java - 如何根据成员对象的字段值而不是成员对象 ID 获取实体

Question

我有三个类，下面显示相同的代码

查询类：

@Entity
public class Enquiry 
{
  @Id
  @GeneratedValue
  private int id;

  private String name;

  private String discription;

  private int status;

  @Temporal(TemporalType.DATE)
  private Date enquiryDate;
}

用户等级：

 @Entity
 public class User 
 {
   @Id
   @GeneratedValue
   private int id;

   private String name;

   private String userId;

   private String password;
 }

UserEnquiryUserEnquiryMapping 类：

@Entity
public class UserEnquiryMapping 
{
   @Id
   @GeneratedValue
   private int id;

   @ManyToOne
   private User user;

   @ManyToOne
   private Enquiry enquiry;
}

现在假设如果我们想要获取Enquiry(s)一个特定User的，我们可以通过传递一个对象轻松获取它，User并且休眠将通过使用id来自User对象的字段生成查询，下面提到了相同场景的代码。

EntityManager entityManager = session.getEntityManagerFactory().createEntityManager();
CriteriaBuilder builder = entityManager.getCriteriaBuilder();

CriteriaQuery<UserEnquiryMapping> criteria = builder.createQuery(UserEnquiryMapping.class);
Root<UserEnquiryMapping> root = criteria.from(UserEnquiryMapping.class);
criteria.select(root);
criteria.where(builder.equal(root.get("user"), user));

userEnquiries = entityManager.createQuery(criteria).getResultList();

但我的要求是我想根据用户名获取用户查询，或者我们可以说我想生成这样的查询

 Select * from UserEnquiryMapping inner join Enquiry on UserEnquiryMapping.Enquiry_ID = Enquiry.ID inner join User on UserEnquiryMapping.User_ID = User.ID where User.name="Test";

我怎样才能做到这一点？

标签： javahibernateormhibernate-mappinghibernate-criteria

Answer 1

builder.equal(root.get("user").get("name"),user.getName()); 

很高兴它能帮助你！