代码如下.........
private List<EmployeeAllRec> listg;
private List<Employee> list;
private List<Employee> gridModel;
private Map<String, String> json;
public String showAllRecord() {
records = 30;
rows = 10;
Session session = HibernateUtil.getSessionFactory().openSession();
Transaction transaction = null;
EmployeeAllRec rs = null;
try {
transaction = session.beginTransaction();
listg = new ArrayList();
List emp = session.createQuery("from Employee e").list();
int c=0;
//code for adding the data into the list
for (Iterator iterator = emp.iterator(); iterator.hasNext();) {
Employee e1 = (Employee) iterator.next();
System.out.println(e1.getName());
rs = new EmployeeAllRec();
rs.setName(e1.getName());
rs.setEmail(e1.getEmail());
rs.setDob(e1.getDob());
rs.setAddress(e1.getAddress());
rs.setGender(e1.getGender());
rs.setAge(e1.getAge());
rs.setCountry(e1.getCountry());
rs.setContact(e1.getContact());
rs.setWebsite(e1.getWebsite());
System.out.println("&&&&&&&&&&&&&&&&---i m on it " + rs.getName());
listg.add(rs);
}
setGridModel(listg);
// some stuff
我的问题是我需要先将“listg”变量按降序排序,然后再将其添加到变量setGridModel ....
答案 0 :(得分:1)
由于添加到listg中的对象是从DB中获取的,我建议通过HQL查询对它进行排序.DB排序比内存排序(使用比较器)更快,在这种情况下,您可以应用方法。
因此,HQL查询看起来像“FROM Employee e ORDER BY ... DESC”
答案 1 :(得分:0)
PS:如果可能,请考虑在您的馆藏中使用泛型。