Select all items from a list in hibernate - java

I have a list of product ids and I want to get all the products from my db with a hibernate query. How can I do this?
List<Integer> list = custumor.getCart();
Query query = query("select product from Product product where product.id =: list");
I know this isn't the best way to solve this, but I just want to know how I can test for all the values in a list.


There are two things you'll need to do... The first is to change your HQL to the following (making use of IN), and the second is to bind your parameter list:
Query query = query("select product from Product product where product.id IN :list")
.setParameterList("list", list);


Not sure I get you but check out this link: http://www.coderanch.com/t/217864/ORM/java/Hibernate-retrieve-data-database
Specifically this snippet:
public static void retrieve() {
AnnotationConfiguration config = new AnnotationConfiguration();
config.addAnnotatedClass(User.class);
SessionFactory factory= config.configure().buildSessionFactory();
Session session = factory.getCurrentSession();
session.beginTransaction();
Query queryResult = session.createQuery("from User");
java.util.List allUsers;
allUsers = queryResult.list();
for (int i = 0; i < allUsers.size(); i++) {
User user = (User) allUsers.get(i);
}
System.out.println("Database contents delivered...");
}
Actually I just re-read your question and I see you want a sub select. You should consider using a query like:
List<Integer> list = custumor.getCart();
Query query = query("select product from Product product where product.id IN (: list)");

Related

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());
}
}

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

Hibernate HQL list attributes?

I have a small query something like this:
String sqlStr = "select count(*) from MyTable n where primaryKey.userId = :userId and primaryKey.userType = :userType"
String[] paramNames = {"userId", "userType" };
Object[] paramValues = {userId, userType};
List myObjects = (List) hibernateTemplate.findByNamedQueryAndNamedParam("objectsToDeleteForUser", paramNames, paramValues);
and this works fine. I can make a call with a userid and usertype and get one or more matching records.
What I'd like to do now that is have a list of userid's and usertype's and perform the call but not have to iterate over the list but make one call. So basically when userid(0) and usertype(0), userid(1) and usertype(1) ...................
Is it possible to do this with one call in HQL. Something like the IN clause (X).
Thanks

somwthing like this is possible :
hibernateSession = HibernateUtil.getSessionFactory().getCurrentSession();
hibernateSession.beginTransaction();
List<Object> userId = new ArrayList<Object>();
List<Object> userType= new ArrayList<Object>();
Query q1 = hibernateSession.createQuery("select count(*) from MyTable n where primaryKey.userId IN (:userId) and primaryKey.userType IN (:userType)");
q1.setParameterList("userId", userId);
q1.setparameterList("userType" , userType);
List myObjects = q1.list();
hibernateSession.close();
this should probably work for you

hibernate select distinct values order by one value

I want to get distinct values from my db.
I have 10 fields in this db, and when i try to use such query:
SELECT DISTINCT (IMIE)FROM `przychodzace`
I get 26 results, but hibernate returns me just 17...
Here is list function:
List<String> list = new ArrayList<String>();
SessionFactory sf = HibernateUtil.getSessionFactory();
Session session = sf.openSession();
Criteria criteria = session.createCriteria(PrzychodzaceModel.class);
if (i == 0) {
criteria.setProjection(Projections.distinct(Projections.property("imie")));
criteria.addOrder(Order.asc("imie"));
}
list = criteria.list();
System.out.println(list.size() + "size");
return list;
Have anyone idea how to do it properly, i am trying to correct it for long time.
Thanks in advance.

I think an alternative to your problem would be to use a DAO class with a List method, eg:
public List listMenu() {
String hql = "FROM Menu";
org.hibernate.Query query = session.getCurrentSession().createQuery(hql);
query.setFirstResult(0);
query.setMaxResults(5);
List results = query.list();
return results;
}

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