我有以下的數據庫模型:JPA2標準的API:選擇...在(從那裏選擇)
A
aId
AB
aId
bId
B
bId
status
在Spring數據規範,我想回到A,此時B.status的情況下,是'X'。 的JPQL代碼如下:
select a from A a where a in
(select ab.id.a from AB ab where ab.id.b.status= :status)
這些模型類:
@Entity
public class A {
private Long aId;
@OneToMany(fetch = FetchType.EAGER, cascade = CascadeType.ALL, mappedBy = "id.a")
private Set<AB> ab;
}
@Entity
public class B {
private Long bId;
private String Status;
@OneToMany(fetch = FetchType.EAGER, cascade = CascadeType.ALL, mappedBy = "id.b")
private Set<AB> ab;
}
@Entity
public class AB {
private ABPK id;
}
public class ABPK {
@ManyToOne
@JoinColumn(name="aId")
private A a;
@ManyToOne
@JoinColumn(name="bId")
private B b;
}
如何將是JPA標準在Spring規範?
public class ASpecifications {
public static Specification<A> test(final String status) {
return new Specification<Party>() {
@Override
public Predicate toPredicate(Root<A> a, CriteriaQuery<?> query, CriteriaBuilder cb) {
return null;
}
};
}
}
謝謝codeturner。不幸的是,該鏈接中的示例並不能幫助我使用embeddedId。在我的情況下,我沒有subQuery的映射屬性。我需要aId,但ABPK有A,而不是aId。 – Techky