休眠多对多like子句

Posted 2023-04-12

【中文标题】休眠多对多like子句

【英文标题】：Hibernate many to many like clause

【发布时间】：2017-03-19 14:59:41

【问题描述】：

电影

@Entity
@Table(name = "MOVIE")
public class MovieEntity 
    @Id
    @Column(name = "id")
    @GeneratedValue(generator="increment")
    @GenericGenerator(name="increment", strategy = "increment")
    private int id;

    @Column(name = "RELEASE_DATE")
    private Date releaseDate;

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

    @Column(name = "rating")
    private double rating;

    @Column(name = "length")
    private int length;

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

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

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

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

    @ManyToMany(
            cascade = CascadeType.PERSIST, CascadeType.MERGE
    )
     @LazyCollection(LazyCollectionOption.FALSE)
     @JoinTable(name = "CATEGORY",
     joinColumns = 
         @JoinColumn(
             name = "MOVIE_ID", 
             referencedColumnName = "id"
         )
     ,
     inverseJoinColumns = 
         @JoinColumn(
             name = "GENRE_ID", 
             referencedColumnName = "id"
         )
     
 )
    private List<GenreEntity> genre = new ArrayList<>();

    @ManyToMany(
            cascade = CascadeType.PERSIST, CascadeType.MERGE
    )
    @LazyCollection(LazyCollectionOption.FALSE)
     @JoinTable(name = "MOVIE_REPLY",
     joinColumns = 
         @JoinColumn(
             name = "MOVIE_ID", 
             referencedColumnName = "id"
         )
     ,
     inverseJoinColumns = 
         @JoinColumn(
             name = "REPLY_ID", 
             referencedColumnName = "id"
         )
     
 )
    private List<ReplyEntity> reply = new ArrayList<>();

    public List<ReplyEntity> getReply() 
        return reply;
    

    public List<GenreEntity> getGenre() 
        return genre;
    

    public MovieEntity()

    

    public int getId() 
        return id;
    

    public Date getReleaseDate() 
        return releaseDate;
    

    public void setReleaseDate(Date releaseDate) 
        this.releaseDate = releaseDate;
    

    public String getName() 
        return name;
    

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

    public double getRating() 
        return rating;
    

    public void setRating(double rating) 
        this.rating = rating;
    

    public int getLength() 
        return length;
    

    public void setLength(int length) 
        this.length = length;
    

    public String getCasting() 
        return casting;
    

    public void setCasting(String casting) 
        this.casting = casting;
    

    public String getDirector() 
        return director;
    

    public void setDirector(String director) 
        this.director = director;
    

    public String getDescription() 
        return description;
    

    public void setDescription(String description) 
        this.description = description;
    

    public String getWriter() 
        return writer;
    

    public void setWriter(String writer) 
        this.writer = writer;
    

类型

@Entity
@Table(name = "GENRE")
public class GenreEntity 

    @Id
    @Column(name = "id")
    @GeneratedValue(generator="increment2")
    @GenericGenerator(name="increment2", strategy = "increment") 
    private int id;

    @Column(unique = true,name="name")
    private String name;

    public int getId() 
        return id;
    

    @ManyToMany(mappedBy = "genre")
    private List<MovieEntity> movies = new ArrayList<>();

    public GenreEntity( String string)
        this.name=string;
    

    public GenreEntity()

    

    public String getName() 
        return name;
    

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

这是我用来按电影名称搜索的内容

movies = session.createQuery("FROM MovieEntity m where m.name like :name order by m.id ASC")
                            .setParameter("name", "%"+input+"%")
                    .list();

它有效。但是当我尝试按这样的流派名称搜索时：

movies = session.createQuery("FROM MovieEntity m join m.genre g where g.name like '%Ale%'").list();

它不起作用。我得到了页面上显示的正确数量的电影，但没有属性（名称、流派、导演都是空的）。 如果我尝试像这样循环它们：

for(MovieEntity movie:movies)
            System.out.print(movie.getName());
        

我明白了：

  [Ljava.lang.Object; cannot be cast to MovieEntity

【参考方案1】：

使用有效的 JPQL，并始终指定一个选择子句：

select distinct m from MovieEntity ...

否则，如您所见，查询返回List<Object[]>，其中每个数组包含电影和流派。