Jay*_*Jay 5 java generics methods design-patterns factory
我的代码如下所示:
public interface BaseDAO{
// marker interface
}
public interface CustomerDAO extends BaseDAO{
public void createCustomer();
public void deleteCustomer();
public Customer getCustomer(int id);
// etc
}
public abstract class DAOFactory {
public BaseDAO getCustomerDAO();
public static DAOFactory getInstance(){
if(system.getProperty("allowtest").equals("yes")) {
return new TestDAOFactory();
}
else return new ProdDAOFactory();
}
public class TestDAOFactory extends DAOFactory{
public BaseDAO getCustomerDAO() {
return new TestCustomerDAO(); // this is a concrete implementation
//that extends CustomerDAO
//and this implementation has dummy code on methods
}
public class ProdDAOFactory extends DAOFactory {
public BaseDAO getCustomerDAO() {
return new ProdCustomerDAO(); // this implementation would have
// code that would connect to the database and do some stuff..
}
}
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现在,我知道这段代码闻起来有很多原因.但是,此代码也在这里:http: //java.sun.com/blueprints/corej2eepatterns/Patterns/DataAccessObject.html,请参阅9.8
我打算这样做:1)根据环境(系统属性)在运行时切换我的DAO实现.2)使用java泛型,以便我可以避免类型转换...例如,做这样的事情:
CustomerDAO dao = factory.getCustomerDAO();
dao.getCustomer();
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相反:
CustomerDAO dao = (CustomerDAO) factory.getCustomerDAO();
dao.getCustomer();
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请你的意见和建议.
Mih*_*der 10
您应该像这样定义工厂:
public abstract class DAOFactory<DAO extends BaseDAO> {
public DAO getCustomerDAO();
public static <DAO extends BaseDAO> DAOFactory<DAO> getInstance(Class<DAO> typeToken){
// instantiate the the proper factory by using the typeToken.
if(system.getProperty("allowtest").equals("yes")) {
return new TestDAOFactory();
}
else return new ProdDAOFactory();
}
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getInstance应该返回一个正确的类型DAOFactory.
工厂变量将具有以下类型:
DAOFactory<CustomerDAO> factory = DAOFactory<CustomerDAO>.getInstance(CustomerDAO.class);
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并且将正确输入用法:
CustomerDAO dao = factory.getCustomerDAO();
dao.getCustomer();
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唯一的问题可能是getInstance方法中需要的强制转换.
有一堆文章详细说明了你的需求:
请注意,与您的示例不同,没有理由DAOFactory不应该返回实际的子类(即CustomerDAO getCustomerDAO()).此外,使用通用DAO的主要好处是使实体类型为"通用化",因此您不必使用load()/get()/find()类似方法进行转换.
您的示例并未证明需要BaseDAO,并且没有理由DAOFactory.getCustomerDAO()不应声明为什么不返回CustomerDAO.所以,我并不认为那里需要仿制药.但是,请考虑以下事项:
interface DataAccess<T> {
void store(T entity);
T lookup(Serialiable identifier);
void delete(Serializable identifier);
Collection<? extends T> find(Criteria query);
}
abstract class DataAccessFactory {
abstract DataAccess<T> getDataAccess(Class<T> clz);
static DataAccessFactory getInstance() {
...
}
}
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我在几个项目中使用了类似这种方法的东西,编写一个适用于模型中每个实体的DAO非常好.弱点是"发现者"方法.有一些巧妙的方法,JPA即将开展的工作是标准化"Criteria"API,但是现在通常最容易公开底层持久性机制的标准.