[英]Persist PK Object with JPA (ManyToMany)
如果您知道更好的名稱,請更改標題,因為我真的不知道如何表達問題。
我有三節課:
@Entity
public class User {
@Id
@GeneratedValue(strategy = GenerationType.IDENTITY)
@Column(name = "id")
private Integer id;
@NotNull
@Size(min = 1, max = 45)
@Column(name = "name")
private String name;
@JoinTable(name = "usuer_has_contact", joinColumns = {
@JoinColumn(name = "user_id", referencedColumnName = "id")}, inverseJoinColumns = {
@JoinColumn(name = "contact_id", referencedColumnName = "id")})
@ManyToMany(cascade = CascadeType.ALL)
private List<Contato> contactList;
//Getters and Setters
}
DB Table:
Table Name: User
Columns: id (int pk), name (varchar(45) not null).
@Entity
private class Contact {
@EmbeddedId
protected UserHasContact userHasContact;
@NotNull
@Size(min = 1, max = 45)
@Column(name = "value")
private String value;
@ManyToMany(mappedBy = "contactList")
private List<User> userList;
//Getters and Setters
}
DB Table:
Table Name: Contact
Columns: id (int pk), value (varchar(45) not null).
@Embeddable
private class UserHasContact {
@NotNull
@Column(name = "id")
private Integer id;
//Getters and Setters
}
DB Table:
Table Name: UserHasContact
Columns: userId (int pk), contactId (int pk).
我想做的是,當我堅持用戶本身時,要堅持一切。 例如:
User user = new User();
user.setContactList(new ArrayList<Contact>());
Contact contact = new Contact();
contact.setValue("555-5555");
user.getContactList().add(contact);
// Here I'd call another class, passing User so it would only do a
// entityManager.persist(user), and it would persist it all and
// take care of all tables for me. What I don't want to do is to
// fill up the values myself, I want let JPA do it for me.
我希望這樣做后保存,但是它說contactId為null,不能為null。 我能做什么?
為什么要創建一個可嵌入的UserHasContact
類以僅存儲一個Integer? 您正在使它變得不必要。 只需使用Integer ID作為聯系人主鍵即可。 但是,這不是問題的原因。
您試圖將包含聯系人的用戶保留在其聯系人列表中。 您的聯系人ID不會自動生成,並且您沒有為此聯系人ID分配任何ID。 JPA如何將該聯系人保存在數據庫中? 此外,您沒有保持聯系,因此是暫時的。
你必須
這是Contact實體的代碼:
@Entity
private class Contact {
@Id
@GeneratedValue(strategy = GenerationType.IDENTITY) // this makes the ID auto-generated
@Column(name = "id")
private Integer id;
@NotNull
@Size(min = 1, max = 45)
@Column(name = "value")
private String value;
@ManyToMany(mappedBy = "contactList")
private List<User> userList;
//Getters and Setters
}
在創建用戶和聯系人的代碼中:
User user = new User();
user.setContactList(new ArrayList<Contact>());
entityManager.persist(user);
Contact contact = new Contact();
contact.setValue("555-5555");
entityManager.persist(contact);
user.getContactList().add(contact);
// you should also make sure that the object graph is consistent, so
// the following line should lso be added, (though not strictly necessary)
contact.getUserList().add(user);
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