JPA 2.1中的NamedSubgraph在Hibernate中忽略子类。

Question

我可以协助您进行翻译。以下是您需要翻译的内容：

我正在使用Hibernate 4.3.8.FINAL，并拥有以下模型：一个部门有多个员工，而一个员工可以成为经理。

员工实体：

@Entity
@Table(name = "employee", schema = "payroll")
@Inheritance(strategy = InheritanceType.JOINED)
public class Employee
{
    @Id
    private Long id;

    @Basic(optional = false)
    @Column(name = "name")
    private String name;

    @JoinColumn(name = "department_id", referencedColumnName = "id")
    @ManyToOne(optional = false, fetch = FetchType.LAZY)
    private Department department;
}

经理实体：

@Entity
@Table(name = "manager", schema = "payroll")
@Inheritance(strategy = InheritanceType.JOINED)
@PrimaryKeyJoinColumn(name = "employee_id", referencedColumnName = "id")
public class Manager extends Employee
{
    @Basic(optional = false)
    @Column(name = "car_allowance")
    private boolean carAllowance;
}

部门实体：

@NamedEntityGraph(
        name = "Graph.Department.FetchManagers",
        includeAllAttributes = false,
        attributeNodes = {
                @NamedAttributeNode(value = "name"),
                @NamedAttributeNode(value = "employees", subgraph = "FetchManagers.Subgraph.Managers")
        },
        subgraphs = {
                @NamedSubgraph(
                        name = "FetchManagers.Subgraph.Managers",
                        type = Employee.class,
                        attributeNodes = {
                                @NamedAttributeNode(value = "name")
                        }
                ),
                @NamedSubgraph(
                        name = "FetchManagers.Subgraph.Managers",
                        type = Manager.class,
                        attributeNodes = {
                                @NamedAttributeNode(value = "carAllowance"),
                        }
                )
        }
)
@Entity
@Table(name = "department", schema = "payroll")
public class Department
{

    @Id
    private Long id;

    @Basic(optional = false)
    @Column(name = "name")
    private String name;

    @OneToMany(cascade = CascadeType.ALL, mappedBy = "department", fetch = FetchType.LAZY)
    private Set<Employee> employees;
}

如 Department 实体所示，我正在尝试创建一个@NamedSubgraph，以加载所有员工并获取 Manager.carAllowance。但是我收到以下错误：

Unable to locate Attribute  with the the given name [carAllowance] on this ManagedType [com.nemea.hydra.model.test.Employee]

据我了解，@NamedSubgraph.type 应该用于指定要获取的实体子类属性。Hibernate 是否忽略 @NamedSubgraph 注释中 type=Manager.class 属性，还是我漏掉了什么？

- edmallia

Answer 1

这可能是Hibernate 4.3.8.FINAL的缺陷，例如EclipseLink 2.5.1在使用subgraphs属性时不会抛出异常。无论如何，在Manager类型的情况下，应该使用subclassSubgraphs而不是subclass来指定。

@NamedEntityGraph(
    name = "Graph.Department.FetchManagers", 
    includeAllAttributes = false,
    attributeNodes = {
        @NamedAttributeNode(value = "name"),
        @NamedAttributeNode(value = "employees", subgraph = "FetchManagers.Subgraph.Managers")
    },
    subgraphs = {
        @NamedSubgraph(
            name = "FetchManagers.Subgraph.Managers",
            type = Employee.class,
            attributeNodes = {
                @NamedAttributeNode(value = "name")
            }
        )
    },
    subclassSubgraphs = {
        @NamedSubgraph(
            name = "FetchManagers.Subgraph.Managers",
            type = Manager.class,
            attributeNodes = {
                @NamedAttributeNode(value = "carAllowance"),
            }
        )
    }
)

- wypieprz

---

我在另一个问题中遇到了有关嵌套子图的问题。 如果您也能看一下，我将不胜感激。http://stackoverflow.com/questions/29645613/jpa-2-1-namedsubgraph-in-hibernate-ignoring-nested-subgraphs - edmallia