Spring数据:JPA存储库返回映射而不是列表？

Question

我有这样的查询，我想从中提取电影标题及其评级

   @Query(value ="SELECT m.title, avg(r.rating) " +
        "FROM movies m " +
        "INNER JOIN " +
        "ratings r " +
        "ON m.movieid = r.movieid " +
        "WHERE m.genres LIKE %:genre% " +
        "GROUP BY m.title " +
        "ORDER BY avg(r.rating) DESC " +
        "LIMIT 10",
        nativeQuery = true)
 List<Movies> topTenMoviesByGenreLikeIgnoreCase(@Param("genre") String genre);

但此列表显然不能存储具有2个值的结果集

在了解它之后，我发现例如这个链接Spring Data： JPA repository findAll（）返回*Map而不是List？但我似乎不清楚它们的实现。

我不明白他们从哪里得到第一个链接中的长值

default Map<Long, TransactionModel> findByClientIdMap(Long id) {
    return findByClientId(id).stream().collect(Collectors.toMap(TransactionModel::getId, v -> v));
}

我如何创建我的地图？

default Map<Movies, Float> topTenMoviesByGenreLikeIgnoreCaseMap(@Param("genre") String genre) {
    return topTenMoviesByGenreLikeIgnoreCase(genre).stream().collect(Collectors.toMap());
}

电影类:

@Entity
public class Movies {
@Id
@GeneratedValue(strategy = GenerationType.AUTO)
private Integer movieid;

private String title;
private String genres;

@OneToMany(mappedBy = "movieid")
private Set<Ratings> rates = new HashSet<>();

public Movies() {
}

public Movies(String title, String genres) {
    this.title = title;
    this.genres = genres;
}

/*Getters and Setters for variables*/

public Set<Ratings> getRates() {
    return rates;
}

public void setRates(Set<Ratings> rates) {
    this.rates = rates;
}
}

评级等级：

@Entity
public class Ratings {
@Id
@GeneratedValue(strategy = GenerationType.AUTO)
private Integer rateid;

@ManyToOne
@JoinColumn(name = "movieid")
private Movies movieid;

@ManyToOne
@JoinColumn(name = "userid")
private Usrs userid;

private float rating;
private Integer timestamp;

public Ratings() {
}

public Ratings(Movies movieid, Usrs userid, float rating, Integer timestamp)
{
    this.movieid = movieid;
    this.userid = userid;
    this.rating = rating;
    this.timestamp = timestamp;
}
/*Getters and Setters for variables*/
}

Answer 1

试试这个：

@Query(value ="SELECT m.title as title, r.avg(rating) as rates " +
    "FROM movies m " +
    "INNER JOIN " +
    "ratings r " +
    "ON m.movieid = r.movieid " +
    "WHERE m.genre = LIKE %:genre% " +
    "GROUP BY r.avg(rating) " +
    "ORDER BY r.avg(rating) DESC " +
    "LIMIT 10",
    nativeQuery = true)

List<Movies> topTenMoviesByGenreLikeIgnoreCase(@Param("genre") String genre);    



default Map<Movies, Set<Ratings>> topTenMoviesByGenreLikeIgnoreCaseMap(@Param("genre") String genre) {
    return topTenMoviesByGenreLikeIgnoreCase(genre).stream().collect(
                                                            Collectors.toMap(
                                                            tuple -> (tuple.getTitle(),
                                                            tuple -> (tuple.getRates())
                                                         ));
}

Answer 2

您可以通过这种方式使用＜code＞对象[]

@Query(value ="SELECT m.title, avg(r.rating) " +
        "FROM movies m " +
        "INNER JOIN " +
        "ratings r " +
        "ON m.movieid = r.movieid " +
        "WHERE m.genres LIKE %:genre% " +
        "GROUP BY m.title " +
        "ORDER BY avg(r.rating) DESC " +
        "LIMIT 10",
        nativeQuery = true)
 List<Object[]> topTenMoviesByGenreLikeIgnoreCase(@Param("genre") String genre);

所以，然后

default Map<String, Float> topTenMoviesByGenreLikeIgnoreCaseMap(@Param("genre") String genre) {
    return topTenMoviesByGenreLikeIgnoreCase(genre)
            .stream()
            .collect(Collectors.toMap((arr->arr[0].toString()), (arr->Float.valueOf(arr[1].toString()))));
}

如您所见，topTenMoviesByGenreLikeIgnoreCaseMap返回地图，其中电影标题是关键。要使电影实体成为键，您应该使用JPQL查询而不是本机查询。