尝试在Hibernate中插入记录时获取SQLGrammarException

Question

我收到以下异常-
org.hibernate.exception.sqlgrammarexception：无法插入：[Employee]
以下是详细信息:

hibernate.cfg.xml

    <hibernate-configuration>
    <session-factory>
    <property name="hibernate.dialect">org.hibernate.dialect.SQLServerDialect</property>
            <property name="hibernate.connection.driver_class">sun.jdbc.odbc.JdbcOdbcDriver</property>
            <property name="hibernate.connection.url">jdbc:odbc:Driver={Microsoft Access Driver (*.mdb, *.accdb)};DBQ=D:/Hibernate.accdb</property>
            <property name="connection.username"></property>
            <property name="connection.password"></property>
            <property name="connection.pool_size">1</property>
            <property name="current_session_context_class">thread</property>
            <property name="cache.provider_class">org.hibernate.cache.NoCacheProvider</property>
            <property name="show_sql">true</property>
            <property name="hbm2ddl.auto">update</property>
            <mapping resource="Employee.hbm.xml" />
    </session-factory>
    </hibernate-conf

employee.hbm.xml

<hibernate-mapping>
   <class name="Employee" table="EMPLOYEE">
      <meta attribute="class-description">
         This class contains the employee detail. 
      </meta>
      <id name="id" type="int" column="id">
         <generator class="native"/>
      </id>
      <property name="firstName" column="first_name" type="string"/>
      <property name="lastName" column="last_name" type="string"/>
      <property name="salary" column="salary" type="int"/>
   </class>
</hibernate-mapping>

public class Employee {
   private int id;
   private String firstName; 
   private String lastName;   
   private int salary;  

   public Employee() {}
   public Employee(String fname, String lname, int salary) {
      this.firstName = fname;
      this.lastName = lname;
      this.salary = salary;
   }
   public int getId() {
      return id;
   }
   public void setId( int id ) {
      this.id = id;
   }
   public String getFirstName() {
      return firstName;
   }
   public void setFirstName( String first_name ) {
      this.firstName = first_name;
   }
   public String getLastName() {
      return lastName;
   }
   public void setLastName( String last_name ) {
      this.lastName = last_name;
   }
   public int getSalary() {
      return salary;
   }
   public void setSalary( int salary ) {
      this.salary = salary;
   }
}

import java.util.List; 
import java.util.Date;
import java.util.Iterator; 

import org.hibernate.HibernateException; 
import org.hibernate.Session; 
import org.hibernate.Transaction;
import org.hibernate.SessionFactory;
import org.hibernate.cfg.Configuration;

public class ManageEmployee {
   private static SessionFactory factory; 
   public static void main(String[] args) {
      try{
         factory = new Configuration().configure().buildSessionFactory();
      }catch (Throwable ex) { 
         System.err.println("Failed to create sessionFactory object." + ex);
         throw new ExceptionInInitializerError(ex); 
      }
      ManageEmployee ME = new ManageEmployee();

      /* Add few employee records in database */
      Integer empID1 = ME.addEmployee("Zara", "Ali", 1000);
      Integer empID2 = ME.addEmployee("Daisy", "Das", 5000);
      Integer empID3 = ME.addEmployee("John", "Paul", 10000);
}
 public Integer addEmployee(String fname, String lname, int salary){
      Session session = factory.openSession();
      Transaction tx = null;
      Integer employeeID = null;
      try{
         tx = session.beginTransaction();
         Employee employee = new Employee(fname, lname, salary);
         employeeID = (Integer) session.save(employee); 
         tx.commit();
      }catch (HibernateException e) {
         if (tx!=null) tx.rollback();
         e.printStackTrace(); 
      }finally {
         session.close(); 
      }
      return employeeID;
   }

}

请帮助解决异常。提前道谢！

Answer 1

Hibernate库中不存在MS Access方言。这会导致Hibernate无法检测标识列，也无法按照MS访问所需的方式操作。

org.hibernate.exception.sqlgrammarexception：无法插入:[Employee]hibernate:插入到Employee(first_name,last_name,salary）值（？,？,？）选择scope_identity()

select scope_identity()是MS Access不支持的TRANSACT-SQL实例。

null

有关如何使用hibernate与MS Access？