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Getting the JPQL/SQL String Representations for a Criteria Query

How to Getting the SQL String Representations for a Criteria Query with parmetervalue ?
I tried this but it returns a string to me without the parameter values:
String QueryString = 'SELECT * FROM CUSTOMERS WHERE lastname = ?'
query = entityManager.createNativeQuery(QueryString);
query.setParameter(1, "toto");
System.out.print(query.unwrap(JpaQuery.class).getDatabaseQuery().getSQLString());
But returns "SELECT * FROM CUSTOMERS WHERE lastname = ?"
instead of "SELECT * FROM CUSTOMERS WHERE lastname = 'toto'"

Thanks to #GrootCfor helping me find the solution.
If it helps other people here is the solution:
String QueryString = 'SELECT * FROM CUSTOMERS WHERE lastname = ?';
Query query = entityManager.createNativeQuery(QueryString);
Session session = entityManager.unwrap(JpaEntityManager.class).getActiveSession();
DatabaseQuery databaseQuery = query.unwrap(org.eclipse.persistence.jpa.JpaQuery.class).getDatabaseQuery();
DatabaseRecord recordWithValues= new DatabaseRecord();
query.setParameter(1, "toto");
recordWithValues.add(new DatabaseField(Integer.toString(1)), "toto");
databaseQuery.prepareCall(session, recordWithValues);
String sqlStringWithArgs = databaseQuery.getTranslatedSQLString(session, recordWithValues);
System.out.print(sqlStringWithArgs);
====SELECT * FROM CUSTOMERS WHERE lastname = 'toto'====

Get entity by email in hibernate

I am trying to get the user from his email , the email is unique in the database.
I write this code :
session.beginTransaction();
User user = (User) session.createQuery("select * from `user` where email = '"+email+"'");
session.getTransaction().commit();
Is this code right ? or there is some function in hibernate to get entity by column value ?

I see two problems with your current code. First, you appear to be running a native SQL query, not HQL (or JPQL). Second, your query is built using string concatenation, leaving it prone to attack by SQL injection
Consider the following code:
Query query = session.createQuery("from User u where u.email = :email ");
query.setParameter("email", email);
List list = query.list();

Without writting any SQL:
public static Person getPersonByEmail(String email) {
SessionFactory sessionFactory = new Configuration().configure().buildSessionFactory();
Session session = sessionFactory.openSession();
session.beginTransaction();
CriteriaBuilder cb = session.getCriteriaBuilder();
CriteriaQuery<Person> cr = cb.createQuery(Person.class);
Root<Person> root = cr.from(Person.class);
cr.select(root).where(cb.equal(root.get("email"), email)); //here you pass a class field, not a table column (in this example they are called the same)
Query<Person> query = session.createQuery(cr);
query.setMaxResults(1);
List<Person> result = query.getResultList();
session.close();
return result.get(0);
}
example of use:
public static void main(String[] args) {
Person person = getPersonByEmail("test#mail.com");
System.out.println(person.getEmail()); //test#mail.com
}

Delete all objects from database in hibernate Spring java

I want to delete all those rows from xyz table where id = 1 using hibernate spring.
I have tried following code but its not giving error but not deleting rows -
Session session = (Session) getEm().getDelegate();
String sql ="Delete from xyz where id=:id" ;
SQLQuery query = session.createSQLQuery(sql);
query.setParameter("id", "1");
int flg = query.executeUpdate();
Can you please help me to delete all rows using hibernate query.

Try wrapping your code within a transaction like this:
Session session = (Session) getEm().getDelegate();
Transaction tx = session.beginTransaction();
String sql ="Delete from xyz where id=:id" ;
SQLQuery query = session.createSQLQuery(sql);
query.setParameter("id", "1");
int flg = query.executeUpdate();
tx.commit();

Try
query.setParameter("id", Long.valueOf(1));
if your entity is of type Long (which ideally should be).
Reference: http://www.codejava.net/frameworks/hibernate/hibernate-basics-3-ways-to-delete-an-entity-from-the-datastore
Note: The link is just for your reference.

public void deleteById(Class clazz,Integer id) {
String hql = "delete " + clazz.getName() + " where id = :id";
Query q = session.createQuery(hql).setParameter("id", id);
q.executeUpdate();
}

Writing sql query in hibernate

I have a sql query:
select COUNT (distinct agentG) as count from Test_CPView where kNum = ‘test k1’ and pName = ‘test p1’
I'm trying to write into criteria query but it hasn't worked for me:
statelessSession = sessionFactory.openStatelessSession();
Criteria crit = statelessSession.createCriteria(APRecord.class, "apr");
ProjectionList projList = Projections.projectionList();
projList.add(Projections.groupProperty("pName"));
projList.add(Projections.groupProperty("kNum"));
projList.add(Projections.countDistinct("agentG"));
crit.setProjection(projList);
This produces:
Hibernate: select this_.pName as y0_, this_.kNum as y1_, count(distinct this_.agentG) as y2_ from Test_CPView this_ where (lower(this_. pName + '~' + this_. kNum) like ? or lower(this_. pName + '~' + this_. kNum) like ? or lower(this_. pName + '~' + this_. kNum) like ? or lower(this_. pName + '~' + this_. kNum) like ?) group by this_.pName, this_. kNum
and the return results are null.
How can I convert the above sql query into hibernate?

Session.createCriteria from Docs
You have not added Restrictions
statelessSession = sessionFactory.openStatelessSession();
Criteria crit = statelessSession.createCriteria(APRecord.class, "apr");
crit .add(Restrictions.eq("kNum", "test k1"));
crit .add(Restrictions.eq("pName ", "test k1"));
crit.setProjection(Projections.countDistinct("agentG"));
Integer count = crit.uniqueResult();

statelessSession = sessionFactory.openStatelessSession();
Criteria crit = statelessSession.createCriteria(APRecord.class, "apr");
crit.add(Expression.eq("kNum","test k1"));
crit.add(Expression.eq("pName","test p1"));
crit.setProjection(Projections.countDistinct("agentG"));

Query query = session.createQuery("count(agentG) from APRecord where kNum = :kNum and pName = :pName");
query.setParameter("kNum", "test k1");
query.setParameter("pName", "test p1");
return (Integer) query.uniqueResult();

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