休眠查詢多對象選擇

Question

我有一個這樣的實體模型：

public class Facture implements Serializable 
{
@Id @GeneratedValue(strategy=GenerationType.SEQUENCE, generator="SEQ_FACTURE")
private long idFacture;
...

private Panier panier;
    ...
 }

 public class Panier
 {
@Id @GeneratedValue(strategy=GenerationType.SEQUENCE, generator="SEQ_PANIER")
private long idPanier;  

@ManyToOne
private Client client;
@OneToMany
private List<LignePanier> articles = new ArrayList<LignePanier>();
...
 }

 public class Client
 {
@Id @GeneratedValue(strategy=GenerationType.SEQUENCE, generator="SEQ_CLIENT")
private long idClient;
...
  }

因此，我想從客戶端X查詢所有斷裂。我嘗試執行以下操作：

 public List<Facture> listeFacture(Long clientID) {
    List<ParameterMap> parameters = new ArrayList<ParameterMap>();
    parameters.add(new ParameterMap(StandardBasicTypes.LONG, clientID));
    return dao.query("select facture from Facture facture where facture.panier.client.idClient = ?", parameters);
}

我得到這個例外：

  org.hibernate.QueryException: could not resolve property: client of: be.infoserv.web.model.Facture [select facture from be.infoserv.web.model.Facture facture where facture.panier.client.idClient = ?]

我認為不可能查詢這樣的第三個對象，但我不知道如何編寫此查詢...

對不起，我的英語語言，我是法語用戶。

Answer 1

您可能必須使用內部聯接來執行此操作：

select facture
from Facture facture
     inner join facture.panier as panier
     inner join panier.client as client
where client.clientId = ?

或使用可能更安全的標准，因為您不能破壞hql：

Criteria factureCrit = session.createCriteria(Facture.class);
Criteria panierCrit = factureCrit.createCriteria("panier");
Criteria clientCrit = panierCrit.createCriteria("client");
clientCrit.add(Restrictions.idEq(clientId));

return factureCrit.list();