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

Question

我一直在解决这个问题。 我有 3 个类(class)

User 和 Park 有 ManyToMany 关系，Piano 和 Park 有 ManyToOne 关系。

在模型中，有趣的部分是 用户.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;
    }
}

在 Piano.java 中

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

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

当我成对使用这 3 个类(class)时(User with Park 或 Park with Piano)一切顺利，但现在我必须为给定用户获取所有钢琴。

在我的 PianoDaoImpl 中我有这个方法

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 查询正确地为我提供了用户的公园列表。然后我将这个列表传递给第二个查询，我得到了这个异常

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

造成

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

我做错了什么？

Answer 1

问题的原因，即引用类型Integer的idPark不能被赋值给基本类型int的值。
尝试替换它。

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

与:

   @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，因为我认为你有异常(exception):

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