Develop Reference

Delete all rows from a mysql table

I'm trying to delete all the records from a MySQL table (46 records).
The code I have tried. Any suitable answer?
Session hs = connection.NewHibernateUtil.getSessionFactory().openSession();
Criteria cr = hs.createCriteria(Bookmark.class);
Bookmark b;
List<Bookmark> li = cr.list();
for (Bookmark s : li) {
b = new Bookmark();
b.setId(s.getId());
Transaction tr = hs.beginTransaction();
hs.delete(b);
tr.commit();
hs.flush();
hs.close();
}
Error
org.hibernate.NonUniqueObjectException: a different object with the same identifier value was already associated with the session: [mypojos.Bookmark#7]

You cant delete objects like that. You would first have to get the object from db and then you can delete using hs.delete(b); this is usually used when you have to cascade changes to associated objects.
Best approach in this case is to use HQL query something like this.
String stringQuery = "DELETE FROM tablename";
Query query = session.createQuery(stringQuery);
query.executeUpdate();

Getting an error while using "Generics" in Hibernate Query

I'm using Hibernate in my project and using "Query" Like below.
Query query = em.createQuery("delete from User where name=:name");
query.setParameter("name", "Zigi");
int deleted = query.executeUpdate();
I'm getting the result. When i'm using generics Like below
Query<?> query = em.createQuery("delete from User where name=:name");
query.setParameter("name", "Zigi");
int deleted = query.executeUpdate();
getting the result because "?" is something as wildcard (or) it will accept any datatype
when i'm using the code as below getting some error( java.lang.IllegalArgumentException: Update/delete queries cannot be typed ). i'm using Integer datatype because createQuery return number when it's executed
Query<Integer> query = em.createQuery("delete from User where name=:name", Integer.class);
query.setParameter("name", "Zigi");
int deleted = query.executeUpdate();
Any suggestions. I have to use Generics with specific datatype in it like above code.
thanks in advance

Unfortunately executeUpdate does not allow you to use a typed query.
Instead you can use uniqueResult, and cast to the type that you want:
Query query = em.createQuery("delete from User where name=:name");
query.setParameter("name", "Zigi");
int deleted = (Integer) query.uniqueResult();
Referencing this answer

Query not giving solution in DAO class

when i run my query in database visualizer its working perfectly, but i think there are some issues in syntax when i convert it in my DAO class method.
I want to get whole data against the name provided
In Visualizer:
SELECT first_name,last_name,nic,phone,email FROM x_hr_user where (first_name = 'Irum');
Now in Dao
public List<XHrUser> findXHrUserByNameInTable()
{
String name ="Irum";
Query query = em.createQuery("SELECT xHrNewUserObj.firstName,xHrNewUserObj.lastName, xHrNewUserObj.nic, xHrNewUserObj.phone, xHrNewUserObj.emil FROM XHrUser xHrNewUserObj where (xHrNewUserObj.firstName) = (name)");
List<XHrUser> list = query.getResultList();
return list;
}
Instead of showing single row, it displays whole data Table
Thank you

Your current query is not valid JPQL. It appears that you intended to insert the raw name string into your query, which could be done via a native query, but certainly is not desirable. Instead, use a named parameter in your JPQL query and then bind name to it.
String name = "Irum";
Query query = em.createQuery("SELECT x FROM XHrUser WHERE x.firstName = :name")
.setParameter("name", name);
List<XhrUser> list = query.getResultList();

You have to write query as below. where : is used for variable
Query query = em.createQuery("SELECT xHrNewUserObj.firstName,xHrNewUserObj.lastName, xHrNewUserObj.nic, xHrNewUserObj.phone, xHrNewUserObj.emil FROM XHrUser xHrNewUserObj where (xHrNewUserObj.firstName) = :name");

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 write hibernate template query from the sql query?

Actually I had one SQL query that runs very fine with mySql database. but, my problem is that I want to write that query in hibernatetemplate query.
My SQL QUERY IS
SELECT * FROM task, dependency WHERE DEPENDENCY_From IN(10,11,12,13)
and I want it to executed by the hibernate template.
How do I to convert this to a HibernateTemplate Query?

If you want to use HibernateTemplate query then below is example of hibernate query but in your case, we need mapping files to see how the entities are associated. Hope below query will help you.
public List<Object[]> getCities(Integer stateId) {
List<Object[]> cityList = new ArrayList<Object[]>();
String query = "select city.cityId,city.cityName from City city where city.state.stateId=?";
Object[] queryParam = {stateId};
cityList = getHibernateTemplate().find(query, queryParam);
return cityList;
}
or
String query="from City";
List<City> cityList = getHibernateTemplate().find(query);

HQL query
Object[] params = {cid,cid};
List list=getHibernateTemplate().find("select S.stateName,C.countryName from State S,Country C where C.countryId=? and S.countryId=?", params);
SQL Query
select S.stateName,C.countryName from State S,Country C where S.countryId=C.countryId;