Question

您好我正在尝试进行一对多插入，但我遇到了问题。我有两张桌子：

CREATE TABLE users_app (
  user_id int UNSIGNED NOT NULL AUTO_INCREMENT,
  user_number varchar(45) NOT NULL default '0',
  user_password varchar(45) NOT NULL default '0',
  os int(1) unsigned NOT NULL,
  token varchar(500) NOT NULL,
  PRIMARY KEY (`user_id`)
) ENGINE=InnoDB AUTO_INCREMENT=20 DEFAULT CHARSET=utf8;

CREATE TABLE user_app_devices(  
    id int AUTO_INCREMENT PRIMARY KEY,  
    user_id int UNSIGNED NOT NULL,
    device_name varchar(45) NOT NULL,
    FOREIGN KEY (user_id) REFERENCES users_app (user_id)  
)ENGINE=InnoDB CHARSET=utf8;

我的课程：

@Entity
@Table(name="user_app_devices")
public class UserAppDevice implements Serializable {
private static final long serialVersionUID = 1L;

@Id
private int id;

@Column(name="device_name")
private String deviceName;

//bi-directional many-to-one association to UsersApp
@ManyToOne
@JoinColumn(name="user_id")
private UsersApp usersApp;

public UserAppDevice() {
}

public int getId() {
    return this.id;
}

public void setId(int id) {
    this.id = id;
}

public String getDeviceName() {
    return this.deviceName;
}

public void setDeviceName(String deviceName) {
    this.deviceName = deviceName;
}

public UsersApp getUsersApp() {
    return this.usersApp;
}

public void setUsersApp(UsersApp usersApp) {
    this.usersApp = usersApp;
}

}

@Entity
@Table(name="users_app")
public class UsersApp implements Serializable {
private static final long serialVersionUID = 1L;

@Id
@Column(name="user_id")
private int userId;

private int os;

private String token;

@Column(name="user_number")
private String userNumber;

@Column(name="user_password")
private String userPassword;

//bi-directional many-to-one association to UserAppDevice
@OneToMany(mappedBy="usersApp")
private List<UserAppDevice> userAppDevices;

public UsersApp() {
}

public int getUserId() {
    return this.userId;
}

public void setUserId(int userId) {
    this.userId = userId;
}

public int getOs() {
    return this.os;
}

public void setOs(int os) {
    this.os = os;
}

public String getToken() {
    return this.token;
}

public void setToken(String token) {
    this.token = token;
}

public String getUserNumber() {
    return this.userNumber;
}

public void setUserNumber(String userNumber) {
    this.userNumber = userNumber;
}

public String getUserPassword() {
    return this.userPassword;
}

public void setUserPassword(String userPassword) {
    this.userPassword = userPassword;
}

public List<UserAppDevice> getUserAppDevices() {
    return this.userAppDevices;
}

public void setUserAppDevices(List<UserAppDevice> userAppDevices) {
    this.userAppDevices = userAppDevices;
}

public UsersApp(int os, String token, String userNumber, String userPassword) {
    this.os = os;
    this.token = token;
    this.userNumber = userNumber;
    this.userPassword = userPassword;
}

我想添加带设备的新用户

我试试这段代码：

    Session session = (Session) em.getDelegate();
    session.beginTransaction();

    UsersApp user = new UsersApp(os, token, userNumber, userPassword);

    session.save(user);

    UserAppDevice ud = new UserAppDevice();

    ud.setUsersApp(user);
    ud.setDeviceName(device);

    session.save(ud);

    session.getTransaction().commit();

但我面临例外：

13:16:48,516 WARN  [org.hibernate.engine.jdbc.spi.SqlExceptionHelper] (http--0.0.0.0-8080-3) SQL Error: 1452, SQLState: 23000
13:16:48,517 ERROR [org.hibernate.engine.jdbc.spi.SqlExceptionHelper] (http--0.0.0.0-8080-3) Cannot add or update a child row: a foreign key constraint fails (`application`.`user_a
pp_devices`, CONSTRAINT `user_app_devices_ibfk_1` FOREIGN KEY (`user_id`) REFERENCES `users_app` (`user_id`))
13:16:48,520 ERROR [org.jboss.as.ejb3.tx.CMTTxInterceptor] (http--0.0.0.0-8080-3) javax.ejb.EJBTransactionRolledbackException: Cannot add or update a child row: a foreign key const
raint fails (`application`.`user_app_devices`, CONSTRAINT `user_app_devices_ibfk_1` FOREIGN KEY (`user_id`) REFERENCES `users_app` (`user_id`))
13:16:48,524 ERROR [org.jboss.ejb3.invocation] (http--0.0.0.0-8080-3) JBAS014134: EJB Invocation failed on component DeviceRegisterDAOImpl for method public abstract void com.break
id.ejb.model.DeviceRegisterDAO.add(int,java.lang.String,java.lang.String,java.lang.String,java.lang.String): javax.ejb.EJBTransactionRolledbackException: Cannot add or update a chi
ld row: a foreign key constraint fails (`application`.`user_app_devices`, CONSTRAINT `user_app_devices_ibfk_1` FOREIGN KEY (`user_id`) REFERENCES `users_app` (`user_id`))
        at org.jboss.as.ejb3.tx.CMTTxInterceptor.handleInCallerTx(CMTTxInterceptor.java:139) [jboss-as-ejb3-7.1.1.Final.jar:7.1.1.Final]
        at org.jboss.as.ejb3.tx.CMTTxInterceptor.invokeInCallerTx(CMTTxInterceptor.java:204) [jboss-as-ejb3-7.1.1.Final.jar:7.1.1.Final]
        at org.jboss.as.ejb3.tx.CMTTxInterceptor.required(CMTTxInterceptor.java:306) [jboss-as-ejb3-7.1.1.Final.jar:7.1.1.Final]
        at org.jboss.as.ejb3.tx.CMTTxInterceptor.processInvocation(CMTTxInterceptor.java:190) [jboss-as-ejb3-7.1.1.Final.jar:7.1.1.Final]
        at org.jboss.invocation.InterceptorContext.proceed(InterceptorContext.java:288) [jboss-invocation-1.1.1.Final.jar:1.1.1.Final]
        at org.jboss.as.ejb3.remote.EJBRemoteTransactionPropagatingInterceptor.processInvocation(EJBRemoteTransactionPropagatingInterceptor.java:80) [jboss-as-ejb3-7.1.1.Final.jar:
7.1.1.Final]

我错过了什么？

Answer 1

您尚未告知Hibernate UserApp的ID是由数据库自动生成的：

@Id
@GeneratedValue(strategy = IDENTITY)
@Column(name="user_id")
private int userId;

（并为其他实体做同样的事情）

Answer 2

由于您使用的是双向，因此请按以下方式更改客户端代码。

Session session = (Session) em.getDelegate();
session.beginTransaction();
UserAppDevice ud = new UserAppDevice();
ud.setDeviceName(device);
UsersApp user = new UsersApp(os, token, userNumber, userPassword);
user.setUserAppDevices(new ArrayList<UserAppDevice>())
user.getUserAppDevices().add(ud);
session.save(user);
session.getTransaction().commit();

Answer 3

正如JB Nizet所提到的，你错过了自动生成策略。

另一种方法是使用UUID作为id列，并使用

自行创建值

@Id
private UUID id = UUID.randomUUID();

此外，请不要忘记设置equals/hashCode以使用The JPA hashCode() / equals() dilemma

中讨论的死亡id字段

顺便说一下，为什么使用Session（特定于hibernate）而不是坚持使用JPA的API？

一对多休眠

3 个答案: