1
테이블이 Book
이고 column(book_id)
이 하나만있는 테이블이 Serial
인 경우. Informix와 openejb 4.7.2를 사용하고 있습니다. 제가 DB에 새 항목을 만들려면 노력하고 있어요 때 , 나는 오류JPG를 사용하여 단일 열의 테이블을 직렬로 삽입 할 때
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;
}
}
코드 책을
Book book = new Book();
documentTemplateDAO.create(book);
쿼리를 만들 얻고 그 실행될 것입니다
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