Question

我有一张表Book，只有一个column(book_id)为Serial。我正在使用Informix和openejb 4.7.2。当我尝试在DB中创建新条目时，我收到错误

OpenEJB - EjbTransactionUtil.handleSystemException: org.hibernate.exception.SQLGrammarException: could not prepare statement
javax.persistence.PersistenceException: org.hibernate.exception.SQLGrammarException: could not prepare statement
    at org.hibernate.ejb.AbstractEntityManagerImpl.convert(AbstractEntityManagerImpl.java:1387) ~[hibernate-entitymanager-4.2.19.Final.jar:4.2.19.Final]
    at org.hibernate.ejb.AbstractEntityManagerImpl.convert(AbstractEntityManagerImpl.java:1310) ~[hibernate-entitymanager-4.2.19.Final.jar:4.2.19.Final]
    at org.hibernate.ejb.AbstractEntityManagerImpl.convert(AbstractEntityManagerImpl.java:1316) ~[hibernate-entitymanager-4.2.19.Final.jar:4.2.19.Final]
    at org.hibernate.ejb.AbstractEntityManagerImpl.persist(AbstractEntityManagerImpl.java:881) ~[hibernate-entitymanager-4.2.19.Final.jar:4.2.19.Final]
    at org.apache.openejb.persistence.JtaEntityManager.persist(JtaEntityManager.java:149) ~[openejb-core-4.7.2.jar:4.7.2]

Book.java

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

    @Id
    @GeneratedValue(strategy = GenerationType.AUTO)
    @Column(name = "book_id", nullable = false)
    private short bookId;

    public short getBookId() {
        return bookId;
    }

    public void setBookId(short bookId) {
        this.bookId = bookId;
    }

}

public class DocumentTemplateDAO{
    @Override
    public Book create(Book entity) {
        LOG.debug("Entity is created {} ", entity);
        this.entityManager.persist(entity);
        this.entityManager.flush();
        return entity;
    }
}

创建图书的代码

Book book = new Book();
documentTemplateDAO.create(book);

将要执行的查询是

insert into book values ( )

Answer 1

2个可能的解决方案：

Informix SERIAL（如果这是您的图书ID在数据库中）是零触发自动增量，它应该影响@GeneratedValue（strategy = GenerationType.AUTO）。
IBM为Informix开发了一个特定的Hibernate方言，你可以在那里下载：http://members.iiug.org/opensource/。

对不起，我现在没有更多，希望它会让你走上正轨。

SQLGrammarException使用JPA插入具有单列的表作为serial时

1 个答案: