Spring JPA 在嵌入的复合键上使用 @MapsId、@AttributeOverride 时插入重复的选择列

Question

我正在开发一个 Spring Boot REST API 应用程序，但在为 Transact-SQL (SQL Server) 方言生成 SQL 时遇到了问题，我不确定我在哪里做错了。

该应用程序是关于存储管理的，我有两个实体：Part 和 Stock。我已将结构简化为尽可能简单。

我有复合 PK - PartPK:

@Data @Embeddable 
class PartPK {

     @Column(name = "PART_ID")
     private String partId;

     @Column(name = "PART_ORGANIZATION_ID")
     private String orgId;

}

... 实体Part 具有PartPK 作为@EmbeddedId：

@Entity @Table(name = "parts") 
class Part {

    @EmbeddedId
    private PartPK id;
}

然后我有一个Stock 实体，它与Part 实体和商店相关联。该实体具有具有以下结构的复合 PK，其中我将覆盖来自 PartPK 的属性（给它们 STOCK_ 前缀）

 @Data @Embeddable 
 class StockPK {

     @Column(name = "STOCK_STORE_ID")
     private String storeId;

    @Embedded
    @AttributeOverrides({
        @AttributeOverride(name = "partId", column = @Column(name = "STOCK_PART_ID")),
        @AttributeOverride(name = "orgId", column = @Column(name = "STOCK_PART_ORGANIZATION_ID")),
    })
    private PartPK partId;

}

... 并附上 Stock 实体，我尝试使用 @MapsId 引用 Part 实体：

@Entity @Table(name = "stocks") 
class Stock {

    @EmbeddedId
    private StockPK id;

    @MapsId("partId")
    @ManyToOne(fetch = FetchType.LAZY)
    @JoinColumns({
        @JoinColumn(name = "STOCK_PART_ID", referencedColumnName = "PART_ID"),
        @JoinColumn(name = "STOCK_PART_ORGANIZATION_ID", referencedColumnName = "PART_ORGANIZATION_ID"),
    })
    private Part part;

}

编译，但在从存储库执行查询后，它会生成以下查询：

select TOP (?) 
           stockdb0_.stock_part_organization_id as bis_part0_0_,
           stockdb0_.stock_store_id             as bis_stor3_0_,
           stockdb0_.stock_part_organization_id as bis_part5_0_,
           stockdb0_.stock_part_id              as bis_part6_0_
from stocks stockdb0_

如您所见，由于某种原因，它使用了 2 次 stock_part_organization_id 列。实体在持久化映射后具有不正确的值（具有相同 Store 但不同部分的两个 Stock 行被认为是同一实体）。从Stock 实体中删除part 属性时，查询和生成的持久性映射是正确的。

是不是我做错了什么？

我用的是Spring Boot 2.4.5（最新）和同版本的Started Data Jpa。

Answer 1

我认为在这种情况下使用@IdClass 会更好：

class StockPK implements Serializable {

  private String storeId;

  private Part part;

  ...
}

@Entity @Table(name = "stocks") 
@IdClass(StockPK.class)
class Stock {

    @Id
    private String id;

    @Id
    @ManyToOne(fetch = FetchType.LAZY)
    @JoinColumns({
        @JoinColumn(name = "STOCK_PART_ID", referencedColumnName = "PART_ID"),
        @JoinColumn(name = "STOCK_PART_ORGANIZATION_ID", referencedColumnName = "PART_ORGANIZATION_ID"),
    })
    private Part part;

    ...
}

但是如果你想使用@EmbeddedId:

@Embeddable
public static class StockPK implements Serializable {

    @ManyToOne(fetch = FetchType.LAZY)
    @JoinColumns({
        @JoinColumn(name = "STOCK_PART_ID", referencedColumnName = "PART_ID"),
        @JoinColumn(name = "STOCK_PART_ORGANIZATION_ID", referencedColumnName = "PART_ORGANIZATION_ID"),
    })
    private Part part;

    @Column(name = "STOCK_STORE_ID")
    private String storeId;


}

@Entity @Table(name = "stocks") 
class Stock {

    @EmbeddedId
    private StockPK id;
    
    // The association is already defined in the key
}

无论如何，您不必使用@MapsId (that's for something else)，您可以找到这两种方法的示例以及更多详细信息in the Hibernate ORM documentation。