Java 一对一休眠未映射
以下是MySQL数据库中的表的外观:Java 一对一休眠未映射,java,mysql,hibernate,Java,Mysql,Hibernate,以下是MySQL数据库中的表的外观: mysql> DESCRIBE customer; +-------------+----------------------+------+-----+ | Field | Type | Null | Key | +-------------+----------------------+------+-----+ | customer_id | smallint(5) unsigned | NO |
mysql> DESCRIBE customer;
+-------------+----------------------+------+-----+
| Field | Type | Null | Key |
+-------------+----------------------+------+-----+
| customer_id | smallint(5) unsigned | NO | PRI |
| first_name | varchar(45) | NO | |
| last_name | varchar(45) | NO | |
| address_id | smallint(5) unsigned | NO | MUL |
+-------------+----------------------+------+-----+
mysql> DESCRIBE address;
+-------------+----------------------+------+-----+
| Field | Type | Null | Key |
+-------------+----------------------+------+-----+
| address_id | smallint(5) unsigned | NO | PRI |
| address | varchar(50) | NO | |
+-------------+----------------------+------+-----+
这就是我试图实现一对一关系的方式:
@Entity
@Table(name = "customer")
public class Customer {
@Id
@GeneratedValue(strategy = GenerationType.IDENTITY)
private int customer_id;
@Basic
private String first_name;
@Basic
private String last_name;
@OneToOne(targetEntity = Address.class)
@JoinColumn(name = "address_id")
private Address address;
在Address.java中
@Entity
@Table(name = "address")
public class Address {
@Id
@GeneratedValue(strategy = GenerationType.IDENTITY)
private int address_id;
@Basic
private String address;
@OneToOne(mappedBy = "address", targetEntity = Customer.class)
private Customer customer;
我正在通过一个简单的测试进行测试:
@Test
public void testGetAll() {
PersistenceUtil.buildEntityManagerFactory();
AddressDaoImpl addressDao = new AddressDaoImpl();
List all = addressDao.getAll();
System.out.println(all.iterator().next());
}
我得到:
java.lang.IllegalArgumentException: org.hibernate.hql.internal.ast.QuerySyntaxException: address is not mapped [FROM address]
我做错了什么?为什么不为地址实体提取客户
编辑
以下是我的DAO类的代码:
package biz.tugay.saqila.dao;
/* User: koray@tugay.biz Date: 08/08/15 Time: 10:18 */
import javax.persistence.EntityManager;
import javax.persistence.EntityTransaction;
import java.util.List;
public abstract class GenericDaoImpl<T> implements GenericDao<T> {
@Override
public List getAll() {
EntityManager em = PersistenceUtil.getEntityManager();
EntityTransaction transaction = em.getTransaction();
transaction.begin();
List<T> all = em.createQuery("FROM " + getTableName()).getResultList();
transaction.commit();
return all;
}
}
public class AddressDaoImpl extends GenericDaoImpl<Address> {
@Override
public String getTableName() {
return "address";
}
}
package biz.tugay.saqila.dao;
/*用户:koray@tugay.biz日期:2015年8月8日时间:10:18*/
导入javax.persistence.EntityManager;
导入javax.persistence.EntityTransaction;
导入java.util.List;
公共抽象类GenericDaoImpl实现GenericDao{
@凌驾
公共列表getAll(){
EntityManager em=PersistenceUtil.getEntityManager();
EntityTransaction=em.getTransaction();
transaction.begin();
List all=em.createQuery(“FROM”+getTableName()).getResultList();
commit();
全部归还;
}
}
公共类AddressDaoImpl扩展了GenericDaoImpl{
@凌驾
公共字符串getTableName(){
返回“地址”;
}
}
JPQL(HQL)类和属性名称为。因此,在查询中,必须为地址
实体使用正确的类名:
from Address ...
因此,请将实体名称改为以大写字母A开头:
public class AddressDaoImpl extends GenericDaoImpl<Address> {
@Override
public String getTableName() {
return "Address";
}
}
public类AddressDaoImpl扩展了GenericDaoImpl{
@凌驾
公共字符串getTableName(){
返回“地址”;
}
}
你能粘贴getAll()代码吗?@wawek我有,我也解决了这个问题,我在这里复制的测试是错误的。@wawek测试失败的是:AddressDaoImpl addressDao=new AddressDaoImpl();不是CustomerDaoImpl..正如@Dragan Bozanovic回答的,您必须使用类名而不是表名。