Question

我正在使用play框架构建一个Web应用程序，我遇到了使用这个模式创建数据库模型的问题

    **Client** – (many to many) – **Events**
          \                          /
      (one to many)           (one to many)
            \                      /
                  **COMMENTS**

我可以创建和添加关系

Event event = new Event("Party");

Client client = new Client("test@gmail.com","fname","lname","street","phonenum");

Comment comment = new Comment("test comment!");

//Add relationships

client.addEvent(event);
event.addClient(client);

event.addComment(comment);
comment.addEvent(event);

event.addClient(client);
client.addComment(comment);

但是当我试图坚持数据时

client.save();
event.save();
comment.save();

我收到错误

[error] Test CommentModelTest.testClientEventCommentRelationship failed: 
javax.persistence.PersistenceException: org.h2.jdbc.JdbcSQLException: Unique index or primary 
key violation: "PRIMARY_KEY_4A ON PUBLIC.EVENT_CLIENT(EVENT_ID, CLIENT_EMAIL)"; SQL statement:     
[error] insert into event_client (client_email, event_id) values (?, ?) [23505-172]

课程定义如下所示，如果有人能帮助我解决这个问题，我将非常感激。我是否错误地定义了这种关系？提前谢谢！

客户类：

@Entity 
public class Client extends Model {
  @Id
  public String email;
  @Required
  public String firstName;
  @Required
  public String lastName;
  public String homeAddress;
  public String phoneNumber;

  @ManyToMany(mappedBy="clients", cascade=CascadeType.ALL)
  public List<Event> events = new ArrayList<Event>();

  @OneToMany(mappedBy="client", cascade=CascadeType.ALL)
  public List<Comment> comments = new ArrayList<Comment>();

活动类：

@Entity
public class Event extends Model {
  @Id
  public Long id;
  @Required
  public String eventName;

  @ManyToMany(cascade=CascadeType.ALL)
  public List<Client> clients = new ArrayList<Client>();

  @OneToMany(mappedBy="event", cascade=CascadeType.ALL)
  public List<Comment> comments = new ArrayList<Comment>();

评论类：

@Entity
public class Comment extends Model {

  @Id
  public Long id;
  @Required
  public String message;

  @ManyToOne(cascade=CascadeType.PERSIST)
  public Client client;

  @ManyToOne(cascade=CascadeType.PERSIST)
  public Event event;

---编辑：-------------------------------

我意识到我哪里出错了。如果您遇到类似的错误，请检查您是否正确使用了模型！在我的JUnit测试中，我尝试将相同的客户端添加到事件两次，导致主键（电子邮件）被复制。

纠正后的关系代码应如下：

client.addEvent(event);
event.addClient(client);

event.addComment(comment);
comment.addEvent(event);

comment.addClient(client);
client.addComment(comment);

Answer 1

例外情况说你试图两次添加相同的关系。

"PRIMARY_KEY_4A ON PUBLIC.EVENT_CLIENT(EVENT_ID, CLIENT_EMAIL)"

我想这是events和clients之间的表格，因此它会映射ManyToMany关系。

client.addEvent(event);
HERE --> event.addClient(client);

event.addComment(comment);
comment.addEvent(event);

HERE --> event.addClient(client);
client.addComment(comment);

它已添加两次，并且有一个唯一的索引可以阻止它。

Play Framework：JPA三个实体之间的一对多关系？

1 个答案: