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用JPA为啥总创建不了mysql数据库的表

发布时间： 2012-12-26 14:39:28 作者： rapoo

用JPA为什么总创建不了mysql数据库的表


package lee;

import java.util.List;
import javax.persistence.*;

import org.crazyit.model.*;


public class JpaQs
{
//一个持久单元对应一个EntityManagerFactory
private static final EntityManagerFactory emf = 
Persistence.createEntityManagerFactory("qs");

public static void main(String[] args)
{
final EntityManager em = emf.createEntityManager();
News news = new News();
news.setTitle("asd");
news.setContent("zxc");
try
{
//开启事务
em.getTransaction().begin();
//保存实体
em.persist(news);
//提交事务
em.getTransaction().commit();
} 
finally 
{
em.close();
}
}
}


package org.crazyit.model;

import javax.persistence.*;

@Entity
@Table(name="hello")
public class News
{
//消息类的标识属性
@Id /* 用于修饰标识属性 */
/* 指定该主键列的主键生成策略 */
@GeneratedValue(strategy=GenerationType.IDENTITY)
private int id;
//消息标题
/* @Column指定该Field映射的列信息，此处指定了列名、长度 */
@Column(name="name" , length=50)
private String title;
//消息内容
/* @Column指定该Field映射的列信息，此处指定允许为null */
@Column(nullable=true)
private String content;
//构造器
public News()
{
}
//标识属性的setter和getter方法
public void setId(int id) 
{
this.id = id; 
}
public int getId()
{
return (this.id); 
}
//消息标题的setter方法和getter方法
public void setTitle(String title) 
{
this.title = title; 
}
public String getTitle() 
{
return (this.title); 
}
//消息内容的setter方法和getter方法
public void setContent(String content)
{
this.content = content; 
}
public String getContent()
{
return (this.content); 
}
}

<?xml version="1.0" encoding="GBK"?>
<persistence version="1.0" xmlns="http://java.sun.com/xml/ns/persistence"
xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
xsi:schemaLocation="http://java.sun.com/xml/ns/persistence 
http://java.sun.com/xml/ns/persistence/persistence_1_0.xsd">
<!-- 为持久化单元指定名称，并通过transaction-type指定事务类型
transaction-type属性合法的属性值有JTA、RESOURCE_LOCAL两个-->
<persistence-unit name="qs" transaction-type="RESOURCE_LOCAL">
<!-- 指定javax.persistence.spi.PersistenceProvider实现类 -->
<provider>org.hibernate.ejb.HibernatePersistence</provider>
<!-- 列出该应用需要访问的所有的Entity类, 
也可以用 <mapping-file> 或<jar-file>元素来定义 -->
<class>org.crazyit.model.News</class>
<!-- properties元素用于为特定JPA实现包配置属性 -->  
<!-- 下面列举的是Hibernate JPA实现中可以配置的部分属性 -->
<properties>  
<!-- 指定连接数据库的驱动名 -->
<property name="hibernate.connection.driver_class"
value="com.mysql.jdbc.Driver"/>
<!-- 指定连接数据库的URL -->
<property name="hibernate.connection.url"
value="jdbc:mysql://localhost:3306/test"/>
<!-- 指定连接数据库的用户名 -->
<property name="hibernate.connection.username"
value="root"/>
<!-- 指定连接数据库的密码 --> 
 
<property name="hibernate.connection.password"
value="root"/>
<!-- 指定连接数据库的方言 -->
<property name="hibernate.dialect" 
value="org.hibernate.dialect.MySQLInnoDBDialect"/>
<property name="hibernate.show_sql" value="true"/>
<!-- 设置是否格式化SQL语句 -->
<property name="hibernate.format_sql"
value="true"/>
<!-- 设置是否根据要求自动建表 -->
<property name="hibernate.hbm2ddl.auto"
value="update"/>
</properties>
</persistence-unit>
</persistence>



