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

Question

我收到的错误是

java.lang.IllegalArgumentException：参数值[2]与org.hibernate.jpa.spi.basequeryimpl.validateBinding（basequeryimpl.java:885)org.hibernate.jpa.internal.queryimpl.access$000（queryimpl.java:80)org.hibernate.jpa.internal.queryimpl.paraccess$parameterregistrationimpl.bindvalue（queryimpl.java:248)

下面是我的代码片段，我如何解决这个问题？

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;
...

@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;
    }
}

Answer 1

因此，当您向查询传递int时，rent.user_id的类型是用户

from Rent a where a.user_id = :user_id

您正在比较user与int。

相反，你需要写

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