Hibernate - 您的 SQL 语法有错误；检查与您的 MariaDB 服务器版本相对应的手册答案

【问题标题】：Hibernate - You have an error in your SQL syntax; check the manual that corresponds to your MariaDB server versionHibernate - 您的 SQL 语法有错误；检查与您的 MariaDB 服务器版本相对应的手册
【发布时间】：2017-06-30 16:21:09
【问题描述】：

我有三个文件：

消息.java

public class Message {
    @Id
    @GeneratedValue
    private Long id;

    private String text;

    public String getText() {
        return text;
    }

    public void setText(String text) {
       this.text = text;
    }
}

persistence.xml

<?xml version="1.0" encoding="UTF-8"?>

<persistence version="2.1" xmlns="http://xmlns.jcp.org/xml/ns/persistence" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xsi:schemaLocation="http://xmlns.jcp.org/xml/ns/persistence
http://xmlns.jcp.org/xml/ns/persistence_2_1.xsd">
  <persistence-unit name="HelloWorldPU">
    <!-- <jta-data-source>myDB</jta-data-source> -->
    <class>org.jpwh.model.helloworld.Message</class>
    <exclude-unlisted-classes>true</exclude-unlisted-classes>
    <properties>
      <property name="javax.persistence.schema-generation.database.action" value="drop-and-create" />
      <property name="hibernate.format_sql" value="true" />
      <property name="hibernate.use_sql_comments" value="true" />
      <property name="javax.persistence.jdbc.driver" value="com.mysql.jdbc.Driver" />
      <property name="javax.persistence.jdbc.user" value="laki" />
      <property name="javax.persistence.jdbc.password" value="laki" />
      <property name="javax.persistence.jdbc.url" value="jdbc:mysql://localhost:3306/nekretnine?autoreconnect=true" />
      <property name="hibernate.dialect" value="org.hibernate.dialect.HSQLDialect" />
      <property name="hibernate.max_fetch_depth" value="3" />
    </properties>
  </persistence-unit>
</persistence>

Main.java

public class Main {
    public static void main(String[] args) {
        EntityManagerFactory emf = Persistence.createEntityManagerFactory("HelloWorldPU");
       EntityManager em = emf.createEntityManager();
       Message message = new Message();
       message.setText("kurcina");
       em.persist(message);
    }  
}

当我执行主类表时，在 DB 中创建了消息，但我得到了这些异常

WARN: GenerationTarget encountered exception accepting command : Error executing DDL via JDBC Statement
org.hibernate.tool.schema.spi.CommandAcceptanceException: Error executing DDL via JDBC Statement
	at org.hibernate.tool.schema.internal.exec.GenerationTargetToDatabase.accept(GenerationTargetToDatabase.java:67)
	at org.hibernate.tool.schema.internal.SchemaCreatorImpl.applySqlString(SchemaCreatorImpl.java:440)
	at org.hibernate.tool.schema.internal.SchemaCreatorImpl.applySqlStrings(SchemaCreatorImpl.java:424)
	at org.hibernate.tool.schema.internal.SchemaCreatorImpl.createFromMetadata(SchemaCreatorImpl.java:290)
	at org.hibernate.tool.schema.internal.SchemaCreatorImpl.performCreation(SchemaCreatorImpl.java:166)
	at org.hibernate.tool.schema.internal.SchemaCreatorImpl.doCreation(SchemaCreatorImpl.java:135)
	at org.hibernate.tool.schema.internal.SchemaCreatorImpl.doCreation(SchemaCreatorImpl.java:121)
	at org.hibernate.tool.schema.spi.SchemaManagementToolCoordinator.performDatabaseAction(SchemaManagementToolCoordinator.java:155)
	at org.hibernate.tool.schema.spi.SchemaManagementToolCoordinator.process(SchemaManagementToolCoordinator.java:72)
	at org.hibernate.internal.SessionFactoryImpl.<init>(SessionFactoryImpl.java:309)
	at org.hibernate.boot.internal.SessionFactoryBuilderImpl.build(SessionFactoryBuilderImpl.java:452)
	at org.hibernate.jpa.boot.internal.EntityManagerFactoryBuilderImpl.build(EntityManagerFactoryBuilderImpl.java:889)
	at org.hibernate.jpa.HibernatePersistenceProvider.createEntityManagerFactory(HibernatePersistenceProvider.java:58)
	at javax.persistence.Persistence.createEntityManagerFactory(Persistence.java:55)
	at javax.persistence.Persistence.createEntityManagerFactory(Persistence.java:39)
	at tutorial.Main.main(Main.java:19)
Caused by: java.sql.SQLSyntaxErrorException: You have an error in your SQL syntax; check the manual that corresponds to your MariaDB server version for the right syntax to use near 'sequence hibernate_sequence start with 1 increment by 1' at line 1
	at com.mysql.cj.jdbc.exceptions.SQLError.createSQLException(SQLError.java:536)
	at com.mysql.cj.jdbc.exceptions.SQLError.createSQLException(SQLError.java:513)
	at com.mysql.cj.jdbc.exceptions.SQLExceptionsMapping.translateException(SQLExceptionsMapping.java:115)
	at com.mysql.cj.jdbc.ConnectionImpl.execSQL(ConnectionImpl.java:1983)
	at com.mysql.cj.jdbc.ConnectionImpl.execSQL(ConnectionImpl.java:1936)
	at com.mysql.cj.jdbc.StatementImpl.executeInternal(StatementImpl.java:891)
	at com.mysql.cj.jdbc.StatementImpl.execute(StatementImpl.java:795)
	at org.hibernate.tool.schema.internal.exec.GenerationTargetToDatabase.accept(GenerationTargetToDatabase.java:54)
	... 15 more

Jun 30, 2017 5:51:41 PM org.hibernate.tool.schema.internal.SchemaCreatorImpl applyImportSources
INFO: HHH000476: Executing import script 'org.hibernate.tool.schema.internal.exec.ScriptSourceInputNonExistentImpl@4566d049'
Jun 30, 2017 5:51:41 PM org.hibernate.engine.jdbc.spi.SqlExceptionHelper logExceptions
WARN: SQL Error: 1064, SQLState: 42000
Jun 30, 2017 5:51:41 PM org.hibernate.engine.jdbc.spi.SqlExceptionHelper logExceptions
ERROR: You have an error in your SQL syntax; check the manual that corresponds to your MariaDB server version for the right syntax to use near 'value for hibernate_sequence' at line 1
Exception in thread "main" javax.persistence.PersistenceException: org.hibernate.exception.SQLGrammarException: could not extract ResultSet
	at org.hibernate.internal.ExceptionConverterImpl.convert(ExceptionConverterImpl.java:147)
	at org.hibernate.internal.ExceptionConverterImpl.convert(ExceptionConverterImpl.java:155)
	at org.hibernate.internal.ExceptionConverterImpl.convert(ExceptionConverterImpl.java:162)
	at org.hibernate.internal.SessionImpl.firePersist(SessionImpl.java:787)
	at org.hibernate.internal.SessionImpl.persist(SessionImpl.java:765)
	at tutorial.Main.main(Main.java:25)
Caused by: org.hibernate.exception.SQLGrammarException: could not extract ResultSet
	at org.hibernate.exception.internal.SQLExceptionTypeDelegate.convert(SQLExceptionTypeDelegate.java:63)
	at org.hibernate.exception.internal.StandardSQLExceptionConverter.convert(StandardSQLExceptionConverter.java:42)
	at org.hibernate.engine.jdbc.spi.SqlExceptionHelper.convert(SqlExceptionHelper.java:111)
	at org.hibernate.engine.jdbc.spi.SqlExceptionHelper.convert(SqlExceptionHelper.java:97)
	at org.hibernate.engine.jdbc.internal.ResultSetReturnImpl.extract(ResultSetReturnImpl.java:80)
	at org.hibernate.id.enhanced.SequenceStructure$1.getNextValue(SequenceStructure.java:95)
	at org.hibernate.id.enhanced.NoopOptimizer.generate(NoopOptimizer.java:40)
	at org.hibernate.id.enhanced.SequenceStyleGenerator.generate(SequenceStyleGenerator.java:432)
	at org.hibernate.event.internal.AbstractSaveEventListener.saveWithGeneratedId(AbstractSaveEventListener.java:105)
	at org.hibernate.jpa.event.internal.core.JpaPersistEventListener.saveWithGeneratedId(JpaPersistEventListener.java:67)
	at org.hibernate.event.internal.DefaultPersistEventListener.entityIsTransient(DefaultPersistEventListener.java:189)
	at org.hibernate.event.internal.DefaultPersistEventListener.onPersist(DefaultPersistEventListener.java:132)
	at org.hibernate.event.internal.DefaultPersistEventListener.onPersist(DefaultPersistEventListener.java:58)
	at org.hibernate.internal.SessionImpl.firePersist(SessionImpl.java:780)
	... 2 more
Caused by: java.sql.SQLSyntaxErrorException: You have an error in your SQL syntax; check the manual that corresponds to your MariaDB server version for the right syntax to use near 'value for hibernate_sequence' at line 1
	at com.mysql.cj.jdbc.exceptions.SQLError.createSQLException(SQLError.java:536)
	at com.mysql.cj.jdbc.exceptions.SQLError.createSQLException(SQLError.java:513)
	at com.mysql.cj.jdbc.exceptions.SQLExceptionsMapping.translateException(SQLExceptionsMapping.java:115)
	at com.mysql.cj.jdbc.ConnectionImpl.execSQL(ConnectionImpl.java:1983)
	at com.mysql.cj.jdbc.PreparedStatement.executeInternal(PreparedStatement.java:1826)
	at com.mysql.cj.jdbc.PreparedStatement.executeQuery(PreparedStatement.java:1923)
	at org.hibernate.engine.jdbc.internal.ResultSetReturnImpl.extract(ResultSetReturnImpl.java:71)
	... 11 more

有什么想法吗？据我发现，当用户编写 SQL 查询时，会报告 MariaDB 的此错误，但我没有任何查询。

【问题讨论】：

您使用的是 HSQLDB 方言，但您的数据库是 MariaDB/MySQL。使用正确的方言。您告诉 Hibernate 删除并为您创建模式，所以它会这样做。创建模式显然需要执行 SQL 查询。
谢谢，这是正确的。
您需要将此作为答案发布以便我接受吗？
也许是这一行："message.setText("kurcina");" ?

标签： hibernate jpa

【解决方案1】：

您使用的是 HSQLDB 数据库的方言，但您的数据库是 MariaDB/MySQL。

使用正确的方言。

您告诉 Hibernate 删除并为您创建架构，所以它这样做了。创建模式显然需要执行 SQL 查询。

【讨论】：