休眠：无法检索多个字段值

Question

我是Hibernate中的新成员。我成功检索单列和完整的表，但是当我尝试在表中的特定列它给我错误。 以下是错误

[Ljava.lang.Object;@2c7d121c, [Ljava.lang.Object;@65aa6596, [Ljava.lang.Object;@67389cb8, [Ljava.lang.Object;@419a20a6 

这里是实体：MovieRatings.java

package com.hibernate.entity; 
import javax.persistence.Column; 
import javax.persistence.Entity; 
import javax.persistence.GeneratedValue; 
import javax.persistence.GenerationType; 
import javax.persistence.Id; 
import javax.persistence.Table; 

@Entity 
@Table(name = "movie_ratings") 
public class MovieRatings { 
    @Id 
    @GeneratedValue(strategy=GenerationType.IDENTITY) 
    @Column(name = "id") 
    private int id; 
    @Column(name = "movie_id") 
    private int user_id; 
    @Column(name = "user_id") 
    private int movie_id; 
    @Column(name = "rating") 
    private int rating; 

    public MovieRatings() 
    { 

    } 

    public int getId() { 
     return id; 
    } 

    public void setId(int id) { 
     this.id = id; 
    } 
    public int getUser_id() { 
     return user_id; 
    } 
    public void setUser_id(int user_id) { 
     this.user_id = user_id; 
    } 

    public int getMovie_id() { 
     return movie_id; 
    } 

    public void setMovie_id(int movie_id) { 
     this.movie_id = movie_id; 
    } 

    public int getRating() { 
     return rating; 
    } 

    public void setRating(int rating) { 
     this.rating = rating; 
    } 

    @Override 
    public String toString() { 
     return "MovieRatings [id=" + id + ", user_id=" + user_id + ", movie_id=" + movie_id + ", rating=" + rating 
       + "]"; 
    } 

} 

这里的主要功能是

package com.hibernate.getdata; 
import java.util.List; 
import org.hibernate.Session; 
import org.hibernate.SessionFactory; 
import org.hibernate.cfg.Configuration; 
import com.hibernate.entity.MovieRatings;; 

public class GetMovieInfo { 

    public static void main(String[] args) { 
     // TODO Auto-generated method stub 
     SessionFactory factory = new Configuration().configure().addAnnotatedClass(MovieRatings.class).buildSessionFactory(); 
     Session session = factory.getCurrentSession(); 
     try { 
      session.beginTransaction(); 
      String hql="select E.user_id, E.rating from MovieRatings as E"; 
      System.out.println(hql); 
      @SuppressWarnings({ "deprecation", "unchecked" })   
      List<MovieRatings> theRatings = session.createQuery(hql).list(); 
      System.out.println(theRatings); 
      /*for(MovieRatings tempRatings:theRatings) 
      { 
       System.out.println(tempRatings.getUser_id()); 
       System.out.println(tempRatings.getRating()); 
      }*/ 
      session.getTransaction().commit(); 
     } catch (Exception e) { 
      // TODO: handle exception 
     } 

    } 

}

Answer 1

我成功地检索单列和完整的表，但wh en 有史以来我想在表中的特定列它给我错误。

使用Hibernate（真还为JPA），当您在查询中特定的列（S），而不是实体本身选择的Query的list()方法返回与此结构的对象：List<Object[]>。

列表中的每个元素对应一行，每个元素是一个数组，其中包含select使用的顺序中的所选值。
所以，你必须遍历列表中检索值：

List<Object[]> result = (List<Object[]>) query.list(); 
for(Object[] current : result){ 
    Integer userId = (String)current[0]; 
    Float rating = (Float) current[1]; 
    ..... 
} 

Answer 2

尝试使用这样的：

try { 
      session.beginTransaction(); 
      String hql="select E.user_id, E.rating from MovieRatings as E"; 
      System.out.println(hql); 
      @SuppressWarnings({ "deprecation", "unchecked" })  
      Query query = session.createSQLQuery(hql); 
      query.setResultTransformer(Transformers.aliasToBean(GetMovieInfo.class)); 
      List<MovieRatings> theRatings = query.list(); 
       System.out.println(theRatings); 
       session.getTransaction().commit(); 
      } catch (Exception e) { 
       // TODO: handle exception 
      }