java Hibernate/JPA单向OneToMany，连接条件为源表中的常量值

Question

我想使用Hibernate注释来表示使用联接的单向一对多关系。我希望在连接上添加一个条件，以便只有当源表中的一列（“一”）等于常量值时才会发生连接。比如说

SELECT *
FROM buildings b
LEFT JOIN building_floors bf on bf.building_id = b.id AND b.type = 'OFFICE'

我想表示该查询的b.type = 'OFFICE'部分

我的问题与这个问题非常相似，只是我在源表上有一个条件JPA/Hibernate Join On Constant Value

Java实体如下所示：

@Entity
@Table(name = "buildings")
public class Building {

    @Id
    @Column(name = "id")
    private int id;

    @Column(name = "type")
    private String type;

    @OneToMany(mappedBy = "buildingId",
            fetch = FetchType.EAGER,
            cascade = {CascadeType.ALL},
            orphanRemoval = true)
    @Fetch(FetchMode.JOIN)
    // buildings.type = 'OFFICE'   ????
    private Set<BuildingFloors> buildingFloors;

    // getters/setters
}

@Entity
@Table(name = "building_floors")
public class BuildingFloor {

    @Id
    @Column(name = "building_id")
    private int buildingId;

    @Id
    @Column(name = "floor_id")
    private int floorId;

    @Column(name = "description")
    private String description;

    // getters/setters
}

我尝试了一些我有占位符注释的方法：

@其中注释

这不起作用，因为这适用于目标实体

@JoinColumns注释

@JoinColumns({
        @JoinColumn(name = "building_id", referencedColumnName = "id"),
        @JoinColumn(name = "'OFFICE'", referencedColumnName = "type")
})

这不起作用，因为我得到了以下错误（为了清楚起见简化）：Syntax error in SQL statement "SELECT * FROM buildings b JOIN building_floors bf on bf.building_id = b.id AND bf.'OFFICE' = b.type"

不同的@JoinColumns注释

@JoinColumns({
        @JoinColumn(name = "building_id", referencedColumnName = "id"),
        @JoinColumn(name = "buildings.type", referencedColumnName = "'OFFICE'")
})

这不起作用，因为当使用单向OneToMany关系时，referencedColumnName来自源表。所以我得到了错误：org.hibernate.MappingException: Unable to find column with logical name: 'OFFICE' in buildings

提前谢谢

Answer 1

在我看来，您应该创建一个特定的查询来实现您的目标，而不是使用常量参数放置特定的注释。除了Hibernate，我没有看到你提到其他框架，所以我会举一些Hibernate的例子。在Building类中unidirectional映射如下所示：

@OneToMany(fetch = FetchType.Lazy, cascade = {CascadeType.ALL}, orphanRemoval = true)
@JoinTable(name = "building_floors", joinColumns = @JoinColumn(name = "id"), inverseJoinColumns = @JoinColumn(name = "building_id")
private Set<BuildingFloor> buildingFloors;

然后，您可以像这样使用TypedQuery获取数据

TypedQuery<Customer> query = getEntityManager().createNamedQuery("select b from building b inner join fetch b.buildingFloors where b.type = 'OFFICE'", Building.class);
List<Building> result = query.getResultList();

我的解决方案不是特定于Hibernate的，实际上你可以用简单的JPA来实现这一点。希望这能帮助你实现目标

Answer 2

为什么不使用inheritance ？（我和JPA一起使用，我从不直接使用hibernate）

@Entity
@Inheritance
@Table(name = "buildings")
@DiscriminatorColumn(name="type")
public class Building {

    @Id
    @Column(name = "id")
    private int id;

    @Column(name = "type")
    private String type;
}

以及：

@Entity
@DiscriminatorValue("OFFICE")
public class Office extends Building {
    @OneToMany(mappedBy = "buildingId",
        fetch = FetchType.EAGER,
        cascade = {CascadeType.ALL},
        orphanRemoval = true)
    private Set<BuildingFloors> buildingFloors;
}

Answer 3

由于需要筛选源表，所以可以使用@Loader注释

@Entity
@Table(name = "buildings")
@Loader(namedQuery = "building")
@NamedNativeQuery(name="building", 
    query="SELECT * FROM buildings b"
        + " LEFT JOIN building_floors bf on bf.building_id = b.id"
        + " WHERE b.type = 'OFFICE' AND b.id = ?",
    resultClass = Building.class)
class Building

如果在数据库中也可以使用视图，则在数据库中使用视图的方法会更好、更清晰。否则，将建筑重命名为明确表示过滤的内容

另一种方法是：@Filter，@FilterDef