Question

我班上有这个方法，

 public static AccountHolder getAccountHolder(String username, String password) {

        try {
            String verifyQuery = "SELECT COUNT(username) FROM accountholder WHERE username=? AND password=?)";
            connection = getConnection();
            ptmt = connection.prepareStatement(verifyQuery);

            ptmt.setString(4, username);
            ptmt.setString(5, password);

            ResultSet verified = ptmt.executeQuery();

            if (verified.next()) {
                // Only one Result
                if (verified.getInt(1) == 1) {
                    connection.close();
                    return 1;
                } else {
                    connection.close();
                    return null;
                }

            }
        } catch (Exception ex) {
        }
    }

我知道我应该返回AccountHolder类型的对象，它是一个类。 AccountHolder类代码如下：

public class AccountHolder {
   private String firstname;
   private String lastname;
   private String username;
   private String password;
   private String address;
   private String phonenumber;

  // get & set methods();
}

我应该返回匹配的用户名＆amp;使用该方法从数据库中获取密码我怎样才能做到这一点？

Answer 1

您需要将作为答案获得的ResultSet元素转换为AccountHolder的对象，并将其返回而不是行return 1;。

Answer 2

尝试更改代码

PreparedStatement stmt = conn.prepareStatement(
        "SELECT USERNAME FROM accountholder 
                             WHERE USERNAME = ? AND PASSWORD = ?");
stmt.setString(1, username);
stmt.setString(2, password);
ResultSet rs = stmt.executeQuery();
if (rs.next()) {
    // valid credentials
} else {
    // invalid credentials
}

Answer 3

您可以采用一种方法，如果找到针对输入用户名和密码的匹配记录，则返回具体的AccountHolder对象，否则为null。

尝试这样的方法，根据您的架构更改列名：

public static AccountHolder getAccountHolder（String username，String password）{

    AccountHolder accountHolder = null;
    try {
        String verifyQuery = "SELECT * FROM accountholder WHERE username=? AND password=?)";
        connection = getConnection();
        ptmt = connection.prepareStatement(verifyQuery);

        ptmt.setString(1, username);
        ptmt.setString(2, password);

        ResultSet verified = ptmt.executeQuery();

        if (verified.next()) {
           accountHolder = new AccountHolder();
           accountHolder.setFirstName(rs.getString("FIRSTNAME"));
           accountHolder.setLastName(rs.getString("LASTNAME"));
           accountHolder.setUsername(rs.getString("USERNAME"));
           accountHolder.setPassword(rs.getString("PASSWORD"));
           accountHolder.setAddress(rs.getString("ADDRESS"));
           accountHolder.setPhoneNumber(rs.getString("PHONENUMBER"));

        }
    } catch (Exception ex) {
    } finally {
                connection.close();
    }
    return accountHolder ;
}

Answer 4

Change that prepared statement query like this

pm.setString(1,username);
pm.setString(2,password);

and

assign the setters like this
accountholder.setUserName(verified.getString(1));
accountholder.setpassword(verified.getString(2));


get those values in array or any collection for retrieval.

不兼容的类型。必需的类名。找到int。

4 个答案: