java - Hibernate MySQL 无法创建表

Question

我正在学习休眠，我一开始就卡住了。所以问题是我的应用程序无法自动创建表。这是所有代码：

休眠.cfg.xml

<?xml version='1.0' encoding='utf-8'?>
<!DOCTYPE hibernate-configuration PUBLIC
"-//Hibernate/Hibernate Configuration DTD//EN"
"http://hibernate.sourceforge.net/hibernate-configuration-3.0.dtd">

<hibernate-configuration>
    <session-factory>
        <property name="hibernate.connection.driver_class">com.mysql.jdbc.Driver</property>
        <property name="hibernate.connection.url">jdbc:mysql://localhost:3306/hibernatedb</property>
        <property name="hibernate.connection.username">root</property>
        <property name="hibernate.connection.password">password</property>
        <property name="hibernate.connection.pool_size">1</property>
        <property name="hibernate.hbm2dll.auto">create</property>   
        <property name="hibernate.show_sql">true</property>
        <property name="dialect">org.hibernate.dialect.MySQLDialect</property>
        <mapping class="azaro.test.hibernate.UserDetails" />    
    </session-factory>
</hibernate-configuration>

HibernateTest.java

package azaro.test.hibernate;

import org.hibernate.Session;
import org.hibernate.SessionFactory;
import org.hibernate.cfg.Configuration;
import org.hibernate.service.ServiceRegistry;
import org.hibernate.service.ServiceRegistryBuilder;

public class HibernateTest {

    public static void main(String[] args) {
        UserDetails user = new UserDetails();
        user.setUserId(1);
        user.setUserName("First User");

        Configuration configuration = new Configuration();
        configuration.configure();
        ServiceRegistry serviceRegistry = new ServiceRegistryBuilder().applySettings(configuration.getProperties()).buildServiceRegistry();        
        SessionFactory sessionFactory = configuration.buildSessionFactory(serviceRegistry);
        Session session = sessionFactory.openSession();
        session.beginTransaction();
        session.save(user);
        session.getTransaction().commit();
    }

}

用户详细信息.java

package azaro.test.hibernate;

import javax.persistence.Entity;
import javax.persistence.Id;

@Entity
public class UserDetails {
    @Id
    private int userId;
    private String userName;

    public int getUserId() {
        return userId;
    }

    public void setUserId(int userId) {
        this.userId = userId;
    }

    public String getUserName() {
        return userName;
    }

    public void setUserName(String userName) {
        this.userName = userName;
    }

}

和控制台输出：

Hibernate: insert into UserDetails (userName, userId) values (?, ?)
Sep 23, 2013 6:34:23 PM org.hibernate.engine.jdbc.spi.SqlExceptionHelper logExceptions
WARN: SQL Error: 1146, SQLState: 42S02
Sep 23, 2013 6:34:23 PM org.hibernate.engine.jdbc.spi.SqlExceptionHelper logExceptions
ERROR: Table 'hibernatedb.userdetails' doesn't exist
Exception in thread "main" org.hibernate.exception.SQLGrammarException: could not execute statement
    at org.hibernate.exception.internal.SQLExceptionTypeDelegate.convert(SQLExceptionTypeDelegate.java:82)
    at org.hibernate.exception.internal.StandardSQLExceptionConverter.convert(StandardSQLExceptionConverter.java:49)
    at org.hibernate.engine.jdbc.spi.SqlExceptionHelper.convert(SqlExceptionHelper.java:125)
    at org.hibernate.engine.jdbc.spi.SqlExceptionHelper.convert(SqlExceptionHelper.java:110)
    at org.hibernate.engine.jdbc.internal.ResultSetReturnImpl.executeUpdate(ResultSetReturnImpl.java:136)
    at org.hibernate.engine.jdbc.batch.internal.NonBatchingBatch.addToBatch(NonBatchingBatch.java:58)
    at org.hibernate.persister.entity.AbstractEntityPersister.insert(AbstractEntityPersister.java:3067)
    at org.hibernate.persister.entity.AbstractEntityPersister.insert(AbstractEntityPersister.java:3509)
    at org.hibernate.action.internal.EntityInsertAction.execute(EntityInsertAction.java:88)
    at org.hibernate.engine.spi.ActionQueue.execute(ActionQueue.java:377)
    at org.hibernate.engine.spi.ActionQueue.executeActions(ActionQueue.java:369)
    at org.hibernate.engine.spi.ActionQueue.executeActions(ActionQueue.java:286)
    at org.hibernate.event.internal.AbstractFlushingEventListener.performExecutions(AbstractFlushingEventListener.java:339)
    at org.hibernate.event.internal.DefaultFlushEventListener.onFlush(DefaultFlushEventListener.java:52)
    at org.hibernate.internal.SessionImpl.flush(SessionImpl.java:1234)
    at org.hibernate.internal.SessionImpl.managedFlush(SessionImpl.java:404)
    at org.hibernate.engine.transaction.internal.jdbc.JdbcTransaction.beforeTransactionCommit(JdbcTransaction.java:101)
    at org.hibernate.engine.transaction.spi.AbstractTransactionImpl.commit(AbstractTransactionImpl.java:175)
    at azaro.test.hibernate.HibernateTest.main(HibernateTest.java:23)
Caused by: com.mysql.jdbc.exceptions.jdbc4.MySQLSyntaxErrorException: Table 'hibernatedb.userdetails' doesn't exist
    at sun.reflect.NativeConstructorAccessorImpl.newInstance0(Native Method)
    at sun.reflect.NativeConstructorAccessorImpl.newInstance(Unknown Source)
    at sun.reflect.DelegatingConstructorAccessorImpl.newInstance(Unknown Source)
    at java.lang.reflect.Constructor.newInstance(Unknown Source)
    at com.mysql.jdbc.Util.handleNewInstance(Util.java:411)
    at com.mysql.jdbc.Util.getInstance(Util.java:386)
    at com.mysql.jdbc.SQLError.createSQLException(SQLError.java:1054)
    at com.mysql.jdbc.MysqlIO.checkErrorPacket(MysqlIO.java:4190)
    at com.mysql.jdbc.MysqlIO.checkErrorPacket(MysqlIO.java:4122)
    at com.mysql.jdbc.MysqlIO.sendCommand(MysqlIO.java:2570)
    at com.mysql.jdbc.MysqlIO.sqlQueryDirect(MysqlIO.java:2731)
    at com.mysql.jdbc.ConnectionImpl.execSQL(ConnectionImpl.java:2818)
    at com.mysql.jdbc.PreparedStatement.executeInternal(PreparedStatement.java:2157)
    at com.mysql.jdbc.PreparedStatement.executeUpdate(PreparedStatement.java:2460)
    at com.mysql.jdbc.PreparedStatement.executeUpdate(PreparedStatement.java:2377)
    at com.mysql.jdbc.PreparedStatement.executeUpdate(PreparedStatement.java:2361)
    at org.hibernate.engine.jdbc.internal.ResultSetReturnImpl.executeUpdate(ResultSetReturnImpl.java:133)
    ... 14 more

