java中的泛型DAO
发布时间:2020-05-25 13:36:49 所属栏目:Java 来源:互联网
导读:我正在尝试在 java中开发通用DAO.我尝试了以下内容.这是 实现通用DAO的好方法?我不想使用hibernate.我试图使它尽可能通用,这样我就不必一遍又一遍地重复相同的代码. public abstract class AbstractDAOT { protected ResultSet findbyId(String tablename, I
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我正在尝试在
java中开发通用DAO.我尝试了以下内容.这是
public abstract class AbstractDAO<T> {
protected ResultSet findbyId(String tablename,Integer id){
ResultSet rs= null;
try {
// the following lines are not working
pStmt = cn.prepareStatement("SELECT * FROM "+ tablename+ "WHERE id = ?");
pStmt.setInt(1,id);
rs = pStmt.executeQuery();
} catch (SQLException ex) {
System.out.println("ERROR in findbyid " +ex.getMessage() +ex.getCause());
ex.printStackTrace();
}finally{
return rs;
}
}
}
我现在有: public class UserDAO extends AbstractDAO<User>{
public List<User> findbyid(int id){
Resultset rs =findbyid("USERS",id) // "USERS" is table name in DB
List<Users> users = convertToList(rs);
return users;
}
private List<User> convertToList(ResultSet rs) {
List<User> userList= new ArrayList();
User user= new User();;
try {
while (rs.next()) {
user.setId(rs.getInt("id"));
user.setUsername(rs.getString("username"));
user.setFname(rs.getString("fname"));
user.setLname(rs.getString("lname"));
user.setUsertype(rs.getInt("usertype"));
user.setPasswd(rs.getString("passwd"));
userList.add(user);
}
} catch (SQLException ex) {
Logger.getLogger(UserDAO.class.getName()).log(Level.SEVERE,null,ex);
}
return userList;
}
}
解决方法如果你能使用Spring,我会建议以下改进:>让Spring进行异常处理. 独立于使用Spring,我将推荐以下内容: >不要将表名作为参数发送.这应该在初始化阶段完成. 一个改进的AbstractDao与Spring: import java.util.Collection;
import org.springframework.jdbc.core.JdbcTemplate;
import org.springframework.jdbc.core.RowMapper;
public abstract class AbstractDao<T> {
protected final RowMapper<T> rowMapper;
protected final String findByIdSql;
protected final JdbcTemplate jdbcTemplate;
protected AbstractDao(RowMapper<T> rowMapper,String tableName,JdbcTemplate jdbcTemplate) {
this.rowMapper = rowMapper;
this.findByIdSql = "SELECT * FROM " + tableName + "WHERE id = ?";
this.jdbcTemplate = jdbcTemplate;
}
public Collection<T> findById(final String id) {
Object[] params = {id};
return jdbcTemplate.query(findByIdSql,params,rowMapper);
}
}
如您所见,没有异常处理或使用原始SQL类进行黑客攻击.此模板为您关闭ResultSet,我在您的代码中看不到. 而UserDao: import java.sql.ResultSet;
import java.sql.SQLException;
import org.springframework.jdbc.core.JdbcTemplate;
import org.springframework.jdbc.core.RowMapper;
public class UserDao extends AbstractDao<User> {
private final static String TABLE_NAME = "USERS";
public UserDao(JdbcTemplate jdbcTemplate) {
super(new UserRowMapper(),TABLE_NAME,jdbcTemplate);
}
private static class UserRowMapper implements RowMapper<User> {
public User mapRow(ResultSet rs,int rowNum) throws SQLException {
User user = new User();
user.setUserName(rs.getString("username"));
user.setFirstName(rs.getString("fname"));
user.setLastName(rs.getString("lname"));
return user;
}
}
}
更新: 当您知道id和id对应于数据库中的单个行时,您应该考虑返回通用对象而不是集合. public T findUniqueObjectById(final String id) {
Object[] params = {id};
return jdbcTemplate.queryForObject(findByIdSql,rowMapper);
}
这使您的服务代码更具可读性,因为您不需要从列表中检索用户,而只需: User user = userDao.findUniqueObjectById("22"); (编辑:安卓应用网) 【声明】本站内容均来自网络,其相关言论仅代表作者个人观点,不代表本站立场。若无意侵犯到您的权利,请及时与联系站长删除相关内容! |
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