Hibernate多對多連接條件

Question

+-------+    +--------------|     +-------+
| BOY   |    | BOY_GIRL     |     | GIRL  | 
+-------+    +--------------|     +-------+
| id    |    | id           |     | id    |
| name  |    | boy_id       |     | name  |
| birth |    | girl_id      |     | birth |
+-------+    | start_dating |     +-------+
             +--------------|

START_DATING是TIMESTAMP或DATE類型

我有兩個豆子男孩和女孩有多對多的關系

@ManyToMany(fetch = FetchType.LAZY)
@JoinTable(name = "BOY_GIRL", joinColumns = {@JoinColumn(name = "BOY_ID")}, inverseJoinColumns = {@JoinColumn(name = "GIRL_ID")})
public Set<Girl> getGirls() {
    return girls;
}

現在，我如何使用HQL選擇查詢，如果我想獲得有條件的列表女孩：

where boy_id = (some_boy_id) and START_DATING > (some_timestamp)

Answer 1

我認為你的實體模型不正確，你需要一個代表關系屬性的第三個實體，你應該將男孩和女孩同時映射到該實體。 否則，無法在您的情況下將starting_date指定為關系屬性，作為查詢中的條件。 查看此鏈接 ，您可以找到有關如何使用其他屬性映射連接表的詳細說明。

Answer 2

我認為你必須創建一個BoyGirl類，因為表BOY_GIRL不是一個簡單的多對多表（如果是，那么列必須只是boy_id和girl_id ）。 那么你應該做的是創建BoyGirl類然后用BoyGirl映射BOY到BOY_GIRL ，並用一對多映射GIRL到BOY_GIRL

表關系

+-------+               +--------------+               +-------+
| BOY   |               | BOY_GIRL     |               | GIRL  | 
+-------+               +--------------|               +-------+
| id    | 0..* --- 1..1 | id           | 1..1 --- 0..* | id    |
| name  |               | boy_id       |               | name  |
| birth |               | girl_id      |               | birth |
+-------+               | start_dating |               +-------+
                        +--------------+

java類

public class BoyGirl {
  private long id;
  private Boy boy;
  private Girl girl;
  private Date startDating;
}

public class Boy {
  //other attributes omitted
  private Set<BoyGirl> boyGirls;
}

public class Girl {
  //other attributes omitted
  private Set<BoyGirl> boyGirls;
}

您需要的選擇查詢

// I'm using criteria here, but it will have the same result as your HQL

public List getGirls(Boy boy, Date startDating) {
  Criteria c = sessionFactory.getCurrentSession().createCriteria(BoyGirl.class);
  c.add(Restrictions.eq("boy.id", boy.getId());
  c.add(Restrictions.lt("startDating", startDating);

  List<BoyGirl> boyGirls = (List<BoyGirl>) c.list();
  // at this point you have lazily fetch girl attributes
  // if you need the girl  attributes to be initialized uncomment line below
  // for (BoyGirl boyGirl : boyGirls) Hibernate.initialize(boyGirl.getGirl());

  return boyGirls;
}

Answer 3

由於中間表BOY_GIRL具有一些額外的屬性（ start_dating ），您需要在域模型中為它​​創建另一個中間實體，例如：

@Table(name="BOY_GIRL")
class Relationship {

  @Id
  long id;

  @ManyToOne
  Boy boy;

  @ManyToOne;
  Girl girl;

  @Column
  Date startDating;
}

Answer 4

我認為將約會信息直接保存在聯接表中並不是很好（因為這樣做的目的應該只是關聯一個男孩和一個女孩）。

但是，如果你真的想保留當前的結構，並用HQL解決它，你應該能夠做類似的事情

SELECT g FROM GIRL g, BOY_GIRL bg 
WHERE bg.start_dating = :revelantdate 
    AND bg.boy_id = :boyid
    AND g.id = bg.girl_id