如果您需要更多信息，请随时询问。

score 1 · Accepted Answer

检查使用生成的 ddl

SchemaExport(cfg).create(true, true);

使用它，您将了解 ddl 并可以更好地分析。此外，如果架构已经更新。将 hbm2ddl.auto 设置为“create-drop”

score 1 · Accepted Answer

您的文件中有错误cfg.xml。将hbm2 dll更改为hbm2 ddl

"hibernate.hbm2dll.auto">create

那么生活又会变得美好。我知道这一点，因为我犯了同样的错误并花了几个小时才弄清楚。

score 0 · Accepted Answer

0

添加配置文件（hibernate.cfg.xml）更新并使用@Column注释在数据库中创建列

于 2014-03-14T06:59:53.960 回答

score 0 · Accepted Answer

“SQLGrammarException”异常意味着在配置方言时编写了错误的语法。

Sep 23, 2013 6:34:23 PM org.hibernate.engine.jdbc.spi.SqlExceptionHelper logExceptions
ERROR: Table 'hibernatedb.userdetails' doesn't exist
Exception in thread "main" org.hibernate.exception.SQLGrammarException: could not execute statement
~~~~~~~~~~~~

Answer:

- The correct dialect entry is: [<property name="hibernate.dialect">]
  ~~~~~~~~~~
   <property name="hibernate.dialect">org.hibernate.dialect.MySQLDialect</property> 
  ~~~~~~~~~

- The given is found from your XML file. [<property name="dialect">]
  ~~~~~~~~~
        <property name="dialect">org.hibernate.dialect.MySQLDialect</property>
        <mapping class="azaro.test.hibernate.UserDetails" />    
    </session-factory>
    </hibernate-configuration>
  ~~~~~~~~~~~

Root Cause: The hibernate property should start with the 'hibernate.' prefix!

In this case, your code contains the property with the "dialect" prefix and this is wrong. Replace it with the "hibernate.dialect".


- Sarfaraz

score -1 · Accepted Answer

请尝试org.hibernate.dialect.MySQL5InnoDBDialect代替org.hibernate.dialect.MySQLDialect

MySQL5Dialact 指的是 MyISAM 存储类型，而 MySQL5InnoDBDialect 指的是 InnoDB（较新）类型。

score -2 · Accepted Answer

不要初始化用户为null，并将hbm2ddl配置为create-drop，代码将执行结果如下：

休眠：插入 USER_DETAILS（地址、描述、joinedDate、userName、userId）值（？，？，？，？，？）休眠：选择 userdetail0_.userId 作为 userId1_0_0_，userdetail0_.Address 作为 Address2_0_0_，userdetail0_.description 作为 descript3_0_0_，userdetail0_ .joinedDate as joinedDa4_0_0_, userdetail0_.userName as userName5_0_0_ from USER_DETAILS userdetail0_ where userdetail0_.userId=?

检索到的用户名是第一个用户

java - Hibernate MySQL 无法创建表

6 回答 6

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