使用GenerationType.IDENTITY 10.4和eclipselink的java @GeneratedValue问题(策略= GenerationType.IDENTITY)

Question

我正在Java中开发一个REST web服务，我使用的是: Glassfish 5.0 (build 25)、MariaDB 10.4和eclipselink (JPA2.1)--这里是我的代码：

commande_line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;

persistance.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;
    }
}

AbstractFacade

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，我尝试用POST请求插入一条记录，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 (strategy = GenerationType.IDENTITY)时，SEQUANCE会出现问题。当我使用@GeneratedValue (strategy = GenerationType.SEQUENCE)进行更改并使用以下脚本创建表时：

CREATE SEQUENCE SEQUANCE START WITH 1 INCREMENT BY 1;

通过应用在：Table 'customerjpa.sequence' doesn't exist JPA中显示的解决方案，但使用相同的问题

提前谢谢你的帮助。

Answer 1

使用克里斯注释解决了这个问题，我只是在我的persistence.xml文件中添加了以下一行：

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

非常感谢克里斯。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>

我刚才在MySQL中的persistence.xml文件中指定了数据库平台，此时MariaDB是基于它的，因为列表中没有提到MariaDB。如果有其他建议，请不要犹豫，谢谢。

Answer 2

另一种解决方案:将?useMysqlMetadata=true添加到JDBC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将将其检测为MySQL。