Question

我想在entites Movie和Genre之间创建一个ManyToMany关系。因为电影可以有很多类型，而且类型可以有很多电影。但我不知道怎么做这个联想。我必须创建实体“genre_movie”？如果我编译它，则生成更多表：movie_genre和genre_movie。我只想要一个。

@Entity
public class Movie implements Serializable {

    private static final long serialVersionUID = 1L;

    @Id
    @JsonView(Views.Private.class)
    @GeneratedValue(strategy = GenerationType.IDENTITY)
    private Long id;

    @JsonView(Views.Private.class)
    private String name;

    @JsonView(Views.Private.class)
    private Long year;
    @JsonView(Views.Private.class)
    private String sinopsis;
    @JsonView(Views.Private.class)
    private Double puntuation;
    @JsonView(Views.Private.class)
    private Long duration;
    @JsonView(Views.Private.class)
    private String idioma;
    @JsonView(Views.Private.class)
    private String trailer;

    //getters seters

    @ManyToMany(targetEntity=Genre.class)
    private Set teacherSet;
}

类型课

 import com.fasterxml.jackson.annotation.JsonView;

 import javax.persistence.*;
 import java.io.Serializable;
 import java.util.*;

 @Entity
 public class Genre implements Serializable  {

     private static final long serialVersionUID = 1L;

     @Id
     @JsonView(Views.Private.class)
     @GeneratedValue(strategy = GenerationType.IDENTITY)
     private Long id;


     @JsonView(Views.Private.class)
     private String genreName;


     public Genre(){}

     public Genre(String genreName){
     this.genreName = genreName;
     }


     @ManyToMany(targetEntity = Movie.class)
     private Set movieSet;

     public String getName() {
     return genreName;
     }

     public void setName(String name) {
     this.genreName = name;
     }
}

查询：

 insert into Movie (name,year,sinopsis,puntuation,duration,idioma,trailer) values ('Titanic',1997,'Is about a movie ....',0,88,'english','https://www.youtube.com/watch?v=zCy5WQ9S4c0&t=112s');
 insert into Movie (name,year,sinopsis,puntuation,duration,idioma,trailer) values ('Nightcrawler',2014,'Is about a movie dpsvr....',0,175,'english','https://www.youtube.com/watch?v=u1uP_8VJkDQ');
 insert into Movie (name,year,sinopsis,,,idioma,trailer) values ('Iron Man 2',2010,'Is about a movie dpsvr....',0,145,'english','https://www.youtube.com/watch?v=BoohRoVA9WQ');

 insert into Genre (genreName) values ('comedy');
 insert into Genre (genreName) values ('drama');
 insert into Genre (genreName) values ('thriller');
 insert into Genre (genreName) values ('terror');

Answer 1

@Entity
public class Movie implements Serializable {
    @ManyToMany
    private Set<Genre> teacherSet;
}

@Entity
public class Genre implements Serializable  {
     @ManyToMany(mappedBy="teacherSet")
     private Set<Movie> movieSet;

}

在此示例中，将仅生成Movie和Genre之间的一个关系，其中关系的所有者将是Movie。在SQL中，将生成一个关联表Movie_Genre。

多对多关系JPA

1 个答案: