问题描述
首先,我很抱歉如果我的问题看起来有些业余,但我对hibernate / jpa和mysql的知识有限。那么,我有以下实体:用户实体
@Entity
@Table(name =user)
public class User {
@Id
@GeneratedValue(strategy = GenerationType.AUTO)
@Column(name = id_user)
private long idUser;
@Column(nullable = false)
私有字符串名称;
@Column(nullable = false)
private String email;
@ManyToOne(可选= true,fetch = FetchType.EAGER)
@JoinColumn(name =level,referencedColumnName =id_level,nullable = true)
private水平;
@Transient
私人长汇票;
}
从那里开始,我正在尝试执行以下mysql查询: p>
public User findUserByEmail(String email){
Session session = HibernateUtil.getSessionFactory()。openSession();
User user = null;
String hql =SELECT
+u.idUser as idUser,
+u.nome as nome,
+u.level as级别,
+(SELECT count(*)FROM UserLikes likes WHERE likes.user = u.idUser)as totalLikes
+FROM User u WHERE u.email =:email;
尝试{
Query query = session.createQuery(hql);
query.setString(email,email);
query.setResultTransformer(Transformers.aliasToBean(User.class));
user =(Usuario)query.uniqueResult();
} catch(RuntimeException ex){
throw ex;
} finally {
session.close();
}
返回用户;
$ b上面的代码正常返回用户,但是当用户搜索时返回null在数据库中没有设置字段级别的值。也就是说,如果我没有错,hibernate会执行内部连接用户/级别,而不是左连接,从而使返回值为null。
有没有人知道如何获取在这附近?
解决方案我认为你需要使用这个。
@ManyToOne //简单
@JoinColumn(nullable = true)简单
并在您的查询中使用简单的方式。
FROM用户u WHERE u.email =:电子邮件
我在类似的项目中使用它并工作。
干得好。
First I apologize if my question seems a bit amateurish, but my knowledge with hibernate / jpa and mysql are somewhat limited.
Well, I have the following entity:
User Entity
@Entity
@Table(name = "user")
public class User {
@Id
@GeneratedValue(strategy = GenerationType.AUTO)
@Column(name = "id_user")
private Long idUser;
@Column(nullable = false)
private String name;
@Column(nullable = false)
private String email;
@ManyToOne(optional = true, fetch = FetchType.EAGER)
@JoinColumn(name = "level", referencedColumnName = "id_level", nullable = true)
private Level level;
@Transient
private Long totalLikes;
}
From there, I'm trying to do the following mysql query:
public User findUserByEmail(String email) {
Session session = HibernateUtil.getSessionFactory().openSession();
User user = null;
String hql = "SELECT "
+ "u.idUser as idUser, "
+ "u.nome as nome, "
+ "u.level as level, "
+ "(SELECT count(*) FROM UserLikes likes WHERE likes.user = u.idUser) as totalLikes "
+ "FROM User u WHERE u.email = :email";
try {
Query query = session.createQuery(hql);
query.setString("email", email);
query.setResultTransformer(Transformers.aliasToBean(User.class));
user = (Usuario) query.uniqueResult();
} catch (RuntimeException ex) {
throw ex;
} finally {
session.close();
}
return user;
}
The above code returns the User normally, however returns null when the User to search in the database do not have value set in the field level. That is, if I am not wrong, hibernate does an inner join user / level and not a left join, making returns null.
Does anyone have any idea how to get around this?
解决方案 i think you need use this.
@ManyToOne //simple
@JoinColumn(nullable = true) simple
and in your query using simple too.
FROM User u WHERE u.email = :email
i use this in similiar project and worked.
Good job man.
这篇关于Hibernate在用@ManyToOne查询时返回null hql的文章就介绍到这了,希望我们推荐的答案对大家有所帮助,也希望大家多多支持!