[Ljava.lang.Object; cannot be cast to

Question

I want to get value from the database, in my case I use List to get the value from the database but I got this error

Exception in thread "main" java.lang.ClassCastException: [Ljava.lang.Object; cannot be cast to id.co.bni.switcherservice.model.SwitcherServiceSource
at id.co.bni.switcherservice.controller.SwitcherServiceController.LoadData(SwitcherServiceController.java:48)
at id.co.bni.switcherservice.controller.SwitcherServiceController.main(SwitcherServiceController.java:62)

this is my code

    Query LoadSource = session_source.createQuery("select CLIENT,SERVICE,SERVICE_TYPE,PROVIDER_CODE,COUNT(*) FROM SwitcherServiceSource" +
            " where TIMESTAMP between :awal and :akhir" +
            " and PROVIDER_CODE is not null group by CLIENT,SERVICE,SERVICE_TYPE,PROVIDER_CODE order by CLIENT,SERVICE,SERVICE_TYPE,PROVIDER_CODE");
    LoadSource.setParameter("awal", fromDate);
    LoadSource.setParameter("akhir", toDate);

    List<SwitcherServiceSource> result_source = (List<SwitcherServiceSource>) LoadSource.list();
    for(SwitcherServiceSource tes : result_source){
        System.out.println(tes.getSERVICE());
    }

any help will be pleasure :)

@raffian, did you mean like this??

List<Switcher> result = (List<Switcher>) LoadSource.list();
for(Switcher tes : result){
    System.out.println(tes.getSERVICE());
}

Answer 1

Your query execution will return list of Object[].

List result_source = LoadSource.list();
for(Object[] objA : result_source) {
    // read it all
}

Answer 2

You need to add query.addEntity(SwitcherServiceSource.class) before calling the .list() on query.

Answer 3

java.lang.ClassCastException: [Ljava.lang.Object; cannot be cast to id.co.bni.switcherservice.model.SwitcherServiceSource

Problem is

(List<SwitcherServiceSource>) LoadSource.list();

This will return a List of Object arrays (Object[]) with scalar values for each column in the SwitcherServiceSource table. Hibernate will use ResultSetMetadata to deduce the actual order and types of the returned scalar values.

Solution

List<Object> result = (List<Object>) LoadSource.list(); 
Iterator itr = result.iterator();
while(itr.hasNext()){
   Object[] obj = (Object[]) itr.next();
   //now you have one array of Object for each row
   String client = String.valueOf(obj[0]); // don't know the type of column CLIENT assuming String 
   Integer service = Integer.parseInt(String.valueOf(obj[1])); //SERVICE assumed as int
   //same way for all obj[2], obj[3], obj[4]
}

Related link

• Using iterator

Answer 4

I've faced such an issue and dig tones of material. So, to avoid ugly iteration you can simply tune your hql:

You need to frame your query like this

select entity from Entity as entity where ...

Also check such case, it perfectly works for me:

public List<User> findByRole(String role) {

    Query query = sessionFactory.getCurrentSession().createQuery("select user from User user join user.userRoles where role_name=:role_name");
    query.setString("role_name", role);
    @SuppressWarnings("unchecked")
    List<User> users = (List<User>) query.list();
    return users;
}

So here we are extracting object from query, not a bunch of fields. Also it's looks much more pretty.