org.hibernate.exception.GenericJDBCException: could not insert: [db.Ch02]

Question

Hibernate初学者，测试一下mysql数据库操作，这问题解决不了，求助啊，谢谢：
test.java:
package db;import org.hibernate.*;
import org.hibernate.cfg.*;public class test {
/**
 * @param args
 */
public static void main(String[] args) {
try{
SessionFactory sf = new Configuration().configure().buildSessionFactory();
Session session = sf.openSession();
Transaction tx = session.beginTransaction();
Ch02 user = new Ch02(1);
System.out.println(user.getId());
//user.setUserName("fmy");
//user.setPassword("123");
session.save(user);
tx.commit();
session.close();
} catch (HibernateException e){
e.printStackTrace();
}
}}Ch02.java:
package db;/**
 * Ch02 entity. @author MyEclipse Persistence Tools
 */public class Ch02 implements java.io.Serializable { // Fields private int id;
private String userName;
private String password;
private String email; // Constructors /** default constructor */
public Ch02() {
} /** minimal constructor */
public Ch02(int id) {
this.id = id;
} /** full constructor */
public Ch02(int id, String userName, String password, String email) {
this.id = id;
this.userName = userName;
this.password = password;
this.email = email;
} // Property accessors public int getId() {
return this.id;
} public void setId(int id) {
this.id = id;
} public String getUserName() {
return this.userName;
} public void setUserName(String userName) {
this.userName = userName;
} public String getPassword() {
return this.password;
} public void setPassword(String password) {
this.password = password;
} public String getEmail() {
return this.email;
} public void setEmail(String email) {
this.email = email;
}}Ch02.hbm.xml:
<?xml version="1.0" encoding="utf-8"?>
<!DOCTYPE hibernate-mapping PUBLIC "-//Hibernate/Hibernate Mapping DTD 3.0//EN"
                                   "http://hibernate.sourceforge.net/hibernate-mapping-3.0.dtd">
<!-- 
    Mapping file autogenerated by MyEclipse Persistence Tools
-->
<hibernate-mapping>
 <class catalog="ch02" name="db.Ch02" table="ch02">
  <id name="id" type="java.lang.Integer">
   <column name="ID" not-null="true"/>
   <generator class="identity"/>
  </id>
  <property generated="never" lazy="false" name="userName" type="java.lang.String">
   <column length="16" name="UserName"/>
  </property>
  <property generated="never" lazy="false" name="password" type="java.lang.String">
   <column length="16" name="Password"/>
  </property>
  <property generated="never" lazy="false" name="email" type="java.lang.String">
   <column length="32" name="Email"/>
  </property>
 </class>
</hibernate-mapping>hibernate.cfg.xml:
<?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"><!-- Generated by MyEclipse Hibernate Tools.                   -->
<hibernate-configuration><session-factory>
<property name="show_sql">true</property>
<property name="myeclipse.connection.profile">ch02</property>
<property name="connection.url">
jdbc:mysql://localhost:3306/test
</property>
<property name="connection.username">root</property>
<property name="connection.password">mysql</property>
<property name="connection.driver_class">
com.mysql.jdbc.Driver
</property>
<property name="dialect">
org.hibernate.dialect.MySQLDialect
</property>
<mapping resource="db/Ch02.hbm.xml" /></session-factory></hibernate-configuration>异常信息:
Hibernate: insert into ch02.ch02 (UserName, Password, Email) values (?, ?, ?)
org.hibernate.exception.GenericJDBCException: could not insert: [db.Ch02]
at org.hibernate.exception.SQLStateConverter.handledNonSpecificException(SQLStateConverter.java:126)
at org.hibernate.exception.SQLStateConverter.convert(SQLStateConverter.java:114)
at org.hibernate.exception.JDBCExceptionHelper.convert(JDBCExceptionHelper.java:66)
at org.hibernate.id.insert.AbstractReturningDelegate.performInsert(AbstractReturningDelegate.java:64)
at org.hibernate.persister.entity.AbstractEntityPersister.insert(AbstractEntityPersister.java:2176)
at org.hibernate.persister.entity.AbstractEntityPersister.insert(AbstractEntityPersister.java:2656)
at org.hibernate.action.EntityIdentityInsertAction.execute(EntityIdentityInsertAction.java:71)
at org.hibernate.engine.ActionQueue.execute(ActionQueue.java:279)
at org.hibernate.event.def.AbstractSaveEventListener.performSaveOrReplicate(AbstractSaveEventListener.java:321)
at org.hibernate.event.def.AbstractSaveEventListener.performSave(AbstractSaveEventListener.java:204)
at org.hibernate.event.def.AbstractSaveEventListener.saveWithGeneratedId(AbstractSaveEventListener.java:130)
at org.hibernate.event.def.DefaultSaveOrUpdateEventListener.saveWithGeneratedOrRequestedId(DefaultSaveOrUpdateEventListener.java:210)
at org.hibernate.event.def.DefaultSaveEventListener.saveWithGeneratedOrRequestedId(DefaultSaveEventListener.java:56)
at org.hibernate.event.def.DefaultSaveOrUpdateEventListener.entityIsTransient(DefaultSaveOrUpdateEventListener.java:195)
at org.hibernate.event.def.DefaultSaveEventListener.performSaveOrUpdate(DefaultSaveEventListener.java:50)
at org.hibernate.event.def.DefaultSaveOrUpdateEventListener.onSaveOrUpdate(DefaultSaveOrUpdateEventListener.java:93)
at org.hibernate.impl.SessionImpl.fireSave(SessionImpl.java:563)
at org.hibernate.impl.SessionImpl.save(SessionImpl.java:551)
at org.hibernate.impl.SessionImpl.save(SessionImpl.java:547)
at db.test.main(test.java:22)
Caused by: java.sql.SQLException: Field 'ID' doesn't have a default value
at com.mysql.jdbc.SQLError.createSQLException(SQLError.java:1073)
at com.mysql.jdbc.MysqlIO.checkErrorPacket(MysqlIO.java:3609)
at com.mysql.jdbc.MysqlIO.checkErrorPacket(MysqlIO.java:3541)
at com.mysql.jdbc.MysqlIO.sendCommand(MysqlIO.java:2002)
at com.mysql.jdbc.MysqlIO.sqlQueryDirect(MysqlIO.java:2163)
at com.mysql.jdbc.ConnectionImpl.execSQL(ConnectionImpl.java:2624)
at com.mysql.jdbc.PreparedStatement.executeInternal(PreparedStatement.java:2127)
at com.mysql.jdbc.PreparedStatement.executeUpdate(PreparedStatement.java:2427)
at com.mysql.jdbc.PreparedStatement.executeUpdate(PreparedStatement.java:2345)
at com.mysql.jdbc.PreparedStatement.executeUpdate(PreparedStatement.java:2330)
at org.hibernate.id.IdentityGenerator$GetGeneratedKeysDelegate.executeAndExtract(IdentityGenerator.java:94)
at org.hibernate.id.insert.AbstractReturningDelegate.performInsert(AbstractReturningDelegate.java:57)
... 16 more求大神，多谢啊

Answer 1

Caused by: java.sql.SQLException: Field 'ID' doesn't have a default value
这句话的意思是id不是自动增长的

Answer 2

你不要用Ch02 user = new Ch02(1);初始化，因为这样的话，就相当于你给表ID已经赋值了，而你的表又设置的是自增。因此你用Ch02 user = new Ch02();就没有问题了.

Answer 3

还有就是mysql数据库的自增是increment而不是identity(sql server数据库),你可以用native让它自动去找。所以首先把id修改了运行再看一下，如果不行，再用我说的上面的方法。