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我正在使用 Java EE 开发 REST Web 服务：Glassfish 5.0 (build 25)、MariaDB 10.4 和 eclipselink (JPA 2.1) 这是我的代码：

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;

持久性.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
}

命令行网络服务

@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);
    }
}

问题是当我使用 Postman 测试我的 Web 服务并尝试使用 POST 请求插入记录时，我收到的错误消息是：

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 = ?") ...

当我使用@GeneratedValue (strategy = GenerationType.IDENTITY) 时，我不明白为什么 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;

通过应用中所示的解决方案：表'customerjpa.sequence'不存在JPA 但同样的问题

预先感谢您的帮助。

4

2 回答 2

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使用 Chris 注释解决了这个问题，我只需在我的 persistence.xml 文件中添加以下行：

<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>

我只是在MariaDB基于它的那一刻的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。如果有其他建议，请不要犹豫，谢谢。

于 2020-05-13T01:27:28.673 回答
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另一种解决方案：添加?useMysqlMetadata=true到您的 JDBC URL 连接，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 会将其检测为 MySQL。

于 2022-02-24T23:04:51.253 回答