java - 如何持久化实体类中的数据并与数据传输对象结合？

Question

我正在开发身份验证功能，但在保留实体类中的数据时遇到了问题。我可以从数据传输对象访问password_hash 和用户名，但无法从数据库中的User 类中查看电子邮件、firstName、lastName 或phoneNumber。下面的照片是我注册新用户时发生的情况的示例。可以看到有4列没有数据。

当我添加 newUser 时，我使用如下所示的 processRegistrationForm 方法...

    @PostMapping("/register")
    public String processRegistrationForm(@ModelAttribute @Valid RegisterFormDTO registerFormDTO,
                                          Errors errors, HttpServletRequest request,
                                          Model model) {

        // a few conditionals that I removed for brevity

        User newUser = new User(registerFormDTO.getUsername(), registerFormDTO.getPassword(), registerFormDTO.getFirstName(), registerFormDTO.getLastName(), registerFormDTO.getEmail(), registerFormDTO.getPhoneNumber());
        userRepository.save(newUser);
        setUserInSession(request.getSession(), newUser);

        return "redirect:";
    }

registerFromDTO

public class RegisterFormDTO extends LoginFormDTO{

    private String verifyPassword;


    private String firstName;

    private String lastName;

    private String email;

    private String phoneNumber;
//getters and setters.

LoginFormDTO

public class LoginFormDTO {

    @NotNull
    @NotBlank
    @Size(min = 3, max = 20, message = "Invalid username. Must be between 3 and 30 characters.")
    private String username;

    @NotNull
    @NotBlank
    @Size(min = 5, max = 20, message = "Invalid password. Must be between 5 and 30 characters.")
    private String password;
//getters and setters

用户类..

@Entity
public class User extends AbstractEntity {

    @NotBlank
    private String username;

    @NotBlank
    private String pwHash;

    //@Column(name = "first_name")
    private String firstName;
    //@Column(name = "last_name")
    private String lastName;
    //@Column(name = "email")
    private String email;
    //@Column(name = "phone_number")
    private String phoneNumber;

    public User() {}

    private static final BCryptPasswordEncoder encoder = new BCryptPasswordEncoder();

    public User(String username, String password, String firstName, String lastName, String email, String phoneNumber){
        this.username =username;
        this.pwHash = encoder.encode(password);
        this.firstName =firstName;
        this.lastName = lastName;
        this.email = email;
        this.phoneNumber = phoneNumber;
    }

    public boolean isMatchingPassword(String password) {
        return encoder.matches(password, pwHash);
    }
//getters and setters

最后但并非最不重要的是这是我的注册表单 View

<form method="post">
    <div class="form-group">
        <label>Username
            <input class="form-control" th:field="${registerFormDTO.username}" />
        </label>
        <p class="error" th:errors="${registerFormDTO.username}"></p>
    </div>
    <div class="form-group">
        <label>Password
            <input class="form-control" th:field="${registerFormDTO.password}" type="password" />
        </label>
        <p class="error" th:errors="${registerFormDTO.password}"></p>
    </div>
    <div class="form-group">
        <label>Verify Password
            <input class="form-control" th:field="${registerFormDTO.verifyPassword}" type="password" />
        </label>
    </div>
    <div class="form-group">
        <label>First Name
            <input class="form-control" th:field="${registerFormDTO.firstName}" type="firstName" />
        </label>
    </div>
    <div class="form-group">
        <label>Last Name
            <input class="form-control" th:field="${registerFormDTO.lastName}" type="lastName" />
        </label>
    </div>

    <input type="submit" class="btn btn-primary" value="Register" />
</form>

Answer 1

您可以使用MapStruct从DTO映射到模型。

例如:

    @Mapper
public interface EmployeeMapper {
    @Mappings({
      @Mapping(target="employeeId", source="entity.id"),
      @Mapping(target="employeeName", source="entity.name")
    })
    EmployeeDTO employeeToEmployeeDTO(Employee entity);
    @Mappings({
      @Mapping(target="id", source="dto.employeeId"),
      @Mapping(target="name", source="dto.employeeName")
    })
    Employee employeeDTOtoEmployee(EmployeeDTO dto);
}

“Quick Guide to MapStruct”有更多信息。