尝试学习Hibernate，给出错误的sql语法错误

Question

我正在练习JPA/Hibernate，在运行下面的代码时:

package com.newjpa;

import javax.persistence.*;

public class App 
{
    public static void main( String[] args ) {
        EntityManagerFactory emf = Persistence.createEntityManagerFactory("book");
        EntityManager em = emf.createEntityManager();
        EntityTransaction et = em.getTransaction();
        et.begin();
        Book book = new Book();
        book.setAuthor("1423414");
        book.setTitle("#143441");
        book.setRating(4);
        book.setYear(2093);
        book.setDescr("13471863146486348626424372");
        em.merge(book);
        et.commit();
        em.close();
        emf.close();

    }
}

和Book类(没有getters和setters):

package com.newjpa;

import javax.persistence.*;

@Entity
@Table(name = "Books")
public class Book {

    @Id
    @GeneratedValue(strategy = GenerationType.AUTO)
    private long bookId;
    private String author;
    private String title;
    private int year;
    private int rating;
    private String descr;


    public Book(String author, String title, int year, int rating, String descr) {
        this.author = author;
        this.title = title;
        this.year = year;
        this.rating = rating;
        this.descr = descr;
    }

    public Book() {
    }
}

我收到以下错误：

You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near 'value for hibernate_sequence' at line 1

我真的不知道这意味着什么以及如何解决它，我花了几个小时搜索，找不到类似的东西。persistence.xml代码：

<?xml version="1.0" encoding="UTF-8"?>
<persistence xmlns="http://java.sun.com/xml/ns/persistence" version="1.0">
<persistence-unit name="book" transaction-type="RESOURCE_LOCAL">
 <provider>org.hibernate.jpa.HibernatePersistenceProvider
 </provider>
    <class>com.newjpa.Book</class>
    <properties>
        <property name="javax.persistence.jdbc.url  " 
 value="jdbc:mysql://www.remotemysql.com:3306/placeholder" />
        <property name="javax.persistence.jdbc.driver" 
 value="com.mysql.jdbc.Driver" />
        <property name="javax.persistence.jdbc.user" 
 value="placeholder" />
        <property name="javax.persistence.jdbc.password" 
 value="placeholder" />

        <property name="hibernate.dialect" 
 value="org.hibernate.dialect.H2Dialect" />
        <property name="hibernate.show_sql" value="true" />
        <property name="hibernate.format_sql" value="true" />
        <property name="hibernate.hbm2ddl.auto" value="create" />

    </properties>
</persistence-unit>

</persistence>

我在这里真的找不到任何错误——我在互联网上的许多地方看到了非常相似的代码。正如另一位用户所指出的，这是Hibernate生成的sql语句：创建表Books（book Id bigint not null，作者varchar（255），desr varchar（255），评级整数不为空，标题varchar（255），年份整数不为空，主键（book Id））值得指出的是，表是创建的，问题是异常，以及我创建的实体没有插入到表中的事实。重要更新：我发现Hibernate在这一行之前也生成了这一行：

Hibernate: create sequence hibernate_sequence start with 1 increment by 1

在创建表之后：

Hibernate: 
    call next value for hibernate_sequence

这似乎与问题有关

Answer 1

结果发现我有两个问题:我有错误的方言集(org . hibernate . dialect . H2 dialect而不是org . hibernate . dialect . MySQLDialect ),结果发现当使用MySQL 8时，您必须将方言从“MySQL dialect”更改为“MySQL8Dialect”。