Получение SQLGrammarException при попытке вставить запись в Hibernate

Question

Я получаю следующее исключение -
org.hibernate.exception.SQLGrammarException: не удалось вставить: [Employee]
Вот подробности:

спящий режим. 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> 

Employee.java

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; 
    } 
} 

ManageEmployee.java

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; 
    } 

} 

Вот полный исключение:

Выход в консоли : Hibernate: insert into EMPLOYEE (first_name, last_name, pay) values ​​(?,?,?) Select scope_identity() org.hibernate.exception.SQLGrammarException: не удалось вставить: [Employee] Hibernate: insert into EMPLOYEE (first_name , last_name, зарплата) (?,?,?) выберите scope_identity() at org.hibernate.exception.SQLStateConverter.convert (SQLStateConverter.java:92) at org.hibernate.exception.JDBCExceptionHelper.convert (JDBCExceptionHelper.java : 66) на org.hibernate.id.insert.AbstractReturningDelegate.performInsert (AbstractReturningDelegate.java:64) на org.hibernate.persister.entity.AbstractEntityPersister.insert (AbstractEntityPersister.java:2327) в org.hibernate.persister .entity.AbstractEntityPersister.insert (AbstractEntityPersiste r.java:2834) at org.hibernate.action.EntityIdentityInsertAction.execute (EntityIdentityInsertAction.java:71) at org.hibernate.engine.ActionQueue.execute (ActionQueue.java:273) at org.hibernate.event. def.AbstractSaveEventListener.performSaveOrReplicate (AbstractSaveEventListener.java:320) на org.hibernate.event.def.AbstractSaveEventListener.performSave (AbstractSaveEventListener.java:203) на org.hibernate.event.def.AbstractSaveEventListener.saveWithGeneratedId (AbstractSaveEventListener.java: 129) на org.hibernate.event.def.DefaultSaveOrUpdateEventListener.saveWithGeneratedOrRequestedId (DefaultSaveOrUpdateEventListener.java:210) на org.hibernate.event.def.DefaultSaveEventListener.saveWithGeneratedOrRequestedId (DefaultSaveEventListener.java:56) на org.hibernate.event.def.DefaultSaveOrUpdateEventListener.entityIsTransient (DefaultSaveOrUpdateEventListener.java:195) в org.hibernate.event.def.DefaultSaveEventListener.performSaveOrUpdate (DefaultSaveEventListener.java:50) в org.hibernate.event.def. DefaultSaveOrUpdateEventListener.onSaveOrUpdate (DefaultSaveOrUpdateEventListener.java:93) на org.hibernate.impl.SessionImpl.fireSave (SessionImpl.java:713) на org.hibernate.impl.SessionImpl.save (SessionImpl.java:701) в орг. hibernate.impl.SessionImpl.save (SessionImpl.java:697) у ManageEmployee.addEmployee (ManageEmployee.java:47) у ManageEmployee.main (ManageEmployee.java:23) Вызвано: java.sql.SQLException: [Microsoft] [ODBC Microsoft Access Driver] Отсутствует точка с запятой (;) в конце инструкции SQL. в sun.jdbc.odbc.JdbcOdbc.createSQLException (Unknown Source) в sun.jdbc.odbc.JdbcOdbc.standardError (Unknown Source) в sun.jdbc.odbc.JdbcOdbc.SQLExecute (Unknown Source) в sun.jdbc .odbc.JdbcOdbcPreparedStatement.execute (Unknown Source) в org.hibernate.id.IdentityGenerator $ InsertSelectDelegate.executeAndExtract (IdentityGenerator.java:138) в org.hibernate.id.insert.AbstractReturningDelegate.performInsert (AbstractReturningDelegate.java:57) ... 17 более org.hibernate.exception.SQLGrammarException: не удалось вставить: [Employee] в org.hibernate.exception.SQLStateConverter.convert (SQLStateConverter.java:92) в org.hibernate.exception.JDBCExceptionHelper. конвертировать (JDBCEx ceptionHelper.java:66) на org.hibernate.id.insert.AbstractReturningDelegate.performInsert (AbstractReturningDelegate.java:64) в org.hibernate.persister.entity.AbstractEntityPersister.insert (AbstractEntityPersister.java:2327) в орг. hibernate.persister.entity.AbstractEntityPersister.insert (AbstractEntityPersister.java:2834) at org.hibernate.action.EntityIdentityInsertAction.execute (EntityIdentityInsertAction.java:71) at org.hibernate.engine.ActionQueue.execute (ActionQueue.java: 273) на org.hibernate.event.def.AbstractSaveEventListener.performSaveOrReplicate (AbstractSaveEventListener.java:320) на org.hibernate.event.def.AbstractSaveEventListener.performSave (AbstractSaveEventListener.java:203) на org.hibernate.event. def.AbstractSa veEventListener.saveWithGeneratedId (AbstractSaveEventListener.java:129) на org.hibernate.event.def.DefaultSaveOrUpdateEventListener.saveWithGeneratedOrRequestedId (DefaultSaveOrUpdateEventListener.java:210) в org.hibernate.event.def.DefaultSaveEventListener.saveWithGeneratedOrRequestedId (DefaultSaveEventListener.java:56) на org.hibernate.event.def.DefaultSaveOrUpdateEventListener.entityIsTransient (DefaultSaveOrUpdateEventListener.java:195) в org.hibernate.event.def.DefaultSaveEventListener.performSaveOrUpdate (DefaultSaveEventListener.java:50) в org.hibernate.event.def. DefaultSaveOrUpdateEventListener.onSaveOrUpdate (DefaultSaveOrUpdateEventListener.java:93) at org.hibernate.impl.SessionImpl.fireSave (SessionImpl.java:713) at org.hibernate.impl.SessionImpl.sav e (SessionImpl.java:701) at org.hibernate.impl.SessionImpl.save (SessionImpl.java:697) у ManageEmployee.addEmployee (ManageEmployee.java:47) у ManageEmployee.main (ManageEmployee.java:24) Вызвано: java.sql.SQLException: [Microsoft] [ODBC Microsoft Access Driver] Отсутствует точка с запятой (;) в конце инструкции SQL. в sun.jdbc.odbc.JdbcOdbc.createSQLException (Unknown Source) в sun.jdbc.odbc.JdbcOdbc.standardError (Unknown Source) в sun.jdbc.odbc.JdbcOdbc.SQLExecute (Unknown Source) в sun.jdbc .odbc.JdbcOdbcPreparedStatement.execute (Unknown Source) в org.hibernate.id.IdentityGenerator $ InsertSelectDelegate.executeAndExtract (IdentityGenerator.java:138) в org.hibernate.id.insert.AbstractReturningDelegate.performInsert (AbstractReturningDelegate.java:57) ... 17 более org.hibernate.exception.SQLGrammarException: не удалось вставить: [Employee] в org.hibernate.exception.SQLStateConverter.convert (SQLStateConverter.java:92) в org.hibernate.exception.JDBCExceptionHelper. конвертировать (JDBCEx ceptionHelper.java:66) на org.hibernate.id.insert.AbstractReturningDelegate.performInsert (AbstractReturningDelegate.java:64) в org.hibernate.persister.entity.AbstractEntityPersister.insert (AbstractEntityPersister.java:2327) в орг. hibernate.persister.entity.AbstractEntityPersister.insert (AbstractEntityPersister.java: 2834) at org.hibernate.action.EntityIdentityInsertAction.execute (EntityIdentityInsertAction.java:71) at org.hibernate.engine.ActionQueue.execute (ActionQueue.java:273) на org.hibernate.event.def. AbstractSaveEventListener.performSaveOrReplicate (AbstractSaveEventListener.java:320) на org.hibernate.event.def.AbstractSaveEventListener.performSave (AbstractSaveEventListener.java:203) в org.hibernate.event.def.AbstractSaveEventListener.saveWithGeneratedId (AbstractSaveEventListener.java:129) на org.hibernate.event.def.DefaultSaveOrUpdateEventListener.saveWithGeneratedOrRequestedId (DefaultSaveOrUpdateEventListener.java:210) на org.hibernate.event.def.DefaultSaveEventListener.saveWithGeneratedOrRequestedId (DefaultSaveEventListener.java:56) на org.hibernate.event.def.DefaultSaveOrUpdateEventListener.entityIsTransient (DefaultSaveOrUpdateEventListener.java:195) в org.hibernate.event.def.DefaultSaveEventListener.performSaveOrUpdate (DefaultSaveEventListener.java:50) в org.hibernate.event.def. DefaultSaveOrUpdateEventListener.onSaveOrUpdate (DefaultSaveOrUpdateEventListener.java:93) на org.hibernate.impl.SessionImpl.fireSave (SessionImpl.java:713) на org.hibernate.impl.SessionImpl.save (SessionImpl.java:701) в орг. hibernate.impl.SessionImpl.save (SessionImpl.java:697) в ManageEmployee.addEmployee (ManageEmployee.java:47) в ManageEmployee.main (ManageEmployee.java:25) Вызванный: java.sql.SQLException: [Microsoft ] [ODBC Microsoft Access Driver] Отсутствует точка с запятой (;) в конце инструкции SQL. в sun.jdbc.odbc.JdbcOdbc.createSQLException (Unknown Source) в sun.jdbc.odbc.JdbcOdbc.standardError (Unknown Source) в sun.jdbc.odbc.JdbcOdbc.SQLExecute (Unknown Source) в sun.jdbc .odbc.JdbcOdbcPreparedStatement.execute (Unknown Source) в org.hibernate.id.IdentityGenerator $ InsertSelectDelegate.executeAndExtract (IdentityGenerator.java:138) в org.hibernate.id.insert.AbstractReturningDelegate.performInsert (AbstractReturningDelegate.java:57) ... 17 более Hibernate: (??,) вставить в РАБОТНИКА (first_name, last_name, заработной платы) значений выберите SCOPE_IDENTITY()

Пожалуйста, помогите устранить исключения. Спасибо заранее!

Answer 1

Диалог MS Access не существует в библиотеке Hibernate. Это приводит к тому, что Hibernate не способен обнаружить столбец и действовать в соответствии с требованиями MS Access.

org.hibernate.exception.SQLGrammarException: не удалось вставить: [Employee] Hibernate: (??,) Вставить в РАБОТНИКА (first_name, last_name, заработной платы) значений выберите SCOPE_IDENTITY()

select scope_identity() - это экземпляр TRANSACT-SQL, не поддерживаемый MS Access.

Вы могли бы сделать один из этих вариантов:

• Не используйте автогенерируемые столбцы идентификаторов или никаких стандартных SQL полей.
• Использовать базу данных спящего режима, поддерживаемую MS Access Hibernate doc.
• Создайте свой собственный диалект спящего режима для поддержки MS Access или используйте thirdparty one, чтобы обеспечить специальные характеристики MS Access.

относящегося к How can I use hibernate with MS Access?