1
我有一個表格,說Book
只有一個column(book_id)
爲Serial
。 我正在使用Informix和openejb 4.7.2。 當我試圖創建數據庫的新條目,我得到錯誤SQLGrammarException當使用JPA插入單列表作爲序列時使用JPA
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;
}
}
代碼來創建本書
Book book = new Book();
documentTemplateDAO.create(book);
的查詢將要執行的是
insert into book values ()