首页 \ 问答 \ 如何扩展@Entity(How to extend an @Entity)

如何扩展@Entity(How to extend an @Entity)

这是我的超类

@Entity
@Table(name = "utente")
@Component
@Inheritance
public class Utente implements Serializable{

    private static final long serialVersionUID = -7124540331184173742L;

    @Id
    @GeneratedValue
    @Column(name = "id")
    private int id;

    @Column(name = "nome")
    @Size(min = 1, max = 20)
    @Pattern(regexp="^[A-Za-z 'òàùèéì]*$")      
    @NotBlank
    private String nome;

    @Column(name = "cognome")
    @Size(min = 1, max = 20)
    @Pattern(regexp="^[A-Za-z 'òàùèéì]*$")
    @NotBlank
    private String cognome;

    @Column(name = "email")
    @Email
    @Size(min = 1, max = 40)
    private String email;

    @OneToOne(mappedBy = "utente", cascade = CascadeType.ALL)
    @Valid
    private Autenticazione autenticazione;  

    @OneToMany(mappedBy = "utente", fetch = FetchType.EAGER,  orphanRemoval=true, cascade=CascadeType.ALL)  
    private List<Autorizzazioni> autorizzazioniLista;
}

这是扩展上述类的类:

@Entity
public class UtenteSubClass extends Utente{ 

@Transient
private String newField;
}

当我尝试检索对象UtenteSubClass时,我收到此错误"org.hibernate.util.JDBCExceptionReporter - Unknown column 'this_.DTYPE' in 'where clause'"

可能我的“绘图系统”是错误的。 我的错误在哪里?

先谢谢你。


This is my superclass

@Entity
@Table(name = "utente")
@Component
@Inheritance
public class Utente implements Serializable{

    private static final long serialVersionUID = -7124540331184173742L;

    @Id
    @GeneratedValue
    @Column(name = "id")
    private int id;

    @Column(name = "nome")
    @Size(min = 1, max = 20)
    @Pattern(regexp="^[A-Za-z 'òàùèéì]*$")      
    @NotBlank
    private String nome;

    @Column(name = "cognome")
    @Size(min = 1, max = 20)
    @Pattern(regexp="^[A-Za-z 'òàùèéì]*$")
    @NotBlank
    private String cognome;

    @Column(name = "email")
    @Email
    @Size(min = 1, max = 40)
    private String email;

    @OneToOne(mappedBy = "utente", cascade = CascadeType.ALL)
    @Valid
    private Autenticazione autenticazione;  

    @OneToMany(mappedBy = "utente", fetch = FetchType.EAGER,  orphanRemoval=true, cascade=CascadeType.ALL)  
    private List<Autorizzazioni> autorizzazioniLista;
}

This is the class that extend the one above:

@Entity
public class UtenteSubClass extends Utente{ 

@Transient
private String newField;
}

When I try to retrieve an object UtenteSubClass I get this error "org.hibernate.util.JDBCExceptionReporter - Unknown column 'this_.DTYPE' in 'where clause'".

Probably my "mapping system" is wrong. Where is my error?

Thank you in advance.


原文:https://stackoverflow.com/questions/36127632
更新时间:2022-05-18 20:05

最满意答案

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The ASP.NET Sitemap feature is built for that and works well in a lot of cases. If you get in a spot where you want your Menu to look different from your Sitemap, here are some workarounds.

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