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Hibernate createNativeQuery - get more than one Entity

I am using following code to execute native SQL query with hibernate:
Query zonesQuery = session.createNativeQuery(
"Select * \n" +
"FROM dbo.Structure AS S\n" +
"JOIN dbo.StructureLocationType AS SLT ON SLT.StructureId = S.Id\n" +
"WHERE SLT.LocationTypeId = :lc").addEntity(StructureEntity.class);
zonesQuery.setParameter("lc", locationTypeID);
List<StructureEntity> zones = zonesQuery.list();
So it works and it gets me list of StructureEntity
now, because my sql query "join" from StructureLocationType table, is there possibility to get whole StructureLocationType row as well, still using single query?
Thank you.

It can be achieved with the following (notice curly braces in SQL and entities aliases):
Query query = session
.createNativeQuery(
"SELECT {S.*}, {SLT.*} " +
"FROM dbo.Structure AS S " +
"JOIN dbo.StructureLocationType AS SLT ON SLT.StructureId = S.Id " +
"WHERE SLT.LocationTypeId = :lc")
.unwrap(SQLQuery.class)
.addEntity("S", StructureEntity.class)
.addEntity("SLT", StructureLocationTypeEntity.class)
.setParameter("lc", locationTypeID);
List<Pair<StructureEntity, StructureLocationTypeEntity>> result = ((List<Object[]>) query.list())
.stream()
.map(p -> Pair.of((StructureEntity) p[0], (StructureLocationTypeEntity) p[1]))
.collect(Collectors.toList());

You can't get multiple Objects from one query.
But you could either select which columns you want and then iterate the returned Object array:
The query:
SELECT s.id, s.someColumn, slt.id, slt.structureId
FROM dbo.Structure AS s
JOIN dbo.StructureLocationType AS slt on slt.structureId = s.id
WHERE slt.locationTypeId = :lc
Then iterate over the Object array:
List<Object[]> result = query.getResultList();
Or you could create a view on your database and map it to a java entity using the Table annotation like it was a normal table:
#Entity
#Table(name = "STRUCTURE_LOCATION_TYPE_VIEW")
public class StructureAndLocationType {
// ...
}
I thought there is a way to map from a query to an Object without creating a DB view but couldn't find it right now.

Not able to fetch resultset in Hibernate using HQL

I'm triggering a query using HQL, normally it should return empty resultset as it doesn't have any records w.r.t it. But, it throws
org.hibernate.exception.SQLGrammarException: could not extract ResultSet
at org.hibernate.exception.internal.SQLStateConversionDelegate.convert(SQLStateConversionDelegate.java:106)
My code is
String hql = "FROM com.pck.Person where userId = " + userId;
Query query = session.createQuery(hql);
#SuppressWarnings("unchecked")
List<Dashboard> listUserDetails = query.list(); <-- Problem here.
I'm expecting list size is 0 because there are no records w.r.t userId passed.
What changes do I need to do?

Lets say the value of userId was "abc12"
Given your code, the value of the string called hql would become:
"FROM com.pck.Person where userId = abc12"
If you took the value of that string and tried to run it as a query on any database, most of them would fail to understand that abc12 is a string. Normally it would be interpreted as a variable.
As other users mentioned including the single quotes would produce the desired query, but the recommended way to assign parameter values is this:
String hql = "FROM com.pck.Person where userId = :id"
query.setParameter("id", userId);

Looks like you are missing single quotes around userid.
Try with "FROM com.pck.Person where userId = '" + userId + "'";
or
Use named parameters with query.setParameter("userid", userId);
Posting the full stacktrace would help if this doesn't solve.

How to set placeholders in hibernate

Hi this my java code here am using hibernate to check whether this email id and password exist in db or not could anybody plz exp line me how to place the value to this place holders.
Session ses = HibernateUtil.getSessionFactory().openSession();
String query;
query = "from RegisterPojo where email=? and pwd=? ";
List<RegBean> list = ses.createQuery(query).list();
ses.close();
Thanks in advance

Try this one.
Session ses = HibernateUtil.getSessionFactory().openSession();
String query = "from RegisterPojo rp where rp.email = :emailAddress and rp.pwd = :password";
List<RegisterPojo> list = ses.createQuery(query)
.setParameter("emailAddress", Your email address)
.setParameter("password", Your password)
.list();
ses.close();

You should use a prepared statement instead of a string. example here
PreparedStatement preparedStatement = con.prepareStatement ("from RegisterPojo where email=? and pwd=?");
preparedStatement.setString(1, "email");
preparedStatement.setString(2, "password");

You have to modify your query like this,
query = "from RegisterPojo where email =:email and pwd =:password ";
List<RegisterPojo> list = ses.createQuery(query)
.setParameter("email",emailVal)
.setParameter("password",emailVal)
.list();
Read the hql docs here

