在Spring Data REST中具有多对多关系的POST请求

问题描述 投票:5回答:1

我正在尝试将电影添加到MySQL数据库,这是我的数据库模式:

Movie(id, name)
Genre(id, name)
Movie_genre(id_movie, id_genre)

这是我的模型类:

Movie.ts

public class Movie {

    private Short id;
    private String name;
    private List<MovieGenre> movieGenres;

    @Id
    @GeneratedValue(strategy = GenerationType.IDENTITY)
    @Column(name = "id", nullable = false)
    public Short getId() {
        return id;
    }

    public void setId(Short id) {
        this.id = id;
    }

    @Column(name = "name", nullable = false, length = 100)
    public String getName() {
        return name;
    }

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

    @OneToMany(mappedBy = "movie")
    public List<MovieGenre> getMovieGenres() {
        return movieGenres;
    }

    public void setMovieGenres(List<MovieGenre> movieGenres) {
        this.movieGenres = movieGenres;
    }
}

Genre.ts

public class Genre {
    private Short id;
    private String name;
    private List<MovieGenre> movieGenres;

    @Id
    @GeneratedValue(strategy = GenerationType.IDENTITY)
    @Column(name = "id", nullable = false)
    public Short getId() {
        return id;
    }

    public void setId(Short id) {
        this.id = id;
    }

    @Column(name = "name", nullable = false, length = 15)
    public String getName() {
        return name;
    }

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

    @OneToMany(mappedBy = "genre")
    @JsonIgnore
    public List<MovieGenre> getMovieGenres() {
        return movieGenres;
    }

    public void setMovieGenres(List<MovieGenre> movieGenres) {
        this.movieGenres = movieGenres;
    }
}

MovieGenre.ts(该模型代表生成的表)

public class MovieGenre {
    private MovieGenrePK id;
    private Movie movie;
    private Genre genre;

    @EmbeddedId
    @JsonIgnore
    public MovieGenrePK getId() {
        return id;
    }

    public void setId(MovieGenrePK id) {
        this.id = id;
    }

    @MapsId("movieId")
    @ManyToOne
    @JoinColumn(name = "movie_id", referencedColumnName = "id", nullable = false)
    @JsonIgnore
    public Movie getMovie() {
        return movie;
    }

    public void setMovie(Movie movie) {
        this.movie = movie;
    }

    @MapsId("genreId")
    @ManyToOne
    @JoinColumn(name = "genre_id", referencedColumnName = "id", nullable = false)
    public Genre getGenre() {
        return genre;
    }

    public void setGenre(Genre genre) {
        this.genre = genre;
    }
}

因为我们在最后一个模型中有复合键,所以我们需要一个类:

public class MovieGenrePK implements Serializable {
    private Short movieId;
    private Short genreId;

    @Column(name = "movie_id", nullable = false)
    public Short getMovieId() {
        return movieId;
    }

    public void setMovieId(Short movieId) {
        this.movieId = movieId;
    }

    @Column(name = "genre_id", nullable = false)
    public Short getGenreId() {
        return genreId;
    }

    public void setGenreId(Short genreId) {
        this.genreId = genreId;
    }
}

因此,我正在尝试通过发布帖子请求添加带有流派的电影,首先我发布了添加电影的帖子请求和另一个用于添加类型的电影,这很好用,现在我需要将一个类型与电影相关联。 我尝试了以下方法: 我向以下端点发出了一个POST请求:带有application / json标头的http://localhost:8080/api/movieGenres以及以下主体:

{
    "movie": "http://localhost:8080/api/movies/6",
    "genre": "http://localhost:8080/api/genres/1"
}

但我得到了错误:

{
    "timestamp": "2018-08-22T21:10:30.830+0000",
    "status": 500,
    "error": "Internal Server Error",
    "message": "NullPointerException occurred while calling setter of com.movies.mmdbapi.model.MovieGenrePK.genreId; nested exception is org.hibernate.PropertyAccessException: NullPointerException occurred while calling setter of com.movies.mmdbapi.model.MovieGenrePK.genreId",
    "path": "/api/movieGenres"
}
spring spring-boot spring-data-jpa spring-data-rest
1个回答
2
投票

您必须实例化MovieGenrePK,更改:

public class MovieGenre {
    private MovieGenrePK id;
}

public class MovieGenre {
    private MovieGenrePK id = new MovieGenrePK();
}
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