Develop Reference

Pessimistic Lock with Update Query

Session session = sessionFactory.openSession();
Transaction transaction = session.beginTransaction();
Query updateQuery = session.createQuery("UPDATE Lot l SET l.currentRate = l.currentRate + 100, l.lastOwner = :lastowner WHERE l.id = :lotid", null);
updateQuery.setLockMode(LockModeType.PESSIMISTIC_WRITE);
So, here is my code fragment. I got an exception: java.lang.IllegalStateException: Expecting a SELECT query : UPDATE Lot l SET l.currentRate = l.currentRate + 100, l.lastOwner = :lastowner WHERE l.id = :lotid when I try to call setLockMode.
Why? Am I doing something wrong?

See the Java doc of the method you are trying to call: https://docs.oracle.com/javaee/6/api/javax/persistence/Query.html#setLockMode(javax.persistence.LockModeType)
It explicitly says, that an IllegalStateException will be thrown if you execute this on a non-SELECT statement.

Hibernate Exception - could not locate named parameter [:laboratoryId]

I have the following query in Hibernate.I got Hibernate Exception and I don't understand why Hibernate throws this exception. Could anyone help me?
Session session = this.sessionFactory.openSession();
session.createQuery("delete from Laboratory l where l.id=:laboratoryId")
.setParameter("laboratoryId", laboratoryId).executeUpdate();
session.close();

Try to add some spaces between the = sign and the bind name :laboratoryId and remove the alias:
Session session = this.sessionFactory.openSession();
session.createQuery("delete from Laboratory where id = :laboratoryId")
.setParameter("laboratoryId", laboratoryId)
.executeUpdate();
session.close();

Are you sure that laboratoryID has value? For my query builder I used something like this:
if (!laboratoryId.isEmpty()) {
query.setParameter("laboratoryId", laboratoryId());
}
also same thing for query
"delete o from Laboratory o"
if(!laboratoryId.isEmpty()){
query.append("where o.id = (:laboratoryId)")
}
But I used it for String values
Please show code for laboratoryId - is it user input or what?

You can try this one.
DELETE Clause
The DELETE clause can be used to delete one or more objects. Following is the simple syntax of using DELETE clause:
String hql = "DELETE FROM Employee " +
"WHERE id = :employee_id";
Query query = session.createQuery(hql);
query.setParameter("employee_id", 10);
int result = query.executeUpdate();
System.out.println("Rows affected: " + result);

How to query a database using Hibernate?

I understand some might simply answer this question with "Why didn't you just Google it"... But I did, and the more I researched this the more confused I got. I'm trying to query my database with Hibernate, the query has a 'where' clause.
Now creating a database entry is easy enough, in the case where I have a 'User' class, I simply do this:
// Gets a new session
Session session = HibernateUtil.getSessionFactory().openSession();
session.beginTransaction();
// Creates a new User object
User user = new User("John", "p#55w0rd*", "john#doe.com");
// Save and commit
session.save(user);
session.getTransaction().commit();
But what do I do when I what to for instance
select * from Users where id = '3';
My Google searches pointed to something called HQL, which makes me wonder why I couldn't of just used straight JDBC then. Also it doesn't seem very object oriented. And then there's something like
session.createCriteria(.......
But I'm not sure how to use this.. Any help? Thanks guys.

When you use Native Query (non HQL ) you need to tell hibernate explicitely to handle it like below :
In below query createSQLQuery is special function to handle native sql's
String sql = "SELECT * FROM EMPLOYEE WHERE id = :employee_id";
SQLQuery query = session.createSQLQuery(sql);
query.addEntity(User.class);
query.setParameter("employee_id", 3);
List<User> results = query.list();
This can be done using criteria as well for that following is good starting point:
Criteria criteria = sess.createCriteria( User.class);
List<User> users= criteria.list();
http://www.developerhelpway.com/framework/hibernate/criteria/index.php

First of all, you need a hibernate.cfg.xml which contains properties for hibernate. This is e.g url, username and password, the driver and dialect. This file is placed in a package called resources.
You have to choose between using Hibernate Annotations example
or using hbm.xml files example
This is how you tell hibernate what your database is like. It wil automatically create queries for you based on how you annotates or defines in e.g user.hbm.xml.
Create a HibernateUtil.java class which holds the session factory.
You can fetch data from the database with
Criteria crit = getSessionFactory().getCurrentSession().createCriteria(User.class);
Example using queries:
List<?> hibTuppleResultList = currentSession.createQuery(
"from Person p, Employment e "
+ "where e.orgno like ? and p.ssn = e.ssn and p"
+ ".bankno = ?")
.setString(0, orgNo).setString(1, bankNo).list();
for (Object aHibTuppleResultList : hibTuppleResultList)
{
Object[] tuple = (Object[]) aHibTuppleResultList;
Person person = (Person) tuple[0];
hibList.add(person);
}

In the end all I really wanted was to know that if you don't want to use HQL you get something called 'Criteria Queries', and that in my case I'd do something like this:
Criteria cr = session.createCriteria(User);
cr.add(Restrictions.eq("id", 3));
List results = cr.list();
Me: "Thanks!"
Me: "No problem :)"
PS - we can really delete this question.

