example - orm java

java.lang.ClassCastException:[Ljava.lang.Object; no se puede convertir a entidad.UserEntity

Question

Quiero consultar dos tablas en Hibernate. tabla de featch 3 (Usuario - Rol - Perfil) en Entidad de usuario. consulta con hql:

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''";

y ejecutar consulta:

 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();
}
}

clase de uso: esta clase es geteer y 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;

hibernate maping para userEntity.hbm.xml

 <?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>

y clase hibernateutil para crear sesión:

 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;
}
}

Answer 1

Debido a que está utilizando una proyección de selección múltiple, en realidad está buscando una matriz de objetos, por lo que debe cambiar la lógica de procesamiento de resultados de la consulta a:

List<Object[]> tuples = (List<Object[]>) session.createQuery(query).list();
for(Object[] tuple : tuples) {
UserEntity ue = tuple[0];
Number roleId = tuple[1];
}