   @Entity
@Table(name = "app_user")
public class User {

  @Id
  @GeneratedValue(strategy = GenerationType.IDENTITY)
  @Column(name = "user_id", nullable = false, updatable = false, unique = true)
  private Long id;


  @Column(name = "email", nullable = false, unique = true)
  private String email;

  @Column(name = "password_hash", nullable = false)
  private String passwordHash;

  @OneToOne
  @JoinColumn(name = "customer_id")
  private Customer customer;

  @Column(name = "role", nullable = false)
  @Enumerated(EnumType.STRING)
  private Role role;

  // jpa requirement
  public User() {
  }

  public Long getId() {
    return id;
  }

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

  public String getEmail() {
    return email;
  }

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

  public String getPasswordHash() {
    return passwordHash;
  }

  public void setPasswordHash(String passwordHash) {
    this.passwordHash = passwordHash;
  }

  public Customer getCustomer() {
    return customer;
  }

  public void setCustomer(Customer customer) {
    this.customer = customer;
  }

  public Role getRole() {
    return role;
  }

  public void setRole(Role role) {
    this.role = role;
  }
}

然后他可以登录并搜索他感兴趣的书籍。当他想订购时（使用POST和JSON包含他的书），他应该在我的客户实体中输入他的详细信息（使用DTO）：

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

  @Id
  @GeneratedValue(strategy = GenerationType.IDENTITY)
  @Column(name = "customer_id", nullable = false, updatable = false, unique = true)
  private Long id;

  @Column(name = "first_name", nullable = false)
  private String firstName;
  @Column(name = "last_name", nullable = false)
  private String lastName;
  @Column(name = "phone_number", nullable = false)
  private String phoneNumber;

  @ManyToOne
  @JoinColumn(name = "address_id")
  private Address address;

  @OneToOne
  @JoinColumn(name = "user_id")
  private User user;

  // jpa requirement
  public Customer() {
  }

  public Long getId() {
    return id;
  }

  public void setId(Long 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 getPhoneNumber() {
    return phoneNumber;
  }

  public void setPhoneNumber(String phoneNumber) {
    this.phoneNumber = phoneNumber;
  }

  public Address getAddress() {
    return address;
  }

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

  public User getUser() {
    return user;
  }

  public void setUser(User user) {
    this.user = user;
  }

这是一个好方法吗？我不知道如何才能将用户和客户联系起来-

编辑：澄清：

>

用户注册并且只有电子邮件和密码-POST@ /users

用户添加其客户详细信息（无地址）POST@/users/1/customer:找到当前登录的用户-

用户添加其地址详细信息POST@users/1/customer/address：找到当前登录的用户-

所有3个都使用不同的DTO-这是好方法吗？

Answer 1

我认为您可以重新建模您的实体：

客户就是用户

顾客买书

创建具有默认属性的用户类。扩展用户类以创建客户类并在其中添加额外字段。创建一个Book类。将Book类的ArrayList作为books添加到Customer类。

一旦用户登录，您就已经有了用户详细信息。为用户创建客户对象，并在用户作为客户填写时添加额外的详细信息，将选定的书籍添加到客户对象，并将客户对象保存到数据库。