java JPA/Hibernate 加入常量值
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JPA/Hibernate Join On Constant Value
提问by EvilJoe
I am trying to join to different entities from the same table using a constant value in the join statement. In SQL, I would do something like this...
我正在尝试使用 join 语句中的常量值从同一个表中加入不同的实体。在 SQL 中，我会做这样的事情......
SELECT *
FROM owner o
JOIN types t on t.owner_id = o.id AND t.type = 'A'
-- ^^^^^^^^^^^^^^^^ THIS IS WHAT I AM TRYING TO REPLICATE
In Java + JPA/Hibernate, I am trying to do something like this...
在 Java + JPA/Hibernate 中，我正在尝试做这样的事情......
@Entity
@Table(name = "OWNER")
public class Owner {
@Id
@Column(name="ID")
private Long id
@OneToOne(mappedBy = "owner", fetch = FetchType.EAGER, cascade = CascadeType.ALL)
@JoinColumns({
@JoinColumn(name = "ID", referencedColumnName = "ID"),
@JoinColumn(constantValue = "A", referencedColumnName="type")})
// ^^^^^^^^^^^^^^^^^^^ I AM LOOKING FOR SOMETHING THAT DOES THIS.
// constantValue IS NOT A VALID ARGUMENT HERE.
private TypeA inspectionSnapshot;
@OneToOne(mappedBy = "owner", fetch = FetchType.EAGER, cascade = CascadeType.ALL)
@JoinColumns({
@JoinColumn(name = "ID", referencedColumnName = "ID"),
@JoinColumn(constantValue = "B", referencedColumnName="type")})
private TypeB inspectionSnapshot;
/* Getters & Setters ... */
}
@Entity
@Table(name = "TYPES")
@Inheritance(strategy = InheritanceType.SINGLE_TABLE)
@DiscriminatorColumn(name = "TYPE", discriminatorType = DiscriminatorType.STRING)
public abstract class BaseType {
@Id
@OneToOne
@JoinColumn(name = "OWNER_ID", referencedColumnName="ID")
private Owner id;
@Id
@Column(name = "TYPE")
private char type
/* Getters & Setters ... */
}
@Entity
@DiscriminatorValue("A")
public class TypeA extends BaseType {
/* All functionality in BaseType */
}
@Entity
@DiscriminatorValue("B")
public class TypeA extends BaseType {
/* All functionality in BaseType */
}
Thanks in advance!
提前致谢！
采纳答案by Kikin-Sama
You're looking at non-standard joins. Here's the documentation for treating such a case:
您正在查看非标准连接。这是处理这种情况的文档：
http://docs.oracle.com/cd/E13189_01/kodo/docs40/full/html/ref_guide_mapping_notes_nonstdjoins.html
http://docs.oracle.com/cd/E13189_01/kodo/docs40/full/html/ref_guide_mapping_notes_nonstdjoins.html
Hope it helps!
希望能帮助到你！
回答by Dmitry
Try to specify a constant as the value of the formula
尝试指定一个常量作为公式的值
@JoinColumnsOrFormulas({
@JoinColumnOrFormula(formula=@JoinFormula(value="'A'", referencedColumnName="type")),
@JoinColumnOrFormula(column = @JoinColumn("id", referencedColumnName="id"))
})
private TypeA inspectionSnapshot;
回答by David Levesque
If you don't mind using Hibernate-specific annotations you could try with the @WhereJoinTableannotation, e.g.:
如果您不介意使用特定于 Hibernate 的注释，您可以尝试使用@WhereJoinTable注释，例如：
@OneToOne(mappedBy = "owner", fetch = FetchType.EAGER, cascade = CascadeType.ALL)
@JoinColumn(name = "ID", referencedColumnName = "ID")
@WhereJoinTable(clause = "TYPE = 'A'")
private TypeA inspectionSnapshot;
Note that the clauseattribute must contain SQL, not JPQL, so you need to use the database column name instead of the JPA entity field name.
请注意，该clause属性必须包含 SQL，而不是 JPQL，因此您需要使用数据库列名称而不是 JPA 实体字段名称。
