Question

我想在我的存储库界面中构造一个find查询，以选择由多对多关系映射的所有元素。

User.java：

@Entity
@Table(name = "user")
public class User {
    @Id
    @Column(name = "id", columnDefinition = "BINARY(16)")
    private UUID id;

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

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

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

    @ManyToMany(mappedBy = "usersTracking")
    private Set<Show> shows;
...

Show.java：

@Entity
@Table(name = "shows")
public class Show {

    @Id
    @Column(name = "id", columnDefinition = "BINARY(16)")
    private UUID id;

    @Column(name = "imdb_id", length = 9, nullable = false)
    private String imdbID;

    @Column(name = "title", length = 100, nullable = false)
    private String title;

    @Column(name = "poster_url", length = 255, nullable = true)
    private String posterURL;

    @Column(name = "rating", length = 3, nullable = true)
    private double rating;

    @Column(name = "genre", length = 100, nullable = true)
    private String genre;

    @ManyToMany
    @JoinTable(name = "users_shows",
            joinColumns = @JoinColumn(name = "show_id", referencedColumnName = "id"),
            inverseJoinColumns = @JoinColumn(name = "user_id", referencedColumnName = "id"))
    private Set<User> usersTracking;
...

要使用查询构建器实现这一点（我认为它已被调用），我已经编写了一个存储库：

public interface ShowRepository extends CrudRepository<Show, UUID> {
    List<Show> findByUser(User user);
}

但我希望能够从我的UserRepository而不是ShowRepository获取节目，所以我应该写这样的东西？

public interface UserRepository extends CrudRepository<User, UUID> {
    List<Show> findShows();
    //Or maybe?
    List<Show> findShowsByUser(User user);
}

我是否正确理解整个查询构建正确？

Spring Data JPA存储库查询构建器，关系多对多

0 个答案: