rez*_*ami 8 java orm hibernate hql java-ee
我想在hibernate中查询两个表.用户实体中的featch 3表(用户 - 角色 - 配置文件).用hql查询:
query= "select ue, ue.roleEntity.roleId from UserEntity ue ,RoleEntity re fetch all properties where ue.roleEntity.roleId=re.roleId and ue.username ='reza' and ue.password='123456'";
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并运行查询:
try {
sessionFactory = HibernateUtil.getSessionFactory();
session = sessionFactory.getCurrentSession();
transaction = session.beginTransaction();
userEntityList = (List<UserEntity>) session.createQuery(query).list();
transaction.commit();
} catch (HibernateException e) {
try {
throw new DaoException("Fetal exception in", e);
} catch (DaoException e1) {
e1.printStackTrace();
}
}
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userentity class:这个类是geteer和seter:
public class UserEntity {
private int userId;
private long personalCode;
private String username;
private String password;
private short active;
private String question;
private String passive;
private ProfileEntity profileEntity;
private RoleEntity roleEntity;
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用于userEntity.hbm.xml的hibernate maping
<?xml version="1.0" encoding="utf-8" ?>
<!DOCTYPE hibernate-mapping PUBLIC
"-//Hibernate/Hibernate Mapping DTD 3.0//EN"
"http://hibernate.sourceforge.net/hibernate-mapping-3.0.dtd">
<hibernate-mapping package="entity">
<class name="UserEntity" table="TABLE_USER">
<id name="userId" type="java.lang.Integer" column="USER_ID">
<generator class="increment" />
</id>
<property name="personalCode" type="java.lang.Long" column="PERSONALCODE">
</property>
<property name="username" type="java.lang.String" column="USERNAME">
</property>
<property name="password" type="java.lang.String" column="PASSWORD">
</property>
<property name="active" type="java.lang.Short" column="ACTIVE">
</property>
<property name="question" type="java.lang.String" column="QUCTION">
</property>
<property name="passive" type="java.lang.String" column="PASSIVE">
</property>
<many-to-one name="roleEntity" class="entity.RoleEntity" column="ROLE_ID" cascade="none" fetch="select" />
<many-to-one name="profileEntity" class="ProfileEntity" cascade="delete" column="profile_id"/>
</class>
</hibernate-mapping>
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和类hibernateutil用于创建sesstion:
import org.hibernate.SessionFactory;
import org.hibernate.boot.registry.StandardServiceRegistryBuilder;
import org.hibernate.cfg.Configuration;
public class HibernateUtil {
private static SessionFactory sessionFactory;
static {
try {
Configuration configuration = new Configuration().configure();
StandardServiceRegistryBuilder builder = new StandardServiceRegistryBuilder().applySettings(configuration.getProperties());
sessionFactory = configuration.buildSessionFactory(builder.build());
} catch (Throwable th) {
System.err.println("Enitial SessionFactory creation failed" + th);
throw new ExceptionInInitializerError(th);
}
}
/**
* @return
*/
public static SessionFactory getSessionFactory() {
return sessionFactory;
}
}
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Vla*_*cea 13
因为您正在使用多选投影,所以实际上是在获取对象数组,因此您需要将查询结果处理逻辑更改为:
List<Object[]> tuples = (List<Object[]>) session.createQuery(query).list();
for(Object[] tuple : tuples) {
UserEntity ue = tuple[0];
Number roleId = tuple[1];
}
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