JPA表"序列"不存在

Question

数据库:

user_account
id(pk)
email
password
...

user_detail
id(pk fk)
name_first
name_last
...

实体

@Entity
@Table(name="user_account")     
@SecondaryTable(name="user_detail", pkJoinColumns=@PrimaryKeyJoinColumn())
public class UserAccount implements Serializable{
    private static final long serialVersionUID = -2606506548742732094L;

    @Id
    @GeneratedValue(strategy=GenerationType.AUTO)
    private Integer id;
    private String email;
    private String password;
    private String tab;
    private String shortcut;
    private String setting;
    private Integer role;

    @Column(table="user_detail", name="name_first")
    private String nameFirst;
    @Column(table="user_detail", name="name_last")
    private String nameLast;
    @Column(table="user_detail")
    private String occupation;
    @Column(table="user_detail")
    @Temporal(TemporalType.DATE)
    private Date birth;
    ....
}

行动

    try{
        EntityTransaction transaction = em.getTransaction();
        transaction.begin();
        em.persist(currentUser);

        transaction.commit();
    } catch (Exception e){
    }

错误

信息:[EL警告]:2012-01-06 18:45:46.77 - ClientSession(17472935) - 异常[EclipseLink-4002](Eclipse Persistence Services - 2.3.0.v20110604-r9504):org.eclipse.persistence .exceptions.DatabaseException内部异常:com.mysql.jdbc.exceptions.MySQLSyntaxErrorException:表'mazedb.sequence'不存在错误代码:1146调用:UPDATE SEQUENCE SET SEQ_COUNT = SEQ_COUNT +？在哪里SEQ_NAME =？bind => [2个参数绑定]查询:DataModifyQuery(name ="SEQUENCE"sql ="UPDATE SEQUENCE SET SEQ_COUNT = SEQ_COUNT +？WHERE SEQ_NAME =？")

INFO:ERROR:内部异常:com.mysql.jdbc.exceptions.MySQLSyntaxErrorException:表'mazedb.sequence'不存在错误代码:1146调用:UPDATE SEQUENCE SET SEQ_COUNT = SEQ_COUNT +？在哪里SEQ_NAME =？bind => [2个参数绑定]查询:DataModifyQuery(name ="SEQUENCE"sql ="UPDATE SEQUENCE SET SEQ_COUNT = SEQ_COUNT +？WHERE SEQ_NAME =？")

我尝试了另一种方式,有两个不同的实体由@PrimaryKeyJoinColumn合并,但我得到了同样的错误.

Answer 1

如果EclipseLink尝试访问此表,那意味着它应该在那里.GenerationType.AUTO表示EclipseLink为您的数据库(MySQL)选择最合适的生成类型.在这种情况下,选择是使用基于表的生成器,它需要一个表.请参阅http://wiki.eclipse.org/EclipseLink/UserGuide/JPA/Basic_JPA_Development/Entities/Ids/GeneratedValue.

如果您不想使用此策略,请选择其他策略.

Answer 2

尽管做了一百次这样的事情,我以前从来没有遇到过这个错误.我发现这个错误是因为我有一个hibernate属性用于创建表persistence.xml但使用EclipseLink:

它是:

    <property name="hibernate.hbm2ddl.auto" value="create"/>

我更改为以下内容以修复错误:

    <property name="eclipselink.ddl-generation" value="drop-and-create-tables"/>
    <!-- or just value="create-tables" -->