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;
}```
报名(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;
}
自复合主键以来具有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;
}
讲师(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;
}
当我尝试建立查询时,会得到很多结果。 (SQL Server在同一查询上打印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);
该代码是手动生成的,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”无效。
最佳答案
好吧,由于ON ...
无法正常工作(Hibernate 5会支持它),因此您需要沿着路径加入或使用where条件来筛选不合适的行。目前,您正在交叉参加所有的讲座,讲师,招生和学生,即使它们没有关联。
另一个问题是,从Lecture
到Enrollment
没有代码路径,因此您可能想从Enrollment
开始。因此查询看起来可能像这样(我将跳过常见部分):
... 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
最后的建议:“我有四个表描述数据库”-在这里改变您的想法。您有四个描述模型的实体,这些实体已映射到数据库(尽管映射可能会更改)。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提供者支持通用的ON
,否则您需要找到一个入口点和一条要加入的所有内容的路径(例如,从Enrollment
到,因此从Lecture
开始。