我有三个类,相同的代码显示在下面
查询类别:
@Entity
public class Enquiry
{
@Id
@GeneratedValue
private int id;
private String name;
private String discription;
private int status;
@Temporal(TemporalType.DATE)
private Date enquiryDate;
}
用户类别:
@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,如果我们想为特定的Enquiry(s)
获取User
,那么我们可以通过传递User
对象来轻松获取它,并且休眠将使用id
生成查询User
对象中的“字段”,以及下面提到的用于同一场景的代码。
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();
但是我的要求是我想根据用户名获取用户查询,或者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";
我该怎么做?
答案 0 :(得分:0)
ListenerA
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