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I have created a Database Access Object (DAO) class, entity class, and table script, but I am getting an error that cannot be mapped to entity.
I am using hibernate framework and the connections are made properly with the database but still error occurs. please check the code below and help in any ways you can, all the files are provided below.

Table Script

DROP TABLE rmc_user;
CREATE TABLE rmc_user(
user_id VARCHAR(50) NOT NULL,
user_name VARCHAR(20) NOT NULL,
user_email VARCHAR(50) NOT NULL,
user_password VARCHAR(20),
CONSTRAINT rmc_user_user_id_pk PRIMARY KEY (user_id),
CONSTRAINT rmc_user_user_email_un UNIQUE (user_email) 
);

INSERT INTO rmc_user VALUES   ('101','yashik','yas@gmail.com','gulati123');

SELECT * FROM rmc_user;

DAO Class

package rmc.dao;

import java.util.List;
import org.hibernate.Session;
import org.hibernate.SessionFactory;
import org.hibernate.query.Query;
import rmc.bean.User;
import rmc.entity.UserEntity;
import rmc.resources.HibernateUtility;

public class LoginDAOImpl implements LoginDAO {

    @SuppressWarnings("deprecation")
    public User getUserDetails(String userName, String password) {
        SessionFactory sessionFactory = HibernateUtility.createSessionFactory();
        Session session = null;
        User u1 = null;
        session = sessionFactory.openSession();
        session.beginTransaction();
        System.out.println("begin trx");
        Query q1 = session
                .createNativeQuery("select * from rmc_user where user_name=?");
        System.out.println("begin trx");
        q1.setParameter(0, userName);
        System.out.println("begin trx");
        @SuppressWarnings("unchecked")
        List<UserEntity> l1 = q1.list();
        System.out.println("begin trx");
        System.out.println("size is"+l1.size());
        if (l1.size() == 0) {
            System.out.println("no Such user Exist");
        } else if (!(l1.get(0).getPassword().equals(password))) {
            System.out.println("Invalid Password");
        }
        System.out.println("begin trx");
        u1 = new User();
        u1.setEmail(l1.get(0).getEmail());
        u1.setPassword(l1.get(0).getPassword());
        u1.setUserId(l1.get(0).getUserId());
        u1.setUserName(l1.get(0).getUserName());
        session.getTransaction().commit();
        if (session != null) {
            session.close();
        }
        return u1;
    }
}

Entity Class

package rmc.entity;
@Id
@Column(name="user_id")
private String userId;
@Column(name="user_name")
private String userName;
@Column(name="user_email")
private String email;
@Column(name="user_password")
private String password;

//getter and setter 
}

Error Message

Exception in thread "main" java.lang.ClassCastException: [Ljava.lang.Object; cannot be cast to rmc.entity.UserEntity
at rmc.dao.LoginDAOImpl.getUserDetails(LoginDAOImpl.java:32)
at rmc.test.UserInterface.main(UserInterface.java:9)

UPDATED

<?xml version="1.0" encoding="UTF-8"?>
<!DOCTYPE hibernate-configuration SYSTEM 
"http://www.hibernate.org/dtd/hibernate-configuration-3.0.dtd">

<hibernate-configuration>
<session-factory>
    <property name="hibernate.dialect">
        org.hibernate.dialect.MySQLDialect
    </property>
    <property name="hibernate.connection.driver_class">
        com.mysql.jdbc.Driver
    </property>

    <!-- Assume test is the database name -->
    <property name="hibernate.connection.url">
        jdbc:mysql://localhost:3306/rmc
    </property>
    <property name="hibernate.connection.username">
        ******
    </property>
    <property name="hibernate.connection.password">
        ******
    </property>


    <!-- List of XML mapping files -->
    <mapping class="rmc.entity.UserEntity"/>

</session-factory>

Yashik
  • 41
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2 Answers2

2

If your result row is compatible with UserEntity the following modification might solve your problem:

Query q1 = session
        .createNativeQuery("select * from rmc_user where user_name=?", UserEntity.class);
ujulu
  • 3,289
  • 2
  • 11
  • 14
0

The problem probably lies here 

Query q1 = session.createNativeQuery("select * from rmc_user where user_name=?");

When you execute this query, because it's an sql query, the return list of q1.list() will have this format : Object[] { row1col1, row1col2,row1,col3,row2col1,...} that is it spreads the columns , it doesn't map row to entity(UserEntity).

It doesn't map because that's not HQL nor JPQL, that's native SQL and SQL doesn't know about your entities.

You should instead do this : 

Query q1=session.createQuery("select * from UserEntity where user_name=?");

This is HQL and with this Hibernate will map every row to entity so the return list of q1.list() will now have format : UserEntity[]{entity1,entity2,...}.

I hope that solves your issue.

niceman
  • 2,653
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  • 57
  • hi thank you for the response, I did some changes but some error occurs, can you please take a look? `org.hibernate.hql.internal.ast.QuerySyntaxException: UserEntity is not mapped [FROM UserEntity u where u.userName=?]` – Yashik Jul 03 '16 at 08:40
  • also the has been edited `Query q1 = session .createQuery("FROM UserEntity u where u.userName=?"); q1.setParameter(0, userName); List l1 = q1.list();` – Yashik Jul 03 '16 at 08:48
  • @Yashik are you by a chance using `@Entity` annotation from org.hibernate package ? you should use the one coming from javax.persistance package, see [this](http://stackoverflow.com/a/8524836/2397162) – niceman Jul 04 '16 at 13:31
  • hmm strange, see [this](http://stackoverflow.com/questions/14446048/hibernate-table-not-mapped-error) – niceman Jul 06 '16 at 17:02