驱动什么的都添加了 上面test是我数据库的名字 那hello是我想自动创建的表
一运行就出现下面这种错误


Hibernate: 
    insert 
    into
        hello
        (content, name) 
    values
        (?, ?)
Exception in thread "main" javax.persistence.PersistenceException: org.hibernate.exception.SQLGrammarException: could not insert: [org.crazyit.model.News]
at org.hibernate.ejb.AbstractEntityManagerImpl.convert(AbstractEntityManagerImpl.java:1235)
at org.hibernate.ejb.AbstractEntityManagerImpl.convert(AbstractEntityManagerImpl.java:1168)
at org.hibernate.ejb.AbstractEntityManagerImpl.convert(AbstractEntityManagerImpl.java:1174)
at org.hibernate.ejb.AbstractEntityManagerImpl.persist(AbstractEntityManagerImpl.java:674)
at lee.JpaQs.main(JpaQs.java:35)
Caused by: org.hibernate.exception.SQLGrammarException: could not insert: [org.crazyit.model.News]
at org.hibernate.exception.SQLStateConverter.convert(SQLStateConverter.java:92)
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:2329)
at org.hibernate.persister.entity.AbstractEntityPersister.insert(AbstractEntityPersister.java:2836)
at org.hibernate.action.EntityIdentityInsertAction.execute(EntityIdentityInsertAction.java:71)
at org.hibernate.engine.ActionQueue.execute(ActionQueue.java:268)
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.ejb.event.EJB3PersistEventListener.saveWithGeneratedId(EJB3PersistEventListener.java:69)
at org.hibernate.event.def.DefaultPersistEventListener.entityIsTransient(DefaultPersistEventListener.java:179)
at org.hibernate.event.def.DefaultPersistEventListener.onPersist(DefaultPersistEventListener.java:135)
at org.hibernate.event.def.DefaultPersistEventListener.onPersist(DefaultPersistEventListener.java:61)
at org.hibernate.impl.SessionImpl.firePersist(SessionImpl.java:800)
at org.hibernate.impl.SessionImpl.persist(SessionImpl.java:774)
at org.hibernate.impl.SessionImpl.persist(SessionImpl.java:778)
at org.hibernate.ejb.AbstractEntityManagerImpl.persist(AbstractEntityManagerImpl.java:668)
... 1 more
Caused by: java.sql.SQLException: Table 'test.hello' doesn't exist
at com.mysql.jdbc.MysqlIO.checkErrorPacket(MysqlIO.java:2921)
at com.mysql.jdbc.MysqlIO.sendCommand(MysqlIO.java:1570) 
 
at com.mysql.jdbc.MysqlIO.sqlQueryDirect(MysqlIO.java:1665)
at com.mysql.jdbc.Connection.execSQL(Connection.java:2978)
at com.mysql.jdbc.Connection.execSQL(Connection.java:2902)
at com.mysql.jdbc.PreparedStatement.executeInternal(PreparedStatement.java:930)
at com.mysql.jdbc.PreparedStatement.executeUpdate(PreparedStatement.java:1159)
at com.mysql.jdbc.PreparedStatement.executeUpdate(PreparedStatement.java:1076)
at com.mysql.jdbc.PreparedStatement.executeUpdate(PreparedStatement.java:1061)
at org.hibernate.id.IdentityGenerator$GetGeneratedKeysDelegate.executeAndExtract(IdentityGenerator.java:94)
at org.hibernate.id.insert.AbstractReturningDelegate.performInsert(AbstractReturningDelegate.java:57)
... 16 more
Java Result: 1

哪位好心人帮忙给看看？？？？
[解决办法]
属性名content，是部分数据库的关键字，修改属性名吧
[解决办法]
引用:
属性名content，是部分数据库的关键字，修改属性名吧


是我上面定义的那个content么 我试了试改成别的也不行 错误还是那样
[解决办法]
引用:
属性名content，是部分数据库的关键字，修改属性名吧


好吧 我用的mysql 5.5 我换成5.0就OK了
[解决办法]
很诡异的问题啊

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