JPA 加入打印到许多结果

Question

我有四个描述数据库的表：
讲师

@Entity
@Table(name = "LECTURER")

public class Lecturer
{
    private static final long            serialVersionUID = 1L;

    @Id
    @GeneratedValue(strategy = GenerationType.IDENTITY)  
    private int id;

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

    @Column(name = "NOBEL_PRICE_YEAR")
    private int nobel_price_year;
}```

注册（使用讲座和学生）

@Entity
@IdClass(EnrollmentId.class)
@Data
public class Enrollment
{
   @Id
   @JoinColumn(name = "lecture_id", referencedColumnName = "id")
   @ManyToOne
   private Lecture lecture;


   @Id
   @JoinColumn(name = "student_id", referencedColumnName = "id")
   @ManyToOne
   private Student student;

   @Id
   private int year;

   @Column(nullable = false, columnDefinition = "TINYINT(1)")
   private boolean exam_taken;
   @Column(nullable = false, columnDefinition = "TINYINT(1)")
   private boolean exam_passed;
}

自复合主键以来的 ID 类

public class EnrollmentId implements Serializable 
{

    @Id
    @JoinColumn(name = "lecture_id", referencedColumnName = "id")
    @ManyToOne
    private int lecture;

    @Id
    @JoinColumn(name = "student_id", referencedColumnName = "id")
    @OneToOne
    private int student;

    private int year;
}

学生

@Entity
@Table(name="STUDENT")
@Data
@NamedQuery(name = "Student.findAll", query = "SELECT s FROM Student s")
public class Student
{
    private static final long           serialVersionUID = 1L;

    @Id
    @Column(name="ID", nullable=false)
    @GeneratedValue(strategy = GenerationType.IDENTITY)    
    private int id;

    @Column(name="FIRSTNAME", nullable=false)
    private String firstname;

    @Column(name="LASTNAME", nullable=false)
    private String lastname;

    @Column(name="YEAR_OF_BIRTH")
    private int yearOfBirth;

    @Column(name="GENDER")
    @Enumerated(EnumType.STRING)
    private Gender gender;        
}

讲座（使用讲师）

@Entity
@Table(name="LECTURE")
public class Lecture
{
    private static final long serialVersionUID = 1L;

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

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

    @JoinColumn(name="LECTURER_ID", referencedColumnName = "id")
    @ManyToOne
    private Lecturer lecturer;

    @Column(name = "ROOM")
    @Enumerated(EnumType.STRING)
    private Room room;

    @Column(name = "DAY_OF_WEEK")
    private DayOfWeek dow;

    @JoinColumn(name = "PREREQUISITE_ID", referencedColumnName = "id")
    @ManyToOne
    private Lecture prerequisite;             

}

当我尝试构建查询时，我会得到很多结果。 （SQL Server 在同一个查询中打印 129，JPA 90k+）

查询：

Query q = em.createQuery("SELECT"
            + "  l.title, l.dow, l.room "
            + ", t.name, CASE WHEN (t.nobel_price_year != NULL) THEN 'YES' ELSE 'No' END"
            + ", e.year, e.exam_taken, e.exam_passed "
            + ", CONCAT(s.lastname,', ', s.firstname) AS student "
            + "FROM Lecture l "
            + "JOIN Lecturer t "
            + "JOIN Enrollment e "
            + "JOIN Student s "
            + "ORDER BY l.id", EmpireDBExample.class);

代码是手动生成的，因为 SQL Server 连接不适用于 Eclipse（据我所知/发现）。

映射中有错误吗？
我需要添加加入条件吗？如果是这样，我该如何实现？ “ON a.id = b.id”不起作用。

Answer 1

好吧，由于ON ... 不起作用（Hibernate 5 会支持它），您需要沿着路径加入或使用 where 条件来过滤不适合的行。目前，您正在交叉加入所有讲座、讲师、注册和学生，即使他们不相关。

另一个问题是没有从Lecture 到Enrollment 的代码路径，因此您可能希望从Enrollment 开始。因此查询可能如下所示（我将跳过公共部分）：

... FROM Enrollment e  
      JOIN e.lecture l  //join the lecture for the enrollment
      JOIN l.lecturer t //join the lecturer for the lecture
      JOIN e.student s  //join the enrolled student

最后一个建议：“我有四个描述数据库的表”——在这里改变你的想法。您有四个描述您的模型的实体，它们被映射到数据库（尽管映射可能会改变）。因此，在处理 JPA 查询时，您需要考虑实体而不是表，例如因为没有从Lecture 到Enrollment 的路径，并且除非您的JPA 提供程序支持通用ON，否则您需要找到一个入口点和您想要加入的所有内容的路径（例如，有一条从Enrollment 到@987654330 的路径@ 所以你从Enrollment开始）。