java - java.lang.IllegalArgumentException:无法将int字段it.besmart.models.Park.idPark设置为[Ljava.lang.Object;

我正在解决这个问题。
我有3节课

User和Park具有ManyToMany关系，而Piano和Park具有ManyToOne关系。

在模型中，有趣的部分是
User.java

@ManyToMany(fetch = FetchType.EAGER)
    @JoinTable(name="users_parcheggio", joinColumns = {@JoinColumn(name="user_id") }, inverseJoinColumns = {
            @JoinColumn(name="park_id") })
    private Set<Park> parks = new HashSet<Park>();

这是Park.java

 @Entity
@Table(name="parcheggio", catalog="SMARTPARK")
public class Park implements Serializable{

    /**
     *
     */
    private static final long serialVersionUID = -7630704706109692038L;
    @Id
    @GeneratedValue(strategy = IDENTITY)
    @Column(name = "id_park", unique = true, nullable = false)
    private  int idPark;

    @NotEmpty
    @Column(name="nome_park")
    private  String nomePark;

    @Column(name="descrizione")
    private  String descrizione;

    @Column(name="indirizzo")
    private String indirizzo;

    @Column(name="citta")
    private String citta;


    @ManyToMany(fetch = FetchType.EAGER)
    @JoinTable(name="users_parcheggio", joinColumns = {@JoinColumn(name="park_id") }, inverseJoinColumns = {
            @JoinColumn(name="user_id") })

    private Set<User> users = new HashSet<User>();

    @OneToMany(fetch = FetchType.LAZY, cascade=CascadeType.ALL, mappedBy = "park")
    private List<Piano> piano;

    public Park(){
    }

    public Park(int idPark, String nomePark, String descrizione, String indirizzo, String citta, Set<User> users, List<Piano> piano){
        this.idPark = idPark;
        this.nomePark = nomePark;
        this.descrizione = descrizione;
        this.indirizzo = indirizzo;
        this.citta = citta;
        this.users = users;
        this.piano = piano;
    }

    public List<Piano> getPiano() {
        return piano;
    }

    public void setPiano(List<Piano> piano) {
        this.piano = piano;
    }

    public  int getIdPark() {
        return idPark;
    }

    public  void setIdPark(int idPark) {
        this.idPark = idPark;
    }


    public  String getNomePark() {
        return nomePark;
    }

    public  void setNomePark(String nomePark) {
        this.nomePark = nomePark;
    }


    public  String getDescrizione() {
        return descrizione;
    }

    public  void setDescrizione(String descrizione) {
        this.descrizione = descrizione;
    }

    public String getIndirizzo() {
        return indirizzo;
    }

    public void setIndirizzo(String indirizzo) {
        this.indirizzo = indirizzo;
    }

    public String getCitta() {
        return citta;
    }

    public void setCitta(String citta) {
        this.citta = citta;
    }

    public Set<User> getUsers() {
        return users;
    }

    public void setUsers(Set<User> users) {
        this.users = users;
    }

    @Override
    public int hashCode() {
        final int prime = 31;
        int result = 1;
        result = prime * result + idPark;
        return result;
    }

    @Override
    public boolean equals(Object obj) {
        if (this == obj)
            return true;
        if (obj == null)
            return false;
        if (getClass() != obj.getClass())
            return false;
        Park other = (Park) obj;
        if (idPark != other.idPark)
            return false;
        return true;
    }
}

和在Piano.java中

@ManyToOne
    @JoinColumn(name = "id_park")
    public Park getPark() {
        return park;
    }

    public void setPark(Park park) {
        this.park = park;
    }

当我在夫妇中使用这3个 class 时(Park的用户或Park的钢琴)，一切都进行得很好，但是现在我必须为给定的用户获得所有的Piano。

在我的PianoDaoImpl中，我有这种方法

public List<Piano> findPianoByUser() {
        Authentication auth = SecurityContextHolder.getContext().getAuthentication();
        String name = auth.getName();
        User user = userService.findBySso(name);
        int userId = user.getId();
        Query q = getSession().createQuery("from Park p join p.users u where u.id = :userId").setParameter("userId", userId);
        List<Park> parks = q.list();

        Query query = getSession().createQuery("from Piano p where p.park in :parks").setParameterList("parks", parks);
        List piani = query.list();

        return piani;
    }

q查询为我正确提供了公园列表。然后，我将此列表传递给第二个查询，但出现了此异常

org.springframework.web.util.NestedServletException: Request processing failed; nested exception is org.hibernate.PropertyAccessException: could not get a field value by reflection getter of it.besmart.models.Park.idPark

由...引起

java.lang.IllegalArgumentException: Can not set int field it.besmart.models.Park.idPark to [Ljava.lang.Object;

我做错了什么？

最佳答案

问题的原因是引用类型idPark的Integer不能被分配原始类型int的值。
尝试替换它。

    @Column(name = "id_park", unique = true, nullable = false)
     private  Integer idPark;

带有:

   @Column(name = "id_park", unique = true, nullable = false)
    private  int idPark;

尝试检查userId的值是否为Integer而不是int，因为我认为与此不同的是:

  "from Park p join p.users u where u.id = :userId").setParameter("userId", userId);