I am new to hibernate . I want to pass 2 column values and want hibernate to return primary key of that table.
String queryString = "select perId from Permission where document.docId=1 and user.id=2";
return getHibernateTemplate().find(queryString);
But this method return List.
How can i return int value.
Use the uniqueResult() method in Query. see here for an example or read the api here.
Here is an example. Replace the place holders as need to use them.
sessionFactory = getHibernateTemplate().getSessionFactory();
Session session = sessionFactory.getCurrentSession();
Query query = session
.createQuery("select value from table where ...");
query.setParameters("param1", value1);
result = (Type) query.uniqueResult();
You could do something like:
String sql = "select count(*) from table where ...";
BigDecimal count = (BigDecimal) hibernateTemplate.execute(
new HibernateCallback() {
public Object doInHibernate(Session session) throws HibernateException {
SQLQuery query = session.createSQLQuery(sql);
return (BigDecimal) query.uniqueResult();
}});
return count;
Here is another way using addScalar:
Query query = session.createQuery("select value from table where param1 = :param1").addScalar("value", Type);
query.setParameters("param1", value1);
result = (Type) query.uniqueResult();
Example of String:
Query query = session.createQuery("select value from table where param1 = :param1").addScalar("value", StandardBasicTypes.STRING);
query.setParameters("param1", value1);
result = (String) query.uniqueResult();
Related
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
}
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();
}
Code:
public void getDetails() {
try {
Session session = sessionFactory.openSession();
Transaction transaction = session.beginTransaction();
String hql = "select c.mobile, c.password FROM CrbtSubMasterDemo c where rownum<20";
Query query = session.createQuery(hql);
List<CrbtSubMasterDemo> itr = query.list();
session.getTransaction().commit();
for (CrbtSubMasterDemo pojo : itr) {//excepion line
System.out.println("[" + pojo.getMobile() + "]");
}
} catch (Exception e) {
e.printStackTrace();
}
}
CrbtSubMasterDemo is pojo mapped with the db.
When I try to run it, it gives following Exception:
java.lang.ClassCastException: [Ljava.lang.Object; cannot be cast to com.telemune.demoPojo.CrbtSubMasterDemo
at com.telemune.demoHibernate.QueryTester.getDetails(QueryTester.java:57)
at com.telemune.demoHibernate.QueryTester.main(QueryTester.java:23)
The question is query.list() is returning the list of objects of pojo class. Then why is this Exception. I am new to Hibernate, sorry if its a silly question.
When you write this:
String hql = "select c.mobile, c.password FROM CrbtSubMasterDemo c where rownum<20";
Your result set is not a List of CrbtSubMasterDemo
Try to write:
String hql = "select FROM CrbtSubMasterDemo c where rownum<20";
Another way is define a new constructor of CrbtSubMasterDemo where you pass only two fields c.mobile, c.password
so your query becomes:
String hql = "select new " + CrbtSubMasterDemo.class.getName() + "(c.mobile, c.password) FROM CrbtSubMasterDemo c where rownum<20";
If you follow this solution, remeber to add a default constructor too (without parameters), so in your pojo you have:
public CrbtSubMasterDemo(String mobile, String password) {
this.mobile = mobile;
this.password = password
}
and
public CrbtSubMasterDemo() {
}
String hql = "select c.mobile, c.password FROM CrbtSubMasterDemo c where rownum<20";
Query query = session.createQuery(hql);
Result of this will be List<Object[]>
List<Object[]> itr = query.list();
for (Object[] row : itr) {
System.out.println(String.format("mobile:%s, password:%s", row[0], row[1]));
}
if mobile and password are strings, of course. You can use a transformer to transform results directly to CrbtSubMasterDemo.
Hibernate 3.2: Transformers for HQL and SQL
FluentHibernateResultTransformer
Sir, Many times user faces this kinda requirements . Hibernate has ResultTransformer to convert a hql/sql in Object.
public CrbtSubMasterDemo{
private Stirng mobile;
private String password;
public CrbtSubMasterDemo(){
--------------
}
#####after setting the transation set whichever columns you are selecting should be given as name of property of your object
String hql = "select c.mobile as mobile, c.password as password FROM CrbtSubMasterDemo c where rownum<20";
Query query = session.createQuery(hql);
List<CrbtSubMasterDemo> itr = query.setResultTransformer(Transformers.aliasToBean(CrbtSubMasterDemo.class) ).list();
##No need to commit the transaction.
}
It will convert you query into the CrbtSubMasterDemo
Do not directly cast the result of "query.list();" to List of CrbtSubMasterDemo. As query.list() return object list. Iterate over the object list received and cast one by one to put in list List of CrbtSubMasterDemo
I have a table by name Person and it has only one row data. I dont want to take it in list(). Can I get the data of Object type. Below is my code snippet
Session session = this.sessionFactory.openSession();
String sql = "SELECT * FROM PERSON";
SQLQuery query = session.createSQLQuery(sql);
query.addEntity(Person.class);
List<Person> list = query.list();
===================================
I want the above code snippet to work as below:
Session session = this.sessionFactory.openSession();
String sql = "SELECT * FROM PERSON";
SQLQuery query = session.createSQLQuery(sql);
query.addEntity(Person.class);
Person p = query;
which is throwing error. Please help me out on fixing the issue.
query is result of your session select. query.list() is your results. and you can get with query.list().get(0);
You can cast query list to personList;
List<Person> list = query.list();
return list.get(0);
You can just cast your result to person;
Person p = (Person) query.list().get(0);
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();