如何在 hibernate 中找到外鍵的 id

Question

我有兩個實體，即：

State

@Entity
@Table(name = "State")
public class StateEntity {

   @Column(name = "id", length = 36, nullable = false, unique = true)
   private String id;

  @ManyToOne (fetch = FetchType.LAZY)
  @JoinColumn(name = "InsurerId", nullable = false)
  private InsurerEntity insurer;

  @Column(name ="StateName", length = 50, nullable = false)
  private String stateName;

//getters and setters

}

保險公司

@Entity
@Table(name = "Insurer")
public class InsurerEntity {

  @Column(name = "InsurerId", length = 36, nullable = false, unique = true)
  private String insurerId;

  @Column(name = "InsurerName", length = 100, nullable = true)
  private String insurerName;

  @OneToMany(mappedBy = "state", fetch = FetchType.LAZY)
  private List<StateEntity> stateEntityList;

//getters and setters

}

保險公司的 id 保存在state數據庫中，我想使用 hibernate 查詢檢索它，但我似乎找不到解決方案

如何編寫此查詢SELECT InsurerId FROM State; 在 Hibernate 查詢中使用CriteriaBuilder ， CriteriaQuery和Root ..

Answer 1

如果您想 select 所有州的所有保險公司 ID：

String selectionQuery = "SELECT s.insurer.insurerId FROM State s";
List<String> insurersIds = session.createQuery(selectionQuery).list();

如果您想 select 某個 state 的 Insurer's Id：

String selectionQuery = "SELECT s.insurer.insurerId FROM State s WHERE s.id = :stateId";
String insurerId = (String) session.createQuery(selectionQuery).setParameter("stateId", stateId).getSingleResult(); //This should be placed in a try/catch block to handle org.hibernate.NonUniqueResultException

編輯：

正如 Prasad 在他的回答中所寫，您應該更新您的 Insurer 實體。

Answer 2

為此，您還必須在 map 兩個 class 中將 @oneToMany 注釋放入 class InsurerEntity 中

@OneToMany(fetch = FetchType.LAZY,mappedBy="StateEntity", cascade = CascadeType.ALL)

private List< StateEntity > StateEntitys;

當您獲取狀態時，您還將在其中獲得 InsurerEntity 的 object ，您可以使用 getter 訪問它