JPA持有具有一对多关系的实体

Question

配置

• EcliplseLink 2.3.2
• JPA 2.0
• 这些实体是从带有Entity Classes from Database ...向导的netbeans的db模式自动创建的.
• 控制器类是从netbeans自动创建的,具有来自Entity Classes ...向导的JPA Controller Classes

问题的简短版本

在经典场景中,两个表具有一对多关系.我创建父实体,然后创建子实体,并将子项附加到父项的集合.当我创建(控制器方法)父实体时,我希望子实体被创建并与父关联.为什么不发生？

长版

家长班

@Entity
@XmlRootElement
public class Device implements Serializable {
    private static final long serialVersionUID = 1L;
    @Id
    @GeneratedValue(strategy = GenerationType.IDENTITY)
    @Basic(optional = false)
    private Integer id;
    @Column(unique=true)
    private String name;
    @Temporal(TemporalType.TIMESTAMP)
    private Date updated;
    @OneToMany(cascade = CascadeType.ALL, mappedBy = "deviceId")
    private Collection<NetworkInterface> networkInterfaceCollection;

    public Device() {
    }

    public Device(String name) {
        this.name = name;
        updated = new Date();
    }

    // setters and getters...

    @XmlTransient
    public Collection<NetworkInterface> getNetworkInterfaceCollection() {
        return networkInterfaceCollection;
    }

    public void setNetworkInterfaceCollection(Collection<NetworkInterface> networkInterfaceCollection) {
        this.networkInterfaceCollection = networkInterfaceCollection;
    }

    public void addNetworkInterface(NetworkInterface net) {
        this.networkInterfaceCollection.add(net);
    }

    public void removeNetworkInterface(NetworkInterface net) {
        this.networkInterfaceCollection.remove(net);
    }
    // other methods
}

儿童班

@Entity
@Table(name = "NETWORK_INTERFACE")
@XmlRootElement
public class NetworkInterface implements Serializable {
    private static final long serialVersionUID = 1L;
    @Id
    @GeneratedValue(strategy = GenerationType.IDENTITY)
    @Basic(optional = false)
    private Integer id;
    private String name;
    @Temporal(TemporalType.TIMESTAMP)
    private Date updated;
    @JoinColumn(name = "DEVICE_ID", referencedColumnName = "ID")
    @ManyToOne(optional = false)
    private Device deviceId;

    public NetworkInterface() {
    }

    public NetworkInterface(String name) {
        this.name = name;
        this.updated = new Date();
    }

    // setter and getter methods...

    public Device getDeviceId() {
        return deviceId;
    }

    public void setDeviceId(Device deviceId) {
        this.deviceId = deviceId;
    }
}

主要课程

public class Main {
    public static void main(String[] args) {
        EntityManagerFactory emf = Persistence.createEntityManagerFactory("wifi-dbPU");
        DeviceJpaController deviceController = new DeviceJpaController(emf);
        NetworkInterfaceJpaController netController = new NetworkInterfaceJpaController(emf);

        Device device = new Device("laptop");
        NetworkInterface net = new NetworkInterface("eth0");

        device.getNetworkInterfaceCollection().add(net);
        deviceController.create(device);
    }
}

这个类在行中抛出NullPointerException: device.getNetworkInterfaceCollection().add(net);

系统知道有一个新实体,device并且net它的集合中有一个元素.我希望它写入devicedb,获取设备的id,将其附加到netdb 并将其写入db.

