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Hibernate (HQL) select query with list of ids as parameter

I have to requirement to fetch product information from the product table by product_code.
so, I have to pass a list of product_code to my select query with the help of WHERE Clause IN.
Am able to pass list of product_code but its returning only one product information ( select query giving results for the first product_code).
Please find my code below that am having issue.
public JSONArray getUpdatedProductDescription(List<String> product_code_list)
{
Collection<String> produceCollection=product_code_list;
List<String> code = null;
Configuration cfg=new Configuration();
cfg.configure("Hibernate.cfg.xml");
SessionFactory factory=cfg.buildSessionFactory();
Session session=factory.openSession();
Transaction t=session.beginTransaction();
String hql = "from Product where code in(:code)";
Query query = session.createQuery(hql);
query.setParameterList("code", produceCollection);
List<Product> listProducts = query.list();
for(Product prod : listProducts)
{
System.out.println("Product Details:::"+prod.getCode());
}
}

Hibernate many-to-many retrieve list with condition

Im working with hibernate and java. I have an Group class and a User class. They share a many-to-many relationship as shown in this ERD .
What Im trying to achieve is that I want to retrieve a list of groups with the condition that they contain a User with a certain User_id.
In the GroupDao I have defined a function retrieveForUser in which I tried to retrieve the list using hibernate query language:
public List<Group> retrieveForUser(int userid){
Session session = HibernateUtil.getSessionFactory().getCurrentSession();
session.beginTransaction();
String hql = "select distinct g from Group g " +
"join g.allGroupMembers u " +
"where u.id = :id";
Query query = session.createQuery(hql).setResultTransformer(Transformers.ALIAS_TO_ENTITY_MAP);
query.setParameter("id", userid);
List<Group> list = query.list();
session.getTransaction().commit();
return list;
}
When I try to loop throught the resulting list using:
for(Group g : groupDao.retrieveForUser(user1.getId())){
System.out.println(g.getName());
}
I get the following errormessage:
java.lang.ClassCastException: java.util.HashMap cannot be cast to nl.hu.jelo.domain.group.Group
Question
How can I achieve it so that I end up with an List<Group> with only groups that contain a User with an certain User_id

You do not need to set result transformer. ALIAS_TO_ENTITY_MAP is for other purpose.
Simply do
String hql = "select distinct g from Group g " +
"join g.allGroupMembers u " +
"where u.id = :id";
Query query = session.createQuery(hql);
query.setParameter("id", userid);
is good enough.
Something off topic, are you sure you want to handle transaction manually like that?

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

Hibernate simple select query doesn't return anything

I got this simple Hibernate query set up but it returns nothing, here is my code:
EntityManagerFactory emf = javax.persistence.Persistence.createEntityManagerFactory("timereg");
EntityManager em = emf.createEntityManager();
int id = em.createQuery("SELECT emp.id FROM Employee as emp WHERE emp.bsn = '398723916'").getFirstResult();
object.getEmployee().setId(id);
System.out.println("query returns employee id: " + id);
The stupid thing is that id stays zero but when i execute this query in PostgreSQL it returns 37.
I think hibernate does not like my way of implementing a select query, does anyone know what is wrong with my select query ?
THE ANSWER:
There was nothing wrong with the select query i just had to use getSingeResult() instead of getFirstResult();
Change the code into:
EntityManagerFactory emf = javax.persistence.Persistence.createEntityManagerFactory("timereg");
EntityManager em = emf.createEntityManager();
Object ob = em.createQuery("select id from Employee where bsn = '398723917'").getSingleResult();
object.getEmployee().setId(Integer.parseInt(ob.toString()));
System.out.println(ob);
This is the total solution for my problem, but i got inspired by Yanflea so he deserves all the credits.

You are using the method getFirstResult(), which gives you the position of the record in the table. You should use getSingleResult() instead. See http://docs.oracle.com/javaee/6/api/javax/persistence/Query.html.
EDIT
Here it is :
Object ob = em.createQuery("select id from Employee where bsn = '398723917'").getSingleResult();