Hibernate : Could not execute query -- Exception

[Zhixiong Pan]

Hi all,

I have a method like below:

 public void listCustomer() throws HibernateException {
		Session session = HibernateUtil.getSessionFactory().openSession();
		Transaction transaction = null;
		try {
			transaction = session.beginTransaction();
			List customers = session.createQuery("from customer").list();
			for (Iterator iterator = customers.iterator(); iterator.hasNext();) {
				Customer customer = (Customer) iterator.next();
				System.out.println(customer.getUsername());
			}
			transaction.commit();
		} catch (HibernateException e) {
			transaction.rollback();
			e.printStackTrace();
		} finally {
			session.close();
		}
	}   

When the program run that method, there is a exception of:

Hibernate: select from
net.sf.hibernate.exception.SQLGrammarException: Could not execute query

.
There is a table of customer, and the session.save(customer), update, and session.delete(customer) can successfully be executed.
If you have an idea, please guide me. Thanks.

[Mark E Hansen]

That doesn't look like a legal query. A query would look like:

SELECT c FROM customer c

This would select all customer records, as there is no WHERE clause, etc.

[Mark Spritzler]

That's because in HQL/JPAQL you do not use table names and field names. You use Java objects and Java object attributes. So you class name is not customer it is Customer. And in the other reply, because you use HQL you do not need a SELECT portion. In JPAQL you would need the SELECT portion and you would need to alias Customer.

Mark

[Mark Spritzler]

One other quick question. I noticed you were using the old way of doing Hibernate with creating transactions and Java 1.4 syntax.

Are you using Java 5 or higher? And are you using Hibernate 3.x? If so, use generics from Java 5 so the list() would return a List<Customer> instead of just List.

I talk more about a better approach if you are using 3.x after your reply.

Mark

[Mark E Hansen]

Mark Spritzler wrote:That's because in HQL/JPAQL you do not use table names and field names. You use Java objects and Java object attributes. So you class name is not customer it is Customer. And in the other reply, because you use HQL you do not need a SELECT portion. In JPAQL you would need the SELECT portion and you would need to alias Customer.

Mark

Sorry. I'm reading EJB3 in Action, which is dealing with JPQL (I assume that is the same as JPAQL as you call it).

I didn't realize there was yet another QL

[Zhixiong Pan]

Thank you all.
After I changed the from

  List customers = session.createQuery("from customer").list();  

to

  List customers = session.createQuery("from Customer").list();  

.
That method can run successfully.

[Mark Spritzler]

Mark E Hansen wrote:

Mark Spritzler wrote:That's because in HQL/JPAQL you do not use table names and field names. You use Java objects and Java object attributes. So you class name is not customer it is Customer. And in the other reply, because you use HQL you do not need a SELECT portion. In JPAQL you would need the SELECT portion and you would need to alias Customer.

Mark

Sorry. I'm reading EJB3 in Action, which is dealing with JPQL (I assume that is the same as JPAQL as you call it).

I didn't realize there was yet another QL

Yes, I called in JAPQL as in the JPA Query Language. But in his code he is not using JPA, he has Hibernate Session instead of EnitytManager.

Mark