而不是这个,我发现这些是我必须要做的步骤:

deviceController.create(device);
net.setDeviceId(device);
device.getNetworkInterfaceCollection().add(net);
netController.create(net);

为什么我必须在父类知道它的孩子时创建孩子并且应该为我创建它？

来自DeviceJpaController的create方法(抱歉字段中的长名称,它们是自动生成的).

public EntityManager getEntityManager() {
    return emf.createEntityManager();
}

public void create(Device device) {
    if (device.getNetworkInterfaceCollection() == null) {
        device.setNetworkInterfaceCollection(new ArrayList<NetworkInterface>());
    }
    EntityManager em = null;
    try {
        em = getEntityManager();
        em.getTransaction().begin();
        Collection<NetworkInterface> attachedNetworkInterfaceCollection = new ArrayList<NetworkInterface>();
        for (NetworkInterface networkInterfaceCollectionNetworkInterfaceToAttach : device.getNetworkInterfaceCollection()) {
            networkInterfaceCollectionNetworkInterfaceToAttach = em.getReference(networkInterfaceCollectionNetworkInterfaceToAttach.getClass(), networkInterfaceCollectionNetworkInterfaceToAttach.getId());
            attachedNetworkInterfaceCollection.add(networkInterfaceCollectionNetworkInterfaceToAttach);
        }
        device.setNetworkInterfaceCollection(attachedNetworkInterfaceCollection);
        em.persist(device);
        for (NetworkInterface networkInterfaceCollectionNetworkInterface : device.getNetworkInterfaceCollection()) {
            Device oldDeviceIdOfNetworkInterfaceCollectionNetworkInterface = networkInterfaceCollectionNetworkInterface.getDeviceId();
            networkInterfaceCollectionNetworkInterface.setDeviceId(device);
            networkInterfaceCollectionNetworkInterface = em.merge(networkInterfaceCollectionNetworkInterface);
            if (oldDeviceIdOfNetworkInterfaceCollectionNetworkInterface != null) {
                oldDeviceIdOfNetworkInterfaceCollectionNetworkInterface.getNetworkInterfaceCollection().remove(networkInterfaceCollectionNetworkInterface);
                oldDeviceIdOfNetworkInterfaceCollectionNetworkInterface = em.merge(oldDeviceIdOfNetworkInterfaceCollectionNetworkInterface);
            }
        }
        em.getTransaction().commit();
    } finally {
        if (em != null) {
            em.close();
        }
    }
}

Answer 1

我终于理解了坚持一个到多个实体的逻辑.过程是:

1. 创建父类
2. 坚持下去
3. 创建子类
4. 将孩子与父母联系起来
5. 持续孩子(父母集合更新)

使用代码:

public class Main {
    public static void main(String[] args) {
        EntityManagerFactory emf = Persistence.createEntityManagerFactory("wifi-dbPU");
        DeviceJpaController deviceController = new DeviceJpaController(emf);
        NetworkInterfaceJpaController netController = new NetworkInterfaceJpaController(emf);

        Device device = new Device("laptop");                 // 1
        deviceController.create(device);                      // 2

        NetworkInterface net = new NetworkInterface("eth0");  // 3
        net.setDeviceId(device.getId());                      // 4
        netController.create(net);                            // 5 
        // The parent collection is updated by the above create     
    }
}

现在,我可以找到一个设备(例如id),我可以让它的所有孩子都使用

Collection<NetworkInterface> netCollection = device.getNetworkInterfaceCollection()

在我在问题中发布的设备实体类中,不需要方法addNetworkInterface和removeNetwokrInterface.

Answer 2

@Dima K 他们说的是正确的。当你这样做时：

    Device device = new Device("laptop");
    NetworkInterface net = new NetworkInterface("eth0");

    device.getNetworkInterfaceCollection().add(net);
    deviceController.create(device);

设备中的集合尚未初始化，因此在尝试添加时会收到 NPE。在你的Device类中，声明你的 时Collection，你还可以初始化它：

private Collection<NetworkInterface> networkInterfaceCollection = new CollectionType<>();

至于坚持，你的假设是正确的，但我认为执行是错误的。当您创建设备时，立即使用 JPA 使其持久化（在需要时进行事务管理）。

Device device = new Device("laptop");
getEntityManager().persist(device);

对网络接口执行相同操作：

NetworkInterface net = new NetworkInterface("eth0");
getEntityManager().persist(net);

现在，由于您的两个实体都已持久化，因此您可以将一个实体添加到另一个实体中。

device.getNetworkInterfaceCollection().add(net);

JPA 应该处理其余的事情，而无需调用任何其他持久性。