使用MariaDB 10.4和eclipselink的java@GeneratedValue(策略=Generation

Question

我正在Java EE中开发一个REST web服务，我正在使用：Glassfish 5.0(build 25(、MariaDB 10.4和eclipselink(JPA 2.1(这是我的代码：

命令行表

CREATE TABLE IF NOT EXISTS `cooldb`.`commande_line` (
`id` INT NOT NULL AUTO_INCREMENT,
`quantity` INT NULL,
`discount` INT NULL,
`dish` INT NOT NULL,
`commande` INT NOT NULL,
PRIMARY KEY (`id`),
UNIQUE INDEX `id_UNIQUE` (`id` ASC),
INDEX `fk_commande_line_dish1_idx` (`dish` ASC),
INDEX `fk_commande_line_commande1_idx` (`commande` ASC),
CONSTRAINT `fk_commande_line_dish1`
FOREIGN KEY (`dish`)
REFERENCES `cooldb`.`dish` (`id`)
ON DELETE CASCADE
ON UPDATE CASCADE,
CONSTRAINT `fk_commande_line_commande1`
FOREIGN KEY (`commande`)
REFERENCES `cooldb`.`commande` (`id`)
ON DELETE CASCADE
ON UPDATE CASCADE)
ENGINE = InnoDB
DEFAULT CHARACTER SET = utf8;

persistence.xml

<?xml version="1.0" encoding="UTF-8"?>
<persistence version="2.2" xmlns="http://java.sun.com/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/persistence_2_2.xsd">
<!-- Define Persistence Unit -->
<persistence-unit name="my_persistence_unit" transaction-type="JTA">
<jta-data-source>jdbc/mariadb</jta-data-source>
<class>com.yac.model.Address</class>
<class>com.yac.model.Commande</class>
<class>com.yac.model.CommandeLine</class>
<class>com.yac.model.Dish</class>
<class>com.yac.model.Dishtype</class>
<class>com.yac.model.Ingredient</class>
<class>com.yac.model.Payement</class>
<class>com.yac.model.Profil</class>
<class>com.yac.model.Restaurant</class>
<class>com.yac.model.Userapp</class>
<exclude-unlisted-classes>true</exclude-unlisted-classes>
<properties>
</properties>
</persistence-unit>
</persistence>

命令行实体

public class CommandeLine implements Serializable {
private static final long serialVersionUID = 1L;
@Id
@GeneratedValue(strategy = GenerationType.IDENTITY)
@Basic(optional = false)
@Column(name = "id")
private Integer id;
@Column(name = "quantity")
private Integer quantity;
@Column(name = "discount")
private Integer discount;
@JoinColumn(name = "commande", referencedColumnName = "id")
@ManyToOne(optional = false)
private Commande commande;
@JoinColumn(name = "dish", referencedColumnName = "id")
@ManyToOne(optional = false)
private Dish dish;
//Constructor
// Setter and Getter
}

命令行web服务

@Stateless
@Path("commandeline")
public class CommandeLineFacadeREST extends AbstractFacade<CommandeLine> {
@PersistenceContext(unitName = "my_persistence_unit")
private EntityManager em;
public CommandeLineFacadeREST() {
super(CommandeLine.class);
}
@POST
@Override
@Consumes(MediaType.APPLICATION_JSON)
public void create(CommandeLine entity) {
super.create(entity);
}
@Override
protected EntityManager getEntityManager() {
return em;
}
}

抽象立面

public abstract class AbstractFacade<T> {
private Class<T> entityClass;
public AbstractFacade(Class<T> entityClass) {
this.entityClass = entityClass;
}
protected abstract EntityManager getEntityManager();
public void create(T entity) {
getEntityManager().persist(entity);
}
}

问题是当我用Postman测试我的web服务时，我试图插入一个带有POST请求的记录这是我收到的错误消息：

Local Exception Stack: 
Exception [EclipseLink-4002] (Eclipse Persistence Services - 2.7.0.v20170811-d680af5): org.eclipse.persistence.exceptions.DatabaseException
Internal Exception: java.sql.SQLSyntaxErrorException: (conn=158) Table 'cooldb.sequence' doesn't exist
Error Code: 1146
Call: UPDATE SEQUENCE SET SEQ_COUNT = SEQ_COUNT + ? WHERE SEQ_NAME = ?
bind => [2 parameters bound]
Query: DataModifyQuery(name="SEQ_GEN_SEQUENCE" sql="UPDATE SEQUENCE SET SEQ_COUNT = SEQ_COUNT + ? WHERE SEQ_NAME = ?") ...

我不明白为什么当我使用@GeneratedValue(策略=GenerationType.IDENTITY(时，SEQUANCE会出现问题。当我使用@GeneratedValue(策略=GenerationType.SSEQUENCE(进行更改时，我使用以下脚本创建表：

CREATE SEQUENCE SEQUANCE START WITH 1 INCREMENT BY 1;

通过应用表'；customerjpa.sequence'；不会'；不存在JPA但也有同样的问题

提前感谢你的帮助。

Answer 1

使用Chris注释解决了问题，我只在persistence.xml文件中添加了以下行：

<property name="eclipselink.target-database" value="MySQL"/>

非常感谢Chris。所以我的新persistence.xml文件是：

<?xml version="1.0" encoding="UTF-8"?>
<persistence version="2.2" xmlns="http://java.sun.com/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/persistence_2_2.xsd">
<!-- Define Persistence Unit -->
<persistence-unit name="my_persistence_unit" transaction-type="JTA">
<jta-data-source>jdbc/mariadb</jta-data-source>
<class>com.yac.model.Address</class>
<class>com.yac.model.Commande</class>
<class>com.yac.model.CommandeLine</class>
<class>com.yac.model.Dish</class>
<class>com.yac.model.Dishtype</class>
<class>com.yac.model.Ingredient</class>
<class>com.yac.model.Payement</class>
<class>com.yac.model.Profil</class>
<class>com.yac.model.Restaurant</class>
<class>com.yac.model.Userapp</class>
<exclude-unlisted-classes>true</exclude-unlisted-classes>
<properties>
<property name="eclipselink.target-database" value="MySQL"/>
</properties>
</persistence-unit>
</persistence>

我只是在MariaDB基于MySQL的persistence.xml文件中指定了MySQL中的数据库平台，因为列表中没有提到MariaDB。如果还有其他建议，请不要犹豫，谢谢。

Answer 2

另一个解决方案：将?useMysqlMetadata=true添加到JDBC URL连接中，如下所示：

<property name="URL" value="jdbc:mariadb://[HOST]:[PORT]/[DB NAME]?useMysqlMetadata=true"/>

这将使MariaDB使用MySQL元数据，然后eclipselink将其检测为MySQL。