Question

当使用hibernate从数据库填充表时，理想情况下想要使用最少量的查询，以便他们的应用程序很快。理想情况下，您不希望进行比正在构建的表中的列更多的查询，因为按查询分组将允许您填充所有行。因此，如果您有m列和n行，则只需执行m次查询。

但是我发现很难在没有编写新方法的情况下在我的Daos中实现这一点。假设您正在处理一个名为Employee的实体，并且您想计算平均工作小时数，那么您将需要2种方法：

public interface EmployeesDao
{
    public List<Double> averageOfHoursWorked(List<Integer> ids);
    public double averageOfHoursWorked(Integer id);
}

因此，当您想要创建表时，您将使用接受列表的方法，但是当您只处理单个员工时，您将使用带有单个ID的那个。

但是我不想写2个方法，我只想写一个理想的方法。但是如果我写一个方法，我最终会使用一个for循环，它会产生一大堆查询而不只是一个：

    // THIS IS BAD, creates many queries
    List<Integer> ids = Arrays.asList(1,3,5,99, 1921);
    for (Integer id : ids) {
        double avg = dao.averageOfHoursWorked(id);
        // .... doing some table stuff with avg...
    }

    // THIS IS GOOD, only 1 query, yet now I must write more code ):
    List<Double> avgs = dao.averageOfHoursWorked(ids);
    for (Double avg : avgs) {
        // .... doing some table stuff with avg...
    }

所以我知道第二个版本要快得多，但是编码需要更长的时间。但是当我写这篇文章时，我想我可以这样做：

class EmployeesDaoImp implements EmployeesDao {

    @Override
    public double averageOfHoursWorked(Integer id) {
        return averageOfHoursWorked(Arrays.asList(id)).get(0);
    }
}

然后我只需要写List<Double> averageOfHoursWorked(List<Integer> ids)，这是个好主意吗？有人用hibernate制作表格更优雅吗？也许建筑师模式或其他东西会更好：

Table table = dao.buildTable().withIds(ids).columnId().columnAvgHoursWorked().columnDaysOff()

其中columnId（）会给你一列id，columnAvgHoursWorked会给你另一个列有工作小时数的列，依此类推。使用hibernate创建表的最佳方法是什么，需要最少量的查询和代码。

Answer 1

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public abstract class GenericDAOImpl<T> implements GenericDAO<T>{

       private Class<T> entityBeanType;

       @Autowired
       protected SessionFactory sessionFactory;

       public GenericDAOImpl() {

          ParameterizedType genericSuperclass = (ParameterizedType) getClass().getGenericSuperclass();
           Type type = genericSuperclass.getActualTypeArguments()[0];

           @SuppressWarnings("unchecked")
           Class<T> clazz = (Class<T>) type;

           this.entityBeanType = clazz;
       }

       @SuppressWarnings("unchecked")
       public T save(T t) {
            T t1 = (T) sessionFactory.getCurrentSession().merge(t);
            sessionFactory.getCurrentSession().flush();
           return t1;
       }
    }

</code>



Repair that if you need to create another table, you will can extend this class to create a concrete DAO.

Example:

1 - create a table User and use the following code:



public class UserDAOImpl extends GenericDAOImpl<User> implements UserDAO{
    ///other specific methods will be here
}




2 - create a table Receipts and use the following code:



public class ReceiptsDAOImpl extends GenericDAOImpl<Receipts> implements     ReceiptsDAO{
  ///other specific methods will be here
}

泛型类型T可以是您软件的某个实体。

我希望我的回答可以帮助你。

Answer 2

起初，对不起我糟糕的英语。我认为可以创建一个将成为数据对象的DTO对象，并且将包含要在JSP中使用的所有信息，您将把这些对象作为一个唯一方法的返回，并在一个唯一的调用中解决您的问题。

例如：

public List getAllEmployeesWithAverageHoursWorked(int start, int finish){



List<EmployeeDTO> result = new ArrayList<EmployeeDTO>();
///execute query that load registers that start in parameter start and finishes in parameter finish

return result;

List<EmployeeDTO> result = new ArrayList<EmployeeDTO>(); ///execute query that load registers that start in parameter start and finishes in parameter finish return result;

EmployeeDTO类将包含您需要在JSP页面中使用的所有字段，或者，如果此选项不能解决此问题，您将更改方法和类EmployeeDTO以获得所需的字段。

Answer 3

我不明白你的问题，但我会给出三种形式，我认为可以提供帮助。

1 - 使用泛型并对软件中的所有类型的对象使用相同的DAO。

2 - 在我的回答中可以指定你需要阅读的起点和终点，例如：你读了10条记录，现在需要阅读超过，然后你传递10和20，并阅读更多10

3 - 在第一次调用中，您将把所需的记录放在静态变量中并使用（这很可怕）。

如果你的问题不是“写”代码，使用泛型，那么解决这个问题是非常好的。

但是，如果我的答案不正确，请再次提出问题，但换句话说，我可以帮助你。

减少hibernate所需的查询数量

3 个答案: