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Hibernate version:3.1.3

Mapping documents:

Code between sessionFactory.openSession() and session.close():

Full stack trace of any exception that occurs:

Name and version of the database you are using:

The generated SQL (show_sql=true):

Debug level Hibernate log excerpt:


i am new to hibernate and trying to fetch data from tbl_user table.

My code the query way
tempUser = (User) session.createQuery("from User as usr where usr.userName like 'yog@abc.com'").uniqueResult();

My code the criteria way
Criteria criteria = getSession().createCriteria(getPersistentClass());
criteria.add(Restrictions.eq("userName", (String)user.getUserName()));
tempUser = (User)criteria.uniqueResult();

The SQL generated by Hibernate is:
Hibernate: select user0_.USER_ID as USER1_0_, user0_.userName as userName0_, user0_.password as password0_ from TBL_USER user0_ where user0_.userName like 'yog@abc.com'

This throws an error
The exception is : org.hibernate.exception.SQLGrammarException: could not execute query

Web Server: Tomcat
IDE:Eclipse 3.1

Could anyone help plz...

Could you please post the complete stack trace. this looks like a mapping mistake where you eventually mapped a column / table with a typo.

Thanx.
It was a silly mistake :) I had a sequence in the hbm file but didn't have one in the oracle db.