休眠复杂的数据检索？

Question

我的数据库中有九个相关表。我必须在过滤用户请求后检索记录。

我的实体如下，

电影

@Entity
@Table(name = "movie")
public class Movie implements Serializable {

    private static final long serialVersionUID = 1L;

    @Id
    @GeneratedValue(strategy = GenerationType.IDENTITY)
    @Column(name = "movie_id")
    private int movieId;

    @Column(name = "category_id")
    private Integer categoryId;

    @Column(name = "movie_title")
    private String movieTitle;

    @Column(name = "movie_description", columnDefinition = "TEXT")
    private String movieDescription;

    @Column(name = "movie_summary", columnDefinition = "TEXT")
    private String movieSummary;

    private Integer status;

    @Column(name = "language_id")
    private Integer languageId;

    @Column(name = "banner_image_url")
    private String bannerImageUrl;

    @Column(name = "imdb_rating")
    private Integer imdbRating;

    @Column(name = "rotten_tomatoes_rating")
    private Integer rottenTomatoesRating;

    @Column(name = "user_avg_rating")
    private Float userAvgRating;

    @Column(name = "main_genre_id")
    private Integer mainGenreId;

    @Column(name = "secondary_genre_id")
    private Integer secondaryGenreId;

    @Column(name = "created_by_user_id")
    private Integer createdByUserId;
}

类别

@Entity
@Table(name = "category")
public class FetchSubCategory implements Serializable {

    private static final long serialVersionUID = 1L;

    @Id
    @GeneratedValue(strategy = GenerationType.IDENTITY)
    @Column(name = "category_id")
    private Integer categoryId;

    @Column(name = "category_name")
    private String categoryName;
}

电影演员映射

@Entity
@Table(name = "movie_actor_mapping")
public class MovieActorMapping implements Serializable {

    private static final long serialVersionUID = 1L;

    @Id
    @GeneratedValue(strategy = GenerationType.IDENTITY)
    @Column(name = "mapping_id")
    private int mappingId;

    @Column(name = "movie_id")
    private Integer movieId;

    @Column(name = "actor_id")
    private Integer actorId;
}

电影演员

@Entity
@Table(name = "movie_actors")
public class MovieActors implements Serializable {

    private static final long serialVersionUID = 1L;

    @Id
    @GeneratedValue(strategy = GenerationType.IDENTITY)
    @Column(name = "actor_id")
    private int actorId;

    @Column(name = "actor_name")
    private String actorName;
}

电影流派

@Entity
@Table(name = "movie_genre")
public class MovieGenre implements Serializable {

    private static final long serialVersionUID = 1L;

    @Id
    @GeneratedValue(strategy = GenerationType.IDENTITY)
    @Column(name = "genre_id")
    private int genreId;

    @Column(name = "genre_name")
    private String genreName;

    @Column(name = "created_by_user_id")
    private Integer createdByUserId;
}

电影语言

@Entity
@Table(name = "movie_language")
public class MovieLanguage implements Serializable {

    private static final long serialVersionUID = 1L;

    @Id
    @GeneratedValue(strategy = GenerationType.IDENTITY)
    @Column(name = "language_id")
    private int languageId;

    @Column(name = "language_name")
    private String languageName;

    @Column(name = "created_by_user_id")
    private Integer createdByUserId;

    @Column(name = "last_updated_user_id")
    private Integer lastUpdatedUserId;
}

用户将如下要求。all是可选字段，

{
"subCategory":"New Release",
"language":"Malayalam",
"actor":"allu arjun",
"filmGenre":"'Family'"
}

根据请求，我将使用子查询通过检查相应表中的条件来返回movie列表。

方法

public List<Movie> getFilterMovieList(FilterMovieRequest filterMovieRequest) throws SQLException, ClassNotFoundException, IOException {

        List<Movie> movies = null;
        try {
            String subCategory = filterMovieRequest.getSubCategory();
            String language = filterMovieRequest.getLanguage();
            String actor = filterMovieRequest.getActor();
            String filmGenre = filterMovieRequest.getFilmGenre();

            String contained = "where";
            String sql = "from Movie as M ";

            if (actor.length() > 1) {
                sql += contained + " movieId in(select movieId from MovieActorMapping where actorId in(select actorId from MovieActors where actorName='" + actor + "')) ";

                contained = "and";
            }

            if (subCategory.length() > 1) {

                sql += contained + " M.categoryId=(select categoryId from FetchSubCategory where categoryName='" + subCategory + "') ";
                contained = "and";
            }

            if (language.length() > 1) {

                sql += contained + " M.languageId=(select languageId from MovieLanguage where languageName='" + language + "') ";
                contained = "and";
            }

            if (filmGenre.length() > 1) {

                sql += contained + " (M.mainGenreId in(select genreId from MovieGenre where genreName in(" + filmGenre + ")) or M.secondaryGenreId in(select genreId from MovieGenre where genreName in(" + filmGenre + ")))";
                contained = "and";
            }

            if (contained.equals("and")) {
                Session session = sessionFactory.getCurrentSession();
                Query query = session.createQuery(sql);
                movies = query.list();
            }
        } catch (Exception e) {
            e.printStackTrace();
        }

        return movies;