Session ses = HibernateUtil.getSessionFactory().openSession();
String query;
query = "from RegisterPojo where email=? and pwd=? ";
List<RegBean> list = ses.createQuery(query).setParameter(0,emailVal).setParameter(1,emailVal).list();
ses.close();

Try this one.
Session ses = HibernateUtil.getSessionFactory().openSession();
String query = "from RegisterPojo where email = '" + varEmailAddress + "' and pwd = '" + varPassword + "'";
List<RegisterPojo> list = ses.createQuery(query).list();
ses.close();

Sorry This Is not Answer but instead another case, in above case #Grigoriev Nick suggested
query = "from RegisterPojo where email=? and pwd=? ";
List<RegBean> list = ses.createQuery(query).setParameter(0,emailVal).setParameter(1,emailVal).list();
but here the script starts with directly from clause while what if I want want to used sql like below
WITH A (
/// do some selection from many tables using various union as per my requirement
),
B (
/// another set of sqls for different set of data
)
select XXX from A a join B b on a.XYZ = b.xyz
where
/// few more fixed where clause conditions
AND A.SOME_COLUMN = ? // Here Instead Of String Concatenation I want to
//use Query parameters but using hibernate instead of raw Prepares statements

Hibernate retrieve results from database based on condition

I am a bit lost when it comes to retrieving results from the database.
My MemberModel consists of 4 fields: id, username, password and email. I have been able to successfully save it to database.
Now I need to retrieve an id of a member who's username equals "Test".
I tried something along the lines:
SQLQuery query = session.createSQLQuery("SELECT id FROM members WHERE username = :username");
query.setString("username", username);
List<MemberModel> returnedMembers = query.list();
MemberModel member = returnedMembers.get(0);
int id = member.getId();
However I get an error that member.getId() cannot be converted to int, since it is MemberModel... But the getter getId() returns int.
I am quite confused. The question is: what would be the easiest and fastes way to retrieve member id based on condition (value of username)?

You are using a native SQL query, but should use HQL query. That means you have to change the query to:
session.createQuery("SELECT m FROM MemberModel m WHERE m.username = :username")

I would change your code into something like this:
public MemberModel getMember(String username) {
Query query = sessionFactory.getCurrentSession().createQuery("from " + MemberModel.class.getName() + " where username = :username ");
query.setParameter("username", username);
return (MemberModel) query.uniqueResult();
}
Then you should be able to do:
MemberModel model = someInstance.getMember("someUsername");
int id = model.getId();

You can also use criteria and restrictions api.
Criteria criteria = session.createCriteria(MemberModel.class);
criteria.add(Restrictions.eq("username", username));
MemberModel member=(MemberModel)criteria.uniqueResult();

Adding IN clause List to a JPA Query

I have built a NamedQuery that looks like this:
#NamedQuery(name = "EventLog.viewDatesInclude",
query = "SELECT el FROM EventLog el WHERE el.timeMark >= :dateFrom AND "
+ "el.timeMark <= :dateTo AND "
+ "el.name IN (:inclList)")
What I want to do is fill in the parameter :inclList with a list of items instead of one item. For example if I have a new List<String>() { "a", "b", "c" } how do I get that in the :inclList parameter? It only lets me codify one string. For example:
setParameter("inclList", "a") // works
setParameter("inclList", "a, b") // does not work
setParameter("inclList", "'a', 'b'") // does not work
setParameter("inclList", list) // throws an exception
I know I could just build a string and build the whole Query from that, but I wanted to avoid the overhead. Is there a better way of doing this?
Related question: if the List is very large, is there any good way of building query like that?

When using IN with a collection-valued parameter you don't need (...):
#NamedQuery(name = "EventLog.viewDatesInclude",
query = "SELECT el FROM EventLog el WHERE el.timeMark >= :dateFrom AND "
+ "el.timeMark <= :dateTo AND "
+ "el.name IN :inclList")

The proper JPA query format would be:
el.name IN :inclList
If you're using an older version of Hibernate as your provider you have to write:
el.name IN (:inclList)
but that is a bug (HHH-5126) (EDIT: which has been resolved by now).

public List<DealInfo> getDealInfos(List<String> dealIds) {
String queryStr = "SELECT NEW com.admin.entity.DealInfo(deal.url, deal.url, deal.url, deal.url, deal.price, deal.value) " + "FROM Deal AS deal where deal.id in :inclList";
TypedQuery<DealInfo> query = em.createQuery(queryStr, DealInfo.class);
query.setParameter("inclList", dealIds);
return query.getResultList();
}
Works for me with JPA 2, Jboss 7.0.2

You must convert to List as shown below:
String[] valores = hierarquia.split(".");
List<String> lista = Arrays.asList(valores);
String jpqlQuery = "SELECT a " +
"FROM AcessoScr a " +
"WHERE a.scr IN :param ";
Query query = getEntityManager().createQuery(jpqlQuery, AcessoScr.class);
query.setParameter("param", lista);
List<AcessoScr> acessos = query.getResultList();