参数值[2]与预期的类型[com.cityBike.app.model.User不匹配

Question

我收到以下错误

java.lang.IllegalArgumentException: Parameter value [2] did not match expected type [com.cityBike.app.model.User (n/a)] at org.hibernate.jpa.spi.BaseQueryImpl.validateBinding(BaseQueryImpl.java:885) at org.hibernate.jpa.internal.QueryImpl.access$000(QueryImpl.java:80) at org.hibernate.jpa.internal.QueryImpl$ParameterRegistrationImpl.bindValue(QueryImpl.java:248) at org.hibernate.jpa.spi.BaseQueryImpl.setParameter(BaseQueryImpl.java:631) at org.hibernate.jpa.spi.AbstractQueryImpl.setParameter(AbstractQueryImpl.java:180) at org.hibernate.jpa.spi.AbstractQueryImpl.setParameter(AbstractQueryImpl.java:49) at com.cityBike.app.service.RentService.getAllByUser(RentService.java:22)

以下是我的代码段，如何解决此问题？

文件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; 
    ...

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;
...

文件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;
    }
}

我应该添加在控制台上显示的“ 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。

Answer 1

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

您正在将a User与进行比较int。

相反，你需要写

from Rent a where a.user_id.id = :user_id

我建议重命名Rent.user_id以Rent.user避免这种错误。