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

Parameter value [2] did not match expected type [com.cityBike.app.model.User

提问by wero

I receive the error of

我收到错误

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)

java.lang.IllegalArgumentException：参数值 [2] 与 org.hibernate.jpa.spi.BaseQueryImpl.validateBinding(BaseQueryImpl.java:885) 的预期类型 [com.cityBike.app.model.User (n/a)] 不匹配) 在 org.hibernate.jpa.internal.QueryImpl.access$000(QueryImpl.java:80) 在 org.hibernate.jpa.internal.QueryImpl$ParameterRegistrationImpl.bindValue(QueryImpl.java:248) 在 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)

Below is my code snippet, how can I fix this issue?

下面是我的代码片段，我该如何解决这个问题？

File Rent.java

文件租用.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; 
    ... 

File User.java

文件用户.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;
...

File RentService.java

文件租用服务.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;
    }
}

I should add that "user_id" when displayed on the console is correct as it has such a numerical value ex. 2 or 3. Please guidance and assistance.

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。请指导和帮助。

回答by wero

The Type of Rent.user_idis User therefore when you pass a intto the query

Rent.user_id因此，当您将 a 传递int给查询时，类型是用户

from Rent a where a.user_id = :user_id

you are comparing a Userwith an int.

您正在将 aUser与a 进行比较int。

Instead you need to write

相反，你需要写

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

I would recommend to rename Rent.user_idto Rent.userto avoid this kind of error.

我建议重命名Rent.user_id以Rent.user避免这种错误。

回答by karim mohsen

You can use this query :-

您可以使用此查询：-

The user_idin Rentclass isn't int it's an object of type Userso you have to get the Userobject by using it's id

该user_id在Rent类不是INT它的类型的对象User，所以你必须让User使用它的ID对象

List<Rent> result = em.createQuery("from Rent a where a.user_id = :user_id", Rent.class).setParameter("user_id", em.getReference(User.class, user_id)).getResultList();

Or you can write :-

或者你可以写：-

List<Rent> result = em.createQuery("from Rent a where a.user_id.id = :user_id",Rent.class).setParameter("user_id", user_id).getResultList();