JPA-如何使用來自實體繼承的criteriaBuilder.construct填充DTO

Question

我有以下（最小和部分）JPA實體：

@Entity
@Inheritance(strategy = InheritanceType.JOINED)
@DiscriminatorColumn(name = "type")
public abstract class Employee implements Serializable {
    @Id
    protected Long id;
    ...
}

@Entity
...
public class FullTimeEmployee extends Employee implements Serializable {
    private BigDecimal salary;
    ...
}

@Entity
...
public class PartTimeEmployee extends Employee implements Serializable {
   private BigDecimal hourlyWage;
   private BigDecimal maxHoursWeek;
    ...
}

當前，我們正在使用spring-data-jpa進行查詢，如下所示：

public interface EmployeeRepository extends JpaRepository<Employee, Long> {
    List<Employee> findAll();
}

但是，這樣一來，我們就會遇到“ N + 1”問題和很多選擇，因此，我決定使用Criteria API並將其選擇到DTO中，如下所示：

public List<EmployeeDTO> findAll() {

    CriteriaBuilder criteriaBuilder = this.entityManager.getCriteriaBuilder();
    CriteriaQuery<EmployeeDTO> criteriaQuery = criteriaBuilder.createQuery(EmployeeDTO.class);

    Root<Employee> root = criteriaQuery.from(Employee.class);
    Root<FullTimeEmployee> fullTimeEmployeeRoot = criteriaBuilder.treat(root, FullTimeEmployee.class);
    Root<PartTimeEmployee> partTimeEmployeeRoot = criteriaBuilder.treat(root, PartTimeEmployee.class);

    criteriaQuery.select(criteriaBuilder.construct(EmployeeDTO.class,
                          root.get("id"), root.get("name"), 
                          fullTimeEmployeeRoot.get("salary"), 
                          partTimeEmployeeRoot.get("hourlyWage"))
    );

    return this.entityManager
            .createQuery(criteriaQuery).getResultList();
}

這是我們的（示例）DTO

@Getter
@Setter
@AllArgsConstructor
public class EmployeeDTO {
    private Long id;
    private String name;
    private BigDecimal fullTimeEmployeeSalary;
    private BigDecimal partTimeEmployeeHourlyWage;
    private BigDecimal partTimeEmployeeMaxHoursWeek;
    ...
}

但是，我們得到了0個結果。

我們的休眠輸出如下所示：

SELECT employee.id, employee.name, fullTimeEmployee.salary, partTimeEmployee.hourlyWage partTimeEmployee.maxHoursWeek ... FROM employees employee INNER JOIN fullTimeEmployees fullTimeEmployee on fullTimeEmployee.id = employee.id INNER JOIN partTimeEmployees partTimeEmployee on partTimeEmployee.id = employee.id

我的問題是：最好的方法是什么？ 我如何將這些內部聯接轉換為左側聯接？ 存在更好的方法嗎？

謝謝。 :)

Answer 1

首先，我要感謝您提出的格式很好的問題-您做了一個極簡，完整且可驗證的示例，做得很好。 我認為您不想將結果投影到您所描述的此類中。 擁有salary或hourlyWage值的類意味着您一直都在檢查是否為null，這是一個非常糟糕的設計決定。 更好的方法是從employeeRepository獲取不同類型的列表，並使用面向對象的原則來處理混合類型。 這正是OOP發明的目的。

@Entity
@Inheritance(strategy = InheritanceType.JOINED)
@DiscriminatorColumn(name = "type")
public abstract class Employee implements Serializable {
    @Id @GeneratedValue(strategy=GenerationType.IDENTITY)
    protected Long id;
    public abstract BigDecimal getPay();

@Entity
public class FullTimeEmployee extends Employee {
    private BigDecimal salary;
    private int daysWorked;
    @Override
    public BigDecimal getPay() {
        return salary
                .multiply(BigDecimal.valueOf(daysWorked))
                .divide(BigDecimal.valueOf(Year.now().length()), RoundingMode.HALF_DOWN);
    }

@Entity
public class PartTimeEmployee extends Employee {
    private BigDecimal hourlyWage;
    private int hoursWorked;
    @Override
    public BigDecimal getPay() {
        return hourlyWage.multiply(BigDecimal.valueOf(hoursWorked));
    }

接着

BigDecimal sum = employeeRepo.findAll()
                     .stream()
                     .map(e->e.getPay())
                     .reduce(BigDecimal.ZERO, BigDecimal::add);