59

I am using Hibernate 4 and would like to simply list all rows of a table. All solutions I found suggest using something like "from tablename", but I would like to avoid hardcoding tablenames in strings.

Marcus Riemer
  • 7,244
  • 8
  • 51
  • 76

3 Answers3

136

You can use

session.createCriteria(MyEntity.class).list();

for example.

mrab
  • 2,702
  • 1
  • 17
  • 10
8

HQL doesn't use table names. It uses entity names. And entity names are (by default) class names. So you can use

String hql = "select a from " + TheEntity.class.getSimpleName() + " a";

But I would favor readability over type safety here, and use

String hql = "select a from TheEntity a";

You should have automated tests for your queries anyway.

JB Nizet
  • 678,734
  • 91
  • 1,224
  • 1,255
-2

//Hibernate Class

public class CommonDAO<T> {

 Session session = null;
 Transaction transaction = null;
 private Class<T> clazz;

 public CommonDAO(){ //constructor
    session = HibernateUtil.getSessionFactory().openSession();
    transaction = session.beginTransaction();
    Type genericSuperclass = this.getClass().getGenericSuperclass();
    if (genericSuperclass instanceof ParameterizedType) {
      ParameterizedType pt = (ParameterizedType) genericSuperclass;
      Type type = pt.getActualTypeArguments()[0];
      clazz = (Class<T>) type;
    }
 }

 public T listById(Object id) {
  T object = null;
  try {
   object = (T) session.get(clazz, (Serializable) id);
  } catch (Exception e) {
   e.printStackTrace();
  }
  return object;
 }
}

//User Class

public class UserDAO extends CommonDAO<UserMaster> { // Here UserMaster is pojo

 public UserDAO() {
  super();
 }

 public static void main(String ar[]) {
  UserMaster user = new UserDAO().listById(1); // 1 is id
  System.out.println(user.getUserName());
 }
}
Manish
  • 25
  • 1
  • 6