org.hibernate.exception.SQLGrammarException：无法获取下一个序列值

Question

Am using Hibernate 3.6

Below is Employee class code.

public class Employee {

private int id;

public int getId() {
    return id;
}

public void setId(int id) {
    this.id = id;
}

Below is Employee.hbm.xml file,

<class name="com.testing.hibernate.Employee" table="HIB_EMPLOYEE">
      <meta attribute="class-description">
            This class contains the Employee details.
      </meta>
      <id name="id" type="int" column="id">
         <generator class="sequence"></generator>
      </id>
      <property name="firstName" column="first_name" type="string"></property>
      <property name="lastName" column="last_name" type="string"></property>
      <property name="salary" column="salary" type="int"></property>            
  </class>

I have created sequence in Database. Below SS for reference. How can I overcome the exception that am getting?

Answer 1

You have to give the reference of the sequence to hibernate.

<generator class="sequence">
     <param name="sequence">SEQUENCE_NAME</param>
</generator>

Answer 2

What annotation can i use to do this?

i've tried

@GeneratedValue(strategy=GenerationType.SEQUENCE , generator = <SEQUENCE_NAME>") without any success

edit:

it seams that the generator have to also be created

@SequenceGenerator(name="<GEN_NAME>", sequenceName="<SEQUENCE_NAME>")