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Hibernate: Criteria & createSQLQuery: how to get proper JSON?

When I try this I get the proper JSON as a result, but it takes a lot of time:
Criteria c = sessionFactory.getCurrentSession().createCriteria(User.class);
List<User> users = c.list();
List<User> specialUsers = new ArrayList<>();
for (User user : users) {
List<Perm> userPerms = user.getProfile().getPerms();
for (Perm perm : userPerms) {
if (perm.getId().equals(SPECIAL_ID)) {
specialUsers.add(user);
}
}
}
return specialUsers;
and the JSON is like:
[{"id":111,"name":"Name111"},{"id":222,"name":"Name222"}]
In attempt to improve performance I tried code below. In SQL app the results are OK, a few records of users:
String sql = "SELECT u.id, u.name FROM app.user u inner join app.perms p where u.profile = p.profile AND p.right= :rightId";
List<User> specialUsers= (List<User>)sessionFactory.getCurrentSession()
.createSQLQuery(sql)
.setParameter("rightId", SPECIAL_ID)
.list();
return specialUsers;
Now the 'JSON' however looks like this:
[[111,"Name111"],[222,"Name222"]]
I tried several things, like select *, criteria.add(Restrictions...) but to no effect. What I noticed is that in the first case specialUsers.toString returns proper data, in the second case it returns meaningless Strings like Ljava.lang.Object;#23e1469f.
Any hints how to solve this?

I managed to solve this in this way, may not be perfect:
// get ids of all special users
String sql = "SELECT u.id FROM app.user u inner join app.perms p where u.profile = p.profile AND p.right= :rightId";
List<Integer> intIds = sessionFactory.getCurrentSession()
.createSQLQuery(sql)
.setParameter("rightId", SPECIAL_ID)
.list();
// convert to long values
List<Long> longIds = intIds.stream()
.mapToLong(Integer::longValue)
.boxed().collect(Collectors.toList());
// get all special users
Criteria c = sessionFactory
.getCurrentSession()
.createCriteria(User.class)
.add(Restrictions.in("id", longIds));
List<User> specialUsers = c.list();
return specialUsers;
}

SQL Query Hibernate with clause where

how should be this query in hibernate
public Double getValuPrice(int param1, int param2){
Query query = session.createQuery("FROM TableClass WHERE e.product= :param1 and e.type = :param2");
query.setParameter("param1 ",param1);
query.setParameter("param2",param2);
result = query.uniqueResult();
List lista = query.list();
return lista;
}
and i want show the value returned in a Servlet
List list = classDao.getValuPrice(origem, destino);
out.println("<h1>" + list.eq(0) + "</h1>");

In your query you are doing from TableClass and then WHERE e.product = :param1. I would have thought that would give you an error along the lines of Unable to resolve path [e.product], unexpected token [e]. Also you have a rogue space in one of your parameters, which will give you a could not locate named parameter [param1 ] error.
You don't, however, need the select * as suggested by Lenin's answer (see the Hibernate manual (v4.3 §11.4.1 "Executing Queries"))
So correcting these:
remove the space in "param1 "
specify a type for result (and do a cast) - unless result is a field you haven't shown in your sample code
and, of course, remove e. before your properties in the query (or add e as an alias for TableClass)
You are returning a List but the method specifies a Double - I'm not sure what you want to do here so you'll need to work that out yourself.
The code now looks like
public Double getValuPrice(int param1, int param2){
Query query = session.createQuery("FROM TableClass WHERE product= :param1 and type = :param2");
query.setParameter("param1",param1);
query.setParameter("param2",param2);
// assuming result is a field with type TableClass
result = (TableClass)query.uniqueResult();
List lista = query.list();
return lista; // <-- fix the return statement
}
If you want to return a specific field on TableClass, which is a Double (e.g. double price;) then do:
Query q = session.createQuery("select price from TableClass WHERE product= :param1 and type = :param2"
...
Double result = ((Number) q.uniqueResult()).doubleValue();

Seems like you are missing Select clause and table alias.
It should be like
Query query = session.createQuery("Select * FROM TableClass e WHERE e.product= :param1 and e.type = :param2");

Use Criteria to select a particular field from DB?

I have a method like below
public List<String> getSimilarResourceNames(String resourceName){
String searchString = "%"+resourceName+"%";
Session session = getSession();
Criteria criteria = session.createCriteria(Resource.class);
criteria.add(Restrictions.like("name", searchString));
return criteria.list()
}
This will return me the entire resource from the DB, but what i need is just the name of the resource. How can I accomplish that ?

Use Projection, you can find examples in Hibernate documentation.
Criteria criteria = session.createCriteria(Resource.class);
criteria.setProjection(Property.forName("name"))
criteria.add(Restrictions.like("name", searchString));

By using Projection you will get other fields (Which you did not got by Projection) in your Pojo setted to default values. In HQL you can get specified column values as follow:
Query query = session.createQuery("select u.fullname from Users u");
List<Object[]> rows = query.list();
List<String> fullnames = new ArrayList<String>();
for (Object[] row: rows) {
fullnames.add(row[0]);
}
I hope this will help you.

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

Hql Query to Bind a Data Member

Query qry = session.createQuery("From RegistrationBean where ? = ?");
qry.setString(0,searchCriteria);
qry.setString(1,searchField);
searchList =(ArrayList<RegistrationBean>) qry.list();
RegistrationBean Entity class has userName, address, age fields..
I want to search a user by search criteria such as userName, address etc. using the above single query...
But the query is returning me zero results even though the user exist..
what's the problem?

Both parameters are set to 0 position, the second is not set. The position parameter should be set sequentially.
qry.setParameter(0,searchCriteria);
qry.setParameter(1,searchField);
But the field name should pass in the following way
String queryString = "from RegistrationBean as model where model." + propertyName + "= ?";
Query queryObject = getSession().createQuery(queryString);
queryObject.setParameter(0, value);
return queryObject.list();

Try in the following way:
Query qry = session.createQuery("select * from RegistrationBean where :searchCrit = :searchValue");
qry.setString(":searchCrit",searchCriteria);
qry.setString(":searchValue",searchField);
searchList =(ArrayList<RegistrationBean>) qry.list();
It is more appropriate to use it like this instead of setting indexes (your problem is that you set both to 0), because if you change something in your query you might need to change your setters afterwards.