JPA Native Query 映射一对多

Question

我有一个复杂的 Oracle 查询，为了简单起见，它看起来像这样；

SQL> SELECT d.id AS dept_id,
  2    d.name AS dept_name,
  3    e.id AS emp_id,
  4    e.name AS emp_name,
  5    e.dept_id AS emp_dept_id
  6  FROM drs2_dept d, drs2_emp e
  7  WHERE d.id = e.dept_id (+)
  8  /

   DEPT_ID DEPT_NAME           EMP_ID      EMP_NAME      EMP_DEPT_ID
---------- ------------------- ---------- -------------- -----------
         1 SALES                       101 JOHN                    1
         1 SALES                       102 JANE                    1
         2 ADMIN                                                       

我的Department 班级是；

@SqlResultSetMapping(
    name = "Department.employeeMapping",
    classes = {
        @ConstructorResult(
            targetClass = Department.class,
            columns = {
                @ColumnResult(name = "DEPT_ID", type = Integer.class),
                @ColumnResult(name = "DEPT_NAME")
            }
        ),
        @ConstructorResult(
            targetClass = Employee.class,
            columns = {
                @ColumnResult(name = "EMP_ID", type = Integer.class),
                @ColumnResult(name = "EMP_NAME"),
                @ColumnResult(name = "EMP_DEPT_ID", type = Integer.class)
            }
        )
    }
)

@NamedNativeQuery(
    name = "Department.findAllEmployees",
    query = "SELECT d.id AS dept_id, " + 
            "  d.name AS dept_name,  " +
            "  e.id AS emp_id,  " +
            "  e.name AS emp_name " +
            "  e.dept_id AS emp_dept_id, " +
            "FROM drs2_dept d, drs2_emp e " +
            "WHERE d.id = e.dept_id (+)",
    resultSetMapping = "Department.employeeMapping"
)

@Entity
public class Department {
    @Id // JPA will not start without it.
    Integer id;  
    String name;

    @OneToMany // JPA will not start without it.
    List<Employee> employees = new ArrayList<>();

    public Department(Integer id, String name) {
        this.id = id;
        this.name = name;
    }

    public Department() {}

    // getters and setters
}

我的Employee 班级是；

@Entity
public class Employee {
    @Id Integer id;
    Integer departmentId;
    String name;

    public Employee(Integer id, String name, Integer departmentId) {
        this.id = id;
        this.name = name;
        this.departmentId = departmentId;
    }

    public Employee() {}

    // getters and setters
}

因为我使用的是@ConstructorResult，所以我可以得到数据，但它仍然是一个扁平结构，也就是说一个List<Object[]>有三个条目，每个条目都包含[Department, Employee]。所以我必须执行以下操作才能将Employee 记录移动到各自的Department 中；

@Component
public class DepartmentDAO  {

    @PersistenceContext EntityManager entityManager;

    public Collection<Department> getAllDepartments() {
        Query query = entityManager.createNamedQuery("Department.findAllEmployees");

        Map<Integer, Department> map = new HashMap<>();
        List<Object[]> list = query.getResultList();
        for (Object[] tuple: list) {
            Department d = (Department) tuple[0];
            if (! map.containsKey(d.getId())) {
                map.put(d.getId(), d);
            }
            d = map.get(d.getId());
            Employee e = (Employee) tuple[1];
            if (e.getId() != null) {
                d.getEmployees().add(e);
            }
        }
        return map.values();
    }
}

每当我向@OneToMany 添加任何其他属性时，我似乎都会在 Hibernate 日志中生成不正确的虚假 SQL（即不存在的列或表名），但正如我在此问题开头所述，我只想要本机 SQL - 我不希望 Hibernate 弄清楚我想要做什么。

有没有办法让 JPA/Hibernate 将 Employee 对象放入 Department 的列表中？

( 作为子注释，我在这里看到过这个问题，但在 2011 年之前从未回答或回答过，那时 JPA 和 Hibernate 可能已经取得了进展。

我还应该补充一点，在我的项目的其他地方，我已经将 Department 和 Employee 完全映射为 CrudRepository 与 @Table 和 @Column 一起使用，但是他们的 @OneToMany 定义并不能描述我的身份在上面的查询中做，因此在我的示例代码中没有它们。 )

Answer 1

此查询没有任何子句强制它使用本机实现。 事实上，这被认为是一种不好的做法。

试试这个：

CriteriaBuilder cb = em.getCriteriaBuilder();
CriteriaQuery<Departament> cq = cb.createQuery(Departament.class);

Root<Departament> rootDepartament = cq.from(Departament.class);
Join<Departament,Employee> joinEmployee = rootDepartament.join(Departament_.employees,JoinType.Left);

cq.select(rootDepartament);

List<Departament> result = entityManager.createQuery(cq).getResultList();