模型 java class 作为 JPA 中的连接结果

Question

考虑以下数据。这是一个虚构的在线商店系统。 Surcharge 根据不同的商品类别（食物或衣服）和支付方式（现金或信用卡）存储附加费配置。 Packaging 根据商品类别和付款方式存储可用的包装选项。如果您用现金支付食物，抱歉没有优质包装给您。

下划线表示主键。

Surcharge和Packaging是一对多的关系。我的情况是，我不想拥有 4 个 Surcharge 对象实例，每个实例都有一个 List<Packaging>。我想要 7 个对象，每个对象都有 category、paymentMeans、surcharge、option 的副本。就像他们加入时一样，结果将是 7 行。如何为我的 java class 建模以实现此目的？谢谢。

@Entity
@Table(name = "SURCHARGE")
public class Surcharge implements Serializable {

    @EmbeddedId
    private SurchargeId surchargeId;

    @Column(name = "SURCHARGE")
    private int surcharge;

    // how to map this attribute with packaging table as a single instance?
    private Packaging packaging;
}

@Embeddable
public class SurchargeId implements Serializable {

    @Column(name = "CATEGORY")
    private String category;

    @Column(name = "PAYMENT_MEANS")
    private String paymentMeans;

}

@Entity
@Table(name = "PACKAGING")
public class Packaging implements Serializable {

    @EmbeddedId
    private PackagingId packagingId;
}


@Embeddable
public class PackagingId implements Serializable {

    @Column(name = "CATEGORY")
    private String category;

    @Column(name = "PAYMENT_MEANS")
    private String paymentMeans;

    @Column(name = "OPTION")
    private String option;
}

Answer 1

在Surcharge中，总会有Packaging的列表。你可以做的是在 Packaging 中有一个 Surcharge 对象。代码将如下所示；

@ManyToOne
@JoinColumn(name = "CATEGORY", insertable = false, updatable = false)
@JoinColumn(name = "PAYMENT_MEANS", insertable = false, updatable = false)
private Surcharge surcharge;

如果您不想这样，您也可以创建一个 DTO 并使用 JPQL 相应地填充它。

模型 java class 作为 JPA 中的连接结果

Model java class as joined result in JPA

java

rdbms

jpa

one-to-one

one-to-many