用于选择多个值的Hibernate查询

在hibernate中，我可以做到以下

Query q = session.createQuery("from Employee as e); List emps = q.list(); 

现在如果我想获取int和String我该怎么办？

 Query q = session.createQuery(""SELECT E.firstName,E.ID FROM Employee E"; List ans = q.list(); 

现在列表的结构是什么？

Answer 1

这可以。 您需要了解的是它将返回Object []列表，如下所示：

  Query q = session.createQuery("select e.id, e.firstName from Employee e"); List employees= (List)q.list(); for(Object[] employee: employees){ Integer id = (Integer)employee[0]; String firstName = (String)employee[1]; ..... } 

Answer 2

你将得到一个Object数组列表（每个都有两个元素）

 List< Object[] > employees = q.list(); for ( Object[] employee : employees ) { // employee[0] will contain the first name // employee[1] will contail the ID } 

Answer 3

您应该使用新对象来保存这些值，如下所示：

 "SELECT NEW EmpMenu(e.name, e.department.name) " + "FROM Project p JOIN p.students e " + "WHERE p.name = :project " + "ORDER BY e.name").setParameter("project", projectName).getResultList() 

我从http://www.java2s.com/Tutorial/Java/0355__JPA/EJBQLCreatenewObjectInSelectStatement.htm得到了这个例子。

Answer 4

 List is the structure. 

所以你得到这样的每个元素：

 List ans = q.list(); for(Object[] array : ans) { String firstName = (String) array[0]; Integer id = (Integer) array[1]; } 

Answer 5

 Query qry=session.createQuery("select e.employeeId,e.employeeName from Employee e where e.deptNumber=:p1"); qry.setParameter("p1",30); List l2=qry.list(); Iterator itr=l2.iterator(); while(itr.hasNext()){ Object a[]=(Object[])itr.next(); System.out.println(a[0]+"/t"a[1]); } 

Answer 6

没有迭代器：

 @SuppressWarnings( "unchecked" ) public List findByDepartment(long departmentId){ SQLQuery query = session.createSQLQuery("SELECT {emp.*} " + " FROM employee emp " + +"WHERE emp.department_id = :departement_id"); query.setLong("department_id", departmentId); query.addEntity("emp", Employee.class); return (List) = query.list(); }