Spring Data JPA 规范继承

Question

我有三个实体，如下所示：

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

@Entity
public class FullTimeEmployee extends Employee {
    protected Integer salary;
    ...
}

@Entity
public class PartTimeEmployee extends Employee {
    protected Float hourlyWage;
}

我将尝试使用Spring Data规范进行查询，如下所示:

我的Spring数据管理存储库：

@Repository
public interface EmployeeRepository<T extends Employee> extends JpaRepository<T, Long> {
    Page<T> findAll(Specification<T> specification, Pageable pageable);
}

我的示例控制器

@Controller
public class MyController {
    private final EmployeeRepository employeeRepository;

    @Autowired
    public MyController(EmployeeRepository employeeRepository) {
        this.employeeRepository = employeeRepository;
    }

    @GetMapping("/")
    public ModelAndView queryEmployeeExample(EmployeeQueryDTO query) {
        Page response = this.employeeRepository.findAll((root, criteriaQuery, criteriaBuilder) -> {
            ArrayList<Predicate> predicates = new ArrayList<>();
            if(query.getId() != null) {
                predicates.add(
                    criteriaBuilder.equal(root.get("id"), query.getId())
                );
            }
            if(query.getSalary() != null) {
                predicates.add(
                    criteriaBuilder.equal(root.get("salary"), query.getSalary())
                );
            }
            if(query.getHourlyWage() != null) {
                predicates.add(
                    criteriaBuilder.equal(root.get("hourlyWage"), query.getHourlyWage())
                );
            }
            return criteriaBuilder.and(predicates.toArray(new Predicate[predicates.size()]));
        }, new PageRequest(0, 100));

        return new ModelAndView("index")
                    .addObject("result", response);
    }

    @Getter
    @Setter
    static class EmployeeQueryDTO {
        private Long id;
        private Integer salary;
        private Float hourlyWage;
    }
}

我可以使用id执行查询，但是其他参数（如salary和hourlyWage）却不行，我收到了以下异常信息：

java.lang.IllegalArgumentException: Unable to locate Attribute with the given name [salary/hourlyWage] on this ManagedType [Employee]

我理解问题在于在超类中查找字段时它不存在，我想知道是否有解决方法？

祝好。

- Joao Moreno

---

使用treat: https://docs.oracle.com/javaee/7/api/javax/persistence/criteria/CriteriaBuilder.html#treat-javax.persistence.criteria.Root-java.lang.Class-使用treat方法，可以将实体根类型转换为其子类型，这在Criteria查询中非常有用。该方法返回一个新的Root对象，它表示具有给定子类型的实体根。例如，如果有一个Person类和一个Employee类，Employee是Person的子类型，那么可以使用treat方法将Root<Person>转换为Root<Employee>，并以此创建与Employee相关的查询。 - JB Nizet

@JBNizet非常好用！ :) 非常感谢您。 - Joao Moreno

请将您用于解决问题的代码作为答案发布。 - JB Nizet

当然，我现在正在做这件事... - Joao Moreno

Answer 1

根据 @JBNizet 的提示，使用 CriteriaBuilder.treat()：

@GetMapping("/")
public ModelAndView queryEmployeeExample(EmployeeQueryDTO query) {
    Page response = this.employeeRepository.findAll((root, 
                                                    criteriaQuery, 
                                                    criteriaBuilder) -> {

    ArrayList<Predicate> predicates = new ArrayList<>();

    if(query.getId() != null) {
        predicates.add(
            criteriaBuilder.equal(
                    root.get("id"), 
                    query.getId()
            )
        );
    }
    if(query.getSalary() != null) {
        //Downcast Root object to the specified type.
        Root<FullTimeEmployee> rootFullTImeEmployee = criteriaBuilder
                                        .treat(root, FullTimeEmployee.class);
        predicates.add(
            criteriaBuilder.equal(
                    rootFullTImeEmployee.get("salary"), 
                    query.getSalary()
            )
        );
    }
    if(query.getHourlyWage() != null) {
        //Downcast Root object to the specified type.
        Root<PartTimeEmployee> rootPartTimeEmployee = criteriaBuilder
                                        .treat(root, PartTimeEmployee.class);
        predicates.add(
            criteriaBuilder.equal(
                    rootPartTimeEmployee.get("hourlyWage"), 
                    query.getHourlyWage()
            )
        );
    }
    return criteriaBuilder.and(predicates.toArray(new Predicate[predicates.size()]));}, 
                new PageRequest(0, 100));
    
    return new ModelAndView("index")
                .addObject("result", response);
}
    
@Getter
@Setter
static class EmployeeQueryDTO {
    private Long id;
    private Integer salary;
    private Float hourlyWage;
}

- Joao Moreno

---

谢谢，这个答案帮助了我一整天的努力。这应该被接受为正确答案。 - Mubasher