java - JPA/hibernate : mapping a single element in an entity with an one-to-many relationship

Question

我有以下表格:

TABLE Orders (
    ID BIGINT NOT NULL PRIMARY KEY,
    ... other columns ...
)

TABLE Order_States (
    ID         BIGINT NOT NULL PRIMARY KEY,
    Order_ID   BIGINT NOT NULL FOREIGN KEY REFERENCES Orders(ID),
    State_Type VARCHAR(30) NOT NULL,
    State_Date DATETIME NOT NULL,
    ... other columns ...
)

以及以下映射:

@Entity
@Table(name = "orders")
class Order {

    @Id
    @GeneratedValue
    private Long id;

    @OneToMany(mappedBy = "order")
    @OrderBy("stateDate DESC")
    private List<OrderState> orderStates

    public OrderState getCurrentState() {
        return orderStates.get(0);
    }

    public void setCurrentState(OrderState state) {
        state.setStateDate(new Date());
        orderStates.add(state);
    }

    ... other members ...
}

@Entity
@Table(name = "order_states")
class OrderState {

    @Id
    @GeneratedValue
    private Long id;

    @ManyToOne
    @JoinColumn(name = "order_id")
    private Order order;

    @Enumerated(EnumType.STRING)
    @Column(name = "state_type")
    private StateType stateType;

    @Temporal(TemporalType.TIMESTAMP)
    @Column(name = "state_date")
    private Date stateDate;

    ... other members ...
}

我不想在我的映射中包含所有订单的状态(它们只是出于历史原因)，而只是订单的当前状态，它由最新状态的日期决定。

这个有注释吗？例如(我猜):

@Entity
@Table(name = "orders")
class Order {
    ...

    @OneToMany(mappedBy = "order")
    @OrderBy("stateDate DESC")
    @TakeFirst
    private OrderState orderState

    ... getter and setter for orderState ...

    ...
}

Answer 1

您所说的功能称为软删除。 Here是通过使用 Hibernate 的注释 @SQLDelete 和 @Where 来做这些事情的例子。

但是下面的代码很难实现

@TakeFirst
private OrderState orderState

这样写会更容易

@TakeFirst
private Set<OrderState> orderState // this will be a set containing one entry