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java - 一对多 hibernate

转载 作者:行者123 更新时间:2023-11-30 11:42:23 25 4
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您好,我正在尝试进行一对多插入,但我遇到了问题。我有两个表:

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;

我的类(class):

@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;
}

我想用设备添加新用户

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();

但我面临异常:

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?

最佳答案

你还没有告诉 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;

(对另一个实体做同样的事情)

关于java - 一对多 hibernate ,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/11737954/

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