在 JPA CriteriaQuery 中使用 @Embeddable 实体

Question

假设我有以下示例实体 - 一个是 @Embeddable，嵌入另一个 @Entity:

@Embeddable
public class ContactInfoEntity {

    @Column
    private String phone;

    @Column
    private String zipCode;
}

@Entity
@Table(name = "EMPLOYEE")
public class EmployeeEntity {

    @Id
    @Column(name = "EMPLOYEE_ID")
    private Long employeeId;

    @Embedded
    @AttributeOverrides({
        @AttributeOverride(name = "phone",
                           column = @Column(name = "EMPLOYEE_PHONE")),
        @AttributeOverride(name = "zipCode",
                           column = @Column(name = "EMPLOYEE_ZIP_CODE"))
    })
    private ContactInfoEntity employeeContactInfo;
}

由 openjpa-maven-plugin 生成的元模型类仅包含一个 employeeContactInfo 变量，不包含 @AttributeOverride 列。

现在假设我想这样做：

Select the EMPLOYEE_ID and EMPLOYEE_PHONE where the EMPLOYEE_ZIP_CODE is equal to "123456"

如何将其创建为 CriteriaQuery？

CriteriaBuilder cb = entityManager.getCriteriaBuilder();

CriteriaQuery<String> qDef = cb.createQuery(String.class);
Root<EmployeeEntity> e = qDef.from(EmployeeEntity.class);
qDef.select(e.get(EmployeeEntity_.employeeId),
            e.get(????))
    .where(cb.equal(e.get(????), "123456"));

return entityManager.createQuery(qDef).getResultList();

Answer 1

示例方法可能如下所示：

CriteriaBuilder cb = em.getCriteriaBuilder();
CriteriaQuery<Object[]> qDef = cb.createQuery(Object[].class);
Root<EmployeeEntity> e = qDef.from(EmployeeEntity.class);
qDef.multiselect(
     e.get(EmployeeEntity_.employeeId),
     e.get(EmployeeEntity_.employeeContactInfo).get(ContactInfoEntity_.phone));
qDef.where(
     cb.equal(
        e.get(EmployeeEntity_.employeeContactInfo).get(ContactInfoEntity_.zipCode), 
        cb.literal("123456")));

List<Object[]> objects = em.createQuery(qDef).getResultList();
for (Object[] element : objects) {
    System.out.format("%d %s", element[0], element[1]);
}

根据您的喜好，您可能还希望获得如下查询结果：

构造函数表达式

public class EmployeeEntityResult {
    private int id;
    private String phone;

    public EmployeeEntityResult(int id, String phone) {
        this.id = id;
        this.phone = phone;
    }
    ...
}

CriteriaQuery<EmployeeEntityResult> cq = cb.createQuery(EmployeeEntityResult.class);
...
List<EmployeeEntityResult> result = em.createQuery(cq).getResultList();
for (EmployeeEntityResult element : result) {
    System.out.format("%d %s", element.getId(), element.getPhone());
}

元组

CriteriaQuery<Tuple> cq = cb.createTupleQuery();
...
cq.select(
   cb.tuple(
      e.get(EmployeeEntity_.employeeId)
       .alias("id"),
      e.get(EmployeeEntity_.employeeContactInfo).get(ContactInfoEntity_.phone)
       .alias("phone")));
...
List<Tuple> tuple = em.createQuery(cq).getResultList();
for (Tuple element : tuple) {
    System.out.format("%d %s", element.get("id"), element.get("phone"));
}

JPQL 查询如下所示：

SELECT e.id, e.employeeContactInfo.phone
FROM EmployeeEntity e
WHERE e.employeeContactInfo.zipCode = '123456'

在 JPA CriteriaQuery 中使用 @Embeddable 实体

Using an @Embeddable entity in a JPA CriteriaQuery

java

jpa

criteria

openjpa

criteria-api