繼續在GAE上添加子實體到父代

Question

我有一個父實體客戶端。該客戶可以訪問該網站並創建約會。這意味着約會是在另一個時間創建的，然後是客戶端對象。

我的問題是：如何將子對象添加到已經保存的父對象？ 如果調用以下示例中的函數addData1（），則會創建約會表並添加條目。當函數addData2（）被調用時，這不會發生。

是不是因爲當關閉實體管理器後更新持久對象時，這也會在表中更新？

@Entity 
public class Client{ 
    @Id 
    private String name; 
    @OneToMany(cascade = CascadeType.ALL) 
    private Set<Appointment> appointments; 
} 

@Entity 
public class Appointment{ 
    @Id 
    @GeneratedValue(strategy = GenerationType.IDENTITY) 
    private Key id; 
} 

// This works 
public function addData1(){ 
    EntityManager em = EMF.get().createEntityManager(); 
    Client client = new Client(name); 
    for(Appointment a : newAppointments) 
     client.addAppointment(a); 
    em.persist(client); 
    em.close(); 
} 

// This doesn't work. 
public function addData2(){ 
    EntityManager em = EMF.get().createEntityManager(); 
    Client client = new Client(name); 
    em.persist(client); 
    em.close(); 
    for(Appointment a : newAppointments) 
     client.addAppointment(a); 
} 

Answer 1

public function addData1(){ 
    EntityManager em = EMF.get().createEntityManager(); 
    Client client = new Client(name); 
    for(Appointment a : newAppointments) 
     client.addAppointment(a); 
    em.persist(client); 
    em.close(); 
} 

此功能，因爲你堅持下去的所有對象。

// This doesn't work. 
public function addData2(){ 
    EntityManager em = EMF.get().createEntityManager(); 
    Client client = new Client(name); 
    em.persist(client); 
    em.close(); <--- You persist client with no appointment 
    for(Appointment a : newAppointments) 
     client.addAppointment(a); <-- those entitys are detached entity (not persisted) 
} 

此功能不會因爲你試圖持久化對象與分離實體

你應該得到這樣的例外：

Caused by: org.hibernate.PersistentObjectException: detached entity passed to persist: com.test.Appointment 

，如果你想避免這種例外，你可以試試這種方法：

public void addData2(List<Appointment> newAppointments){ 
    Client client = new Client(); 
    client.setName("name1"); 
    entityManager.persist(client); 
    for(Appointment a : newAppointments) 
     client.addAppointment(a); 
    entityManager.merge(client); 
    entityManager.close(); 
} 

順便說一句，嘗試使用名稱作爲ID作爲ID