Exception in thread "main" org.hibernate.exception.SQLGrammarException: ORA-02289: sequence does not exist

Question

I am trying to execute a basic hibernate application.However,I am consistently getting the error that is posted in the question.

Below posted is my project structure code:

Below is the code that is present in app.java

public class app {

/**
 * @param args
 */
public static void main(String[] args) {
    // TODO Auto-generated method stub

    Session session =hibernate_utils.getSessionFactory().openSession();
    session.beginTransaction();
    contact  contact=new contact();
    contact.setFirstname("xxx");
    contact.setLastname("xxx");
    contact.setEmail("[email protected]");
    contact.setTelephone("xxxxxxxxxx");
    session.save(contact);
    session.getTransaction().commit();
    System.out.println("saved");
    }

  }

Below posted is the code that is present in the contact.java file

package net.rishanth.contact.form;

import javax.persistence.Entity;
import javax.persistence.GeneratedValue;
import javax.persistence.GenerationType;
import javax.persistence.Id;
import javax.persistence.SequenceGenerator;
import javax.persistence.Table;
import javax.persistence.Column;
import static javax.persistence.GenerationType.SEQUENCE;

@Entity
@Table(name = "contacts")
public class contact {

@Id
@GeneratedValue(strategy=SEQUENCE)
@Column(name = "id", unique = true, nullable = false)     
public Integer getId() {
    return id;
}
public void setId(Integer id) {
    this.id = id;
}

@Column(name = "firstname", nullable = false)
public String getFirstname() {
    return firstname;
}
public void setFirstname(String firstname) {
    this.firstname = firstname;
}
@Column(name = "lastname", nullable = false)
 public String getLastname() {
    return lastname;
}
public void setLastname(String lastname) {
    this.lastname = lastname;
}
@Column(name = "email", nullable = false)
 public String getEmail() {
    return email;
}
public void setEmail(String email) {
    this.email = email;
}
@Column(name = "telephone", nullable = false)

public String getTelephone() {
    return telephone;
}
public void setTelephone(String telephone) {
    this.telephone = telephone;
}
private String firstname;
private String lastname;
private String email;
private String telephone;
private Integer id;

}

Below posted is the code for my hiber_utils class present in service package.

package net.rishanth.contact.service;

import org.hibernate.SessionFactory;
import org.hibernate.cfg.Configuration;

public class hibernate_utils {

private static final SessionFactory sessionfactory= buildSessionFatory();

@SuppressWarnings("deprecation")
private static SessionFactory buildSessionFatory(){
    // TODO Auto-generated method stub

        return new Configuration().configure().buildSessionFactory();



}
public static SessionFactory getSessionFactory()
{

    return sessionfactory;
}
public static void shutdown()
{

    getSessionFactory().close();
}

}

Below present is the hibernate.cnfg.xml file

<?xml version="1.0" encoding="UTF-8"?> 
<!DOCTYPE hibernate-configuration PUBLIC
 "-//Hibernate/Hibernate Configuration DTD 3.0//EN"
 "http://hibernate.sourceforge.net/hibernate-configuration-3.0.dtd">
 <hibernate-configuration>
 <session-factory>
 <property    
 name="hibernate.bytecode.use_reflection_optimizer">false</property>
 <property   
 name="hibernate.connection.driver_class">oracle.jdbc.driver.OracleDriver
 </property>
 <property   
 name="hibernate.connection.url">jdbc:oracle:thin:@127.0.0.1:1521:XE
 </property>
 <property   name="hibernate.dialect">org.hibernate.dialect.Oracle10gDialect
 </property>
 <property name="hibernate.connection.username">system</property>
 <property name="hibernate.connection.password">xxxxx</property>
 <mapping class="net.rishanth.contact.form.contact"></mapping>
 </session-factory>
 </hibernate-configuration>

Below attached is my oracle screenshot

Any help would be highly appreciated. Thanks!

Please post the error message instead of a screenshot of a part of the message. Anyways, the reason is quite obvious: the SEQUENCE is missing from the DB. — Mick Mnemonic
– Mick Mnemonic, Commented Jan 27, 2016 at 22:03
@MickMnemonic can you please tell me what sequence are you talking about? — mohan babu
– mohan babu, Commented Jan 27, 2016 at 22:04
It's really hard to know without seeing the full stack trace and knowing where the error originates. It might be the SEQUENCE used for generating Contact ids. — Mick Mnemonic
– Mick Mnemonic, Commented Jan 27, 2016 at 22:13

Madhusudana Reddy Sunnapu · Accepted Answer · 2016-01-28 05:51:19Z

1

You have annotated the Contact entity use SEQUENCE as strategy. But you have not specified which sequence should be used. (I believe this is the error you might be getting. If not, posting the exception stack trace will help.)

In this case, by default hibernate looks for a sequence named hibernate_sequence and creating a sequence with this name should help.

Or, In case you want hibernate to use a sequence (say, your_sequence_name) that you have already created then further qualifying the @Id attribute as below should help:

@GeneratedValue(strategy=GenerationType.SEQUENCE, generator="mySeq")
@GenericGenerator(name="mySeq", strategy="sequence", 
                parameters={
                        @Parameter(name="sequence_name", value="your_sequence_name")
                })

answered Jan 28, 2016 at 5:51

Madhusudana Reddy Sunnapu

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Comments

udit khare · Accepted Answer · 2016-01-28 05:26:06Z

0

Usually we need not to create the table/sequence in our DB, hibernate can manage the same.

For that you need to add below tag in your hibernate.cfg.xml

<property name="hbm2ddl.auto">create</property>

hbm2ddl.auto have few possible values:

create : When you create SessionFactory, hibernate drop everything(if exist) and create it again for you.
create-drop : create everything on start-up and drop them on shut-down
update : update the schema.
validate : validate the schema, makes no changes to the database.

Anyway, if you want to create table/sequence explicitly, you can create them.

But then while using @GeneratedValue you have to specify your sequence.

answered Jan 28, 2016 at 5:26

udit khare

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