    }

而且效果很好。问题是我现在必须结合剧院电影和放映时间的结果。

我的剧院相关表格如下，

theatre_movie_mapping

Theatre_list

show_timings

您可以在movie_id的theater_movie_mapping列中theater_movie_mapping与我的基本表movie有关的列。 使用我们可以获取theater_id和show_id为获取剧院和显示timing..note，我有一个电影列表上方conditions.How我可以从show_timings结合剧院从theater_list和放映时间检查后早期牵强？ 作为一名Android开发人员，对我来说似乎很复杂。 任何帮助将不胜感激。正在使用Spring restful webservice 。

现在我得到以下格式的结果，

[
    {
        "movieId": 8,
        "categoryId": 14,
        "movieTitle": "Kanyaka Talkies",
        "movieDescription": "CRITICS 3 out of 5 (Good) 3 out of 5 (Good) The composite emotional weather that the film sports makes it maddening and nurturing at once, rendering it an almost enigmatic feel. And it is this ethereal complexity that 'Kanyaka Talkies' inherently has, that makes the film singular. ",
        "movieSummary": "The concurrence of the three key characters in 'Kanyaka Talkies' isn't of the traditionalist kind; rather, by throwing the three of them together, the film does achieve the ostensibly improbable feat of placing the unlikeliest of players collectively on board, with their fates irrevocably intertwined with each other. ",
        "status": 1,
        "languageId": 1,
        "bannerImageUrl": "0",
        "imdbRating": 1,
        "rottenTomatoesRating": 3,
        "userAvgRating": 2,
        "mainGenreId": 1,
        "secondaryGenreId": 2,
        "createdByUserId": 16
    },
    {
        "movieId": 9,
        "categoryId": 14,
        "movieTitle": "Wonderful Journey",
        "movieDescription": "Wonderful Journey' is one of the most misdirecting titles as yet for a film this year. Anything but wonderful, this is an absolute cinematic misadventure that will have you pulling out your hair strands in no time. ",
        "movieSummary": "Some things in life simply cannot be averted, they say. I do agree, what with the late night show of 'Wonderful Journey' getting cancelled yesterday night and me courageously venturing out for it yet again today noon, only to embark on one of the most horrendous journeys I have ever gone for in my entire life. ",
        "status": 1,
        "languageId": 1,
        "bannerImageUrl": "0",
        "imdbRating": 1,
        "rottenTomatoesRating": 3,
        "userAvgRating": 2,
        "mainGenreId": 1,
        "secondaryGenreId": 1,
        "createdByUserId": 16
    },
    {
        "movieId": 10,
        "categoryId": 14,
        "movieTitle": "Oru New Generation Pani",
        "movieDescription": "Very occasionally does a movie come along that almost makes you vow to stay off the screens for a few weeks, and this year, the one has finally arrived. I'd gladly go ahead with a no-star rating for this one, had it not been for a technical glitch that prevents me from doing so! ",
        "movieSummary": "'Oru New Generation Pani' is an atrocity that shocks you with its attempt to spin out a story line that will have you banging you head against the rails. Inauthentic to the core, the film tells a story that will have an insomniac snoring away in no time. ",
        "status": 1,
        "languageId": 1,
        "bannerImageUrl": "0",
        "imdbRating": 1,
        "rottenTomatoesRating": 3,
        "userAvgRating": 2,
        "mainGenreId": 1,
        "secondaryGenreId": 2,
        "createdByUserId": 16
    }
]

我必须在每个json对象（即每部电影）中添加剧院并放映时间。

Answer 1

假设Theater_movie_mapping映射到其中包含Movie movie Class TheaterMovie

ArrayList<TheaterMovie> list = new ArrayList<TheaterMovie>();
for(int i = 0 ; i < movies.size() ; i++){
list.addAll((ArrayList<TheaterMovie>)createQuery("From TheaterMovie tm where tm.movies = :mo").setParameter("mo",movies.get(i)).getList());
}

现在假设将show_timings映射到其中包含Theater theater Class ShowTiming

ArrayList<ShowTiming> showTimeList = new ArrayList<ShowTiming>();
for(int i = 0 ; i < list.size() ; i++){
showTimeList = (ArrayList<ShowTiming>)createQuery("From ShowTiming st where st.theater = :th").setParameter("th",list.get(i).getTheater()).getList();
}

我希望这对你有用