Query q = session.createQuery("from User as u where u.id = :u.id");
q.setString("id", "3");
List result = q.list();
Query with Criteria:
Criteria cr = session.createCriteria(User.class);
List results = cr.list();
Restrictions with Criteria:
Criteria cr = session.createCriteria(User.class);
cr.add(Restrictions.eq("id", 3));
// You can add as many as Restrictions as per your requirement
List results = cr.list();
You could also use it like this
List results = session.createCriteria(User.class).add(Restrictions.eq("id", 3)).list();
Some example for Crieteria Rsetriction query
Criteria cr = session.createCriteria(Employee.class);
// To get records having salary more than 2000
cr.add(Restrictions.gt("salary", 2000));
// To get records having salary less than 2000
cr.add(Restrictions.lt("salary", 2000));
// To get records having fistName starting with zara cr.add(Restrictions.like("firstName", "zara%"));
// Case sensitive form of the above restriction.
cr.add(Restrictions.ilike("firstName", "zara%"));
// To get records having salary in between 1000 and 2000
cr.add(Restrictions.between("salary", 1000, 2000));
// To check if the given property is null
cr.add(Restrictions.isNull("salary"));
// To check if the given property is not null
cr.add(Restrictions.isNotNull("salary"));
// To check if the given property is empty
cr.add(Restrictions.isEmpty("salary"));
// To check if the given property is not empty
cr.add(Restrictions.isNotEmpty("salary"));
You can create AND or OR conditions using LogicalExpression restrictions as follows:
Criteria cr = session.createCriteria(Employee.class);
Criterion salary = Restrictions.gt("salary", 2000);
Criterion name = Restrictions.ilike("firstNname","zara%");
// To get records matching with OR condistions
LogicalExpression orExp = Restrictions.or(salary, name);
cr.add( orExp );
// To get records matching with AND condistions
LogicalExpression andExp = Restrictions.and(salary, name);
cr.add( andExp );
List results = cr.list();
I think this will help you

How to use Delete query in Hibernate

I'm using Hibernate to delete the records from a table,but this giving an exception, could anyone know how to overcome this problem in the below query?
Session ses = HibernateUtil.getSessionFactory().openSession();
Transaction tx = ses.beginTransaction();
Query q = ses.createQuery("from RegisterPojo where email =:email");
q.setParameter("email", sl_no);
RegisterPojo pojo = (RegisterPojo) q.list().get(0);
ses.delete(pojo);
tx.commit();
ses.close();

Why not ?
Query q = ses.createQuery("delete from RegisterPojo where email =:email");
q.setParameter("email", sl_no);
q.executeUpdate();
Learn HQL
Before proceeding please learn more about hql. That reduces large amount of code.

Cannot Cast HibernateQuery resuly list to Object class

I am using the following query to fetch data from db in hibernate
Session session = HibernateUtil.getSessionFactory().getCurrentSession();
session.beginTransaction();
Query q = session.createSQLQuery("select name,addr1,addr2,postal_code,country,email," +
"tel1,tel2,HeadOffice_id,Subscription_id from Restaurant " +
"where id=" +id);
session.getTransaction().commit();
Restaurant rest = (Restaurant)result.get(0);
But this is returning exception
java.lang.ClassCastException: [Ljava.lang.Object; cannot be cast to com.hibernate.model.Restaurant
I also tried this way as well not sure whats doing
AnnotationConfiguration config = new AnnotationConfiguration();
config.addAnnotatedClass(Restaurant.class);
SessionFactory factory= config.configure().buildSessionFactory();
Session session =factory.getCurrentSession();
session.beginTransaction();
Query q = session.createSQLQuery("select name,addr1,addr2,postal_code,country,email," +
"tel1,tel2,HeadOffice_id,Subscription_id from Restaurant " +
"where id=" +id);
java.util.List<Restaurant> result = (List<Restaurant>)q.list();
session.getTransaction().commit();
Restaurant rest = (Restaurant)result.get(0);
Again I am getting the same exception. How can i do this with hibernate?
Thanks

Your query doesn't return instances of the Restaurant entity. It returns individual fields from this entity. The result of such a query is a List<Object[]>, each Object[] containing all the selected fields.
See http://docs.jboss.org/hibernate/core/3.6/reference/en-US/html_single/#queryhql-select:
Queries can return multiple objects and/or properties as an array of type Object[]:
If you want your query to returninstances of Restaurant, it should be
select r from Restaurant r where id = :id
And please, don't use concatenation to pass your parameter. Use named parameters as the above query.

as simple as:
Restaurant rest = (Restaurant)session.get(Restaurant.class, id);