鉴于以下JPQL语句,我如何修改它,以便结果列表中的小猫按其age属性排序？
SELECT c FROM Cat c left join fetch c.kittens WHERE c.id = :id
我尝试了多个approches,但没有任何运气.这基本上是我想要做的,但它不起作用:
SELECT c FROM Cat c left join fetch c.kittens k WHERE c.id = :id ORDER BY k.age
我在两个实体之间有一个多对多的关系:汽车和经销商.
在原生MySQL中我有:
car (id and other values)
dealership (id and other values)
car_dealership (car_id and dealership_id)
我想在JPQL中创建的查询是:
#Select List of cars in multiple dealerships
SELECT car_id FROM car_dealership WHERE dealership_id IN(1,2,3,56,78,999);
使JPQL等效的正确方法是什么？
我的Java方法签名是:
public List<Car> findByDealership(List<Dealership> dealerships);
我试过了
//TOTALLY WRONG QUERY CALL!!!
Query query = em.createQuery("SELECT c FROM Car c WHERE :dealer_ids IN c.dealerships");
List<Long> dealerIds = new ArrayList<Long>();
for(Dealership d : dealerships) {
dealerIds.add(d.getId());
}
query.setParameter(":dealer_ids", dealerIds);
List<Dealership> result = (List<Dealership>) query.getResultList();
return result;
}
这是我在java中的这种关系的JPA注释:
@Entity
@Table(name = …我试过阅读QueryDSL文档,但我仍然很困惑.我习惯于编写很多SQL,但这是我第一次使用带有JPQL(JPA2)的QueryDSL.
我有以下实体:
@Entity
public class Provider implements Serializable {
@Id
@GeneratedValue(strategy = GenerationType.AUTO)
@Column(name = "id")
private Long id;
@Version
@Column(name = "version")
private Integer version;
private String name;
@ManyToMany(cascade=CascadeType.ALL)
@JoinTable(name = "provider_contact", joinColumns = @JoinColumn(name = "contact_id", referencedColumnName = "id"), inverseJoinColumns = @JoinColumn(name = "provider_id", referencedColumnName = "id"))
@OrderColumn
private Collection<Contact> contact;
}
其中Contact是一个带有idpk 的简单实体.
@Entity
public class Contact {
@Id
@GeneratedValue(strategy = GenerationType.AUTO)
@Column(name = "id")
private Long id;
/**
* User first name
*/
@NotNull …我有一个实体,我使用它ColumnTransformer来进行绑定和提取值:
@Entity
class BPoint {
@Id
private Integer id;
@ColumnTransformer(read = "astext(shape)", write = "toshape(?)")
private Shape shape;
}
道:
class BPointDao {
@Autowired
private EntityManager em;
@Override
public Page<BPoint> findAll(Pageable pageable) {
Query q = em.createQuery("from BPoint");
List<BPoint> r = q.getResultList();
int total = em.createQuery("select count(*) from BPoint").getFirstResult();
return new PageImpl(r, pageable, total);
}
@Override
public Integer save(BPoint hbds) {
em.persist(hbds);
return hbds.getId();
}
}
它工作,但是一旦我必须做一些需要使用sql函数的查询,我遇到一些问题,以这个有效的本机sql为例:
select * from BPoint h where inside(h.shape, 100) = 1;
首先我尝试使用这样的hql: …
有没有办法SELECT NEW在jpql查询中使用多个语句(Hibernate)？
这对我有用:
@Query("SELECT NEW com.test.project.dto.ItemService(g,s,l,r) "
+" FROM Item g, Service s, Service l , Service r"
+" WHERE s.id = g.id"
+" AND s.location = l.name"
+" AND s.serviceType = 'type'"
+" AND l.serviceType = 'Location'"
+" AND l.area = r.name"
+" AND r.serviceType = 'Region'")
public List<Item> getAllItemsWithServices();
我得到了预期的结果DTO.
@Component
public class ItemServiceDTO{
private Item item;
private Service serviceType;
private Service serviceLocation;
private Service serviceRegion;
public ItemServiceDTO(item item, Service serviceType, Service …我无法一起使用Spring Data JPA投影和规范.我有以下设置:
实体:
@Entity
public class Country {
@Id
@GeneratedValue(strategy = GenerationType.AUTO)
private Long id;
@Column(name = "NAME", nullable = false)
private String name;
@Column(name = "CODE", nullable = false)
private String code;
---getters & setters---
}
投影界面:
public interface CountryProjection {
String getName();
}
国家规格:
public class CountrySpecification {
public static Specification<Country> predicateName(final String name) {
return new Specification<Country>() {
@Override
public Predicate toPredicate(Root<Country> eventRoot, CriteriaQuery<?> criteriaQuery, CriteriaBuilder criteriaBuilder) {
return criteriaBuilder.equal(eventRoot.get(Country_.name), name);
}
};
}
}
库: …
我对JPA比较陌生,我想独家使用jpql解决以下问题(注意我使用的实现是Datanucleus):我有一个版本化实体表,我想得到所有实体的最新版本在表中(即我有一个实体类,它有一个id(唯一标识一行,一个entityId(在整个版本中标识实体本身)和一个时间戳;我想得到所有entityId的最新版本实体).我目前的代码如下:
String innerQueryString = "SELECT entity.entityId, max(entity.timestamp) " +
"FROM Entity entity" +
"GROUP BY entity.entityId";
Query getQuery = getEntityManager().createQuery(innerQueryString);
List<Object[]> queryRes = getQuery.getResultList();
List<IEntity> ret = new ArrayList<IEntity>();
for (Object[] res : queryRes) {
ret.add(getEntity((Long)res[0], (Date)res[1]));
}
return ret;
其中getEntity获取指定entityId的实体数据,即时间戳.我已经找到了几个关于这段代码如何在sql中工作的资源http://www.xaprb.com/blog/2006/12/07/how-to-select-the-firstleastmax-row-per-group-in- sql /但我无法设法创建它的jpql版本.非常感谢帮助,谢谢.
我需要一个jpql查询来查找所有LoadFileHistory,其中finishDate大于当前日期(从00:00:00开始).例如大于27/11/2012 00:00:00.
我已经有了这个"从LoadFileHistory o中选择o,其中o.finishDate = CURRENT_DATE",但没有得到任何东西.
如何在JPQL查询中使用数组列表？我想要这样的事情:我正在传递这个
private static final String[] MASKS = {"30109", "30111"};
成
public List<Account> findAccount(String[] Masks){
StringBuilder sb = new StringBuilder("from AccountTable a where SUBSTRING(a.Account,1,5) in :Masks ");
Query q = em.createQuery(sb.toString(), AccountTable.class)
.setParameter("Masks",Masks);
}
目前,错误是
Encountered array-valued parameter binding, but was expecting [java.lang.String (n/a)]; nested exception is java.lang.IllegalArgumentException: Encountered array-valued parameter binding, but was expecting [java.lang.String (n/a)]
I have a base class for persisted entities like this:
@EntityListeners(AuditListener.class)
@MappedSuperclass
public abstract class BaseEntity {
@Id
private String id;
private Instant createdAt;
private String createdBy;
private Instant modifiedAt;
private String modifiedBy;
...
A listener to fill the created/modified fields on persiste/update:
public class AuditListener {
@PrePersist
private void onCreate(BaseEntity entity) {
entity.setCreatedAt(Instant.now());
entity.setCreatedBy(getIdUserLogged());
}
@PreUpdate
private void onUpdate(BaseEntity entity) {
entity.setModifiedAt(Instant.now());
entity.setModifiedBy(getIdUserLogged());
}
}
For updates made by query the listener would now work and I've manually set …