1 java hibernate jpa createquery
我收到了错误
java.lang.IllegalArgumentException:参数值[2]与org.hibernate.jpa.spi.BaseQueryImpl.validateBinding(BaseQueryImpl.java:885)中的预期类型[com.cityBike.app.model.User(n/a)]不匹配)在org.hibernate.jpa.spi的org.hibernate.jpa.internal.QueryImpl.access $ 000(QueryImpl.java:80)atg.hibernate.jpa.internal.QueryImpl $ ParameterRegistrationImpl.bindValue(QueryImpl.java:248) .baseQueryImpl.setParameter(BaseQueryImpl.java:631)org.hibernate.jpa.spi.AbstractQueryImpl.setParameter(AbstractQueryImpl.java:180)org.hibernate.jpa.spi.AbstractQueryImpl.setParameter(AbstractQueryImpl.java:49)at atg.hibernate.jpa.spi.AbstractQueryImpl.setParameter com.cityBike.app.service.RentService.getAllByUser(RentService.java:22)
下面是我的代码片段,我该如何解决这个问题?
File Rent.java
@Entity
@Table(name="Rent")
public class Rent implements Serializable {
@Id
@GeneratedValue(strategy=GenerationType.AUTO)
@Column(name = "id")
private Integer id;
@ManyToOne
@JoinColumn(name = "start_id")
private Station start_id;
@ManyToOne
@JoinColumn(name = "meta_id")
private Station meta_id;
@ManyToOne
@JoinColumn(name = "user_id")
private User user_id;
...
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文件User.java
@Entity
@Table(name="Users")
public class User implements Serializable {
@Id
@GeneratedValue(strategy=GenerationType.AUTO)
@Column(name = "id")
private Integer id;
@Column(name = "login")
private String login;
...
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File RentService.java
@Service
public class RentService {
@PersistenceContext
private EntityManager em;
@Transactional
public List<Rent> getAllByUser(int user_id){
System.out.println(user_id);
List<Rent> result = em.createQuery("from Rent a where a.user_id = :user_id", Rent.class).setParameter("user_id", user_id).getResultList();
System.out.println(result);
return result;
}
}
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我应该在控制台上显示时添加"user_id"是正确的,因为它具有这样的数值ex.2或3.请指导和协助.
Rent.user_id因此,当您将a传递int给查询时,类型为用户
from Rent a where a.user_id = :user_id
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你是一个比较User有int.
相反,你需要写
from Rent a where a.user_id.id = :user_id
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我建议重命名Rent.user_id以Rent.user避免这种错误.
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