如何使用@OneToMany和@ManyToOne嵌套？

Question

我有如下三個實體：

@Data 
@NoArgsConstructor 
@AllArgsConstructor 
@EqualsAndHashCode(exclude = {"id", "bricksets", "themes"}) 
@Entity 
public class Brand { 
    @Id 
    @GeneratedValue(strategy = GenerationType.AUTO) 
    private Long id;    // Unique database primary key 

    @OneToMany(mappedBy = "brand", cascade = CascadeType.ALL) 
    private List<Brickset> bricksets; 

    @OneToMany(mappedBy = "brand", cascade = CascadeType.ALL) 
    private List<Theme> themes; 

    @NotNull 
    private String name; 
} 

@Data 
@NoArgsConstructor 
@AllArgsConstructor 
@EqualsAndHashCode(exclude = {"id"}) 
@ToString(exclude={"brand", "theme"}) 
@Entity 
public class Brickset { 
    @Id 
    @GeneratedValue(strategy = GenerationType.AUTO) 
    private Long id;   // Unique database primary key 

    @NotNull 
    private String number;  // set number 

    @ManyToOne 
    @JoinColumn(name = "brand_id") 
    private Brand brand; 

    @ManyToOne 
    @JoinColumn(name = "theme_id") 
    private Theme theme; 
} 

@Data 
@NoArgsConstructor 
@AllArgsConstructor 
@EqualsAndHashCode(exclude = {"id"}) 
@ToString(exclude={"brand", "brickset"}) 
@Entity 
public class Theme { 
    @Id 
    @GeneratedValue(strategy = GenerationType.AUTO) 
    private Long id;    // Unique database primary key 

    @NotNull 
    private String name; 

    @ManyToOne 
    @JoinColumn(name = "brand_id") 
    private Brand brand; 

    @OneToMany(mappedBy = "Theme", cascade = CascadeType.ALL) 
    private List<Brickset> brickset; 
} 

但我遇到編譯如下錯誤，我嘗試自己解決這個問題，但我在JPA /休眠很新，你能幫助我，我怎麼能下一步呢？我認爲這個問題是我在Entity Theme中使用了ManyToOne和OneToMany，可能會被JPA/Hibernate弄糊塗，我可以理解嗎？

Caused by: org.hibernate.AnnotationException: Illegal attempt to map a non collection as a @OneToMany, @ManyToMany or @CollectionOfElements: com.zangland.entity.Theme.brickset 

Answer 1

您需要使用Collection作爲@OneToMany註釋。

Hibernate要求持久化的集合值域被聲明爲接口類型。

實際的接口可能是java.util.Set，java.util.Collection，java.util.List，java.util.Map，java.util.SortedSet，java.util.SortedMap或任何你喜歡（ 「任何你喜歡的類型」 代表你需要編寫的org.hibernate.usertype.UserCollectionType的實現。）

@OneToMany(mappedBy = "theme", cascade = CascadeType.ALL) 
    private Set<Brickset> brickset;