Question

我正在使用登录程序，用户使用staffID登录，我希望系统显示登录的员工姓名。

如何从staffID检索staffName？我已经尝试过，但我仍然得到staffName为null。

我的代码如下。前两个方法来自DA类，而最后一个方法是UI。

非常感谢任何帮助，在此先感谢：）

private static Staff convertToStaff(ResultSet rs) throws SQLException {
    Staff staff;
    int id = rs.getInt("id");
    String staffID = rs.getString("staffID");
    String staffName = rs.getString("staffName");
    String staffPassword = rs.getString("staffPassword");
    String staffNRIC = rs.getString("staffNRIC");
    String staffGender = rs.getString("staffGender");
    int staffContactNo = rs.getInt("staffContactNo");
    String staffEmail = rs.getString("staffEmail");
    String dob = rs.getString("dob");
    String department = rs.getString("department");
    staff = new Staff(id, staffID, staffName, staffNRIC, staffGender, staffContactNo, staffEmail, dob, department, staffPassword);

    return staff;
}

public static Staff retrieveNameByStaffID(String staffID) {

    ResultSet rs = null;
    DBController db = new DBController();
    String dbQuery;
    PreparedStatement pstmt;
    Staff staffName = null;

    // step 1 - connect to database
    db.getConnection();

    // step 2 - declare the SQL statement
    dbQuery = "SELECT staffName FROM staff WHERE staffID = ?";
    pstmt = db.getPreparedStatement(dbQuery);

    // step 3 - execute query
    try {

        pstmt.setString(1, staffID);
        rs = pstmt.executeQuery();
        if (rs.next()) { // first record found
            staffName = convertToStaff(rs);
        }
    } catch (Exception e) {
        e.printStackTrace();
    }
    db.terminate();
    return staffName;
}

private void actionPerformedStaffLogin() {
    String staffID = txtID.getText();
    String staffPassword = passwordField.getText();
    String adminID = txtID.getText();
    String adminPassword = passwordField.getText();
    Staff staffName = StaffDA.retrieveNameByStaffID(staffID);
    try {
                    //StaffDA.logIn(staffID, staffPassword);
        //StaffController.logIn(staffID, staffPassword);

        if (StaffController.logIn(staffID, staffPassword) == 1) {

            JOptionPane.showMessageDialog(null, "Welcome " + staffName);
            System.out.println(staffName);
            JPanel contentPane = new StaffMenu(HMSFrame);
            HMSFrame.setContentPane(contentPane);
            HMSFrame.setVisible(true);
        } else if (StaffController.adminlogIn(adminID, adminPassword) == 1) {
            JOptionPane.showMessageDialog(null, "Welcome " + adminID);
            JPanel contentPane = new AdminMenu(HMSFrame);
            HMSFrame.setContentPane(contentPane);
            HMSFrame.setVisible(true);
        } else if (StaffController.logIn(staffID, staffPassword) > 1) {

            JOptionPane.showMessageDialog(null, "duplicate");

        } else {
            JOptionPane.showMessageDialog(null, "ID and Password mismatch");
        }
    } catch (Exception ex) {
        JOptionPane.showMessageDialog(null, ex);
    }
}

Answer 1

更改此查询

dbQuery = "SELECT staffName FROM staff WHERE staffID = ?";

您只使用此查询获得staffName，但使用此

dbQuery = "SELECT * FROM staff WHERE staffID = ?";

你可以获得所有属性。

Answer 2

您的重写代码必须像这样

    private static Staff convertToStaff(ResultSet rs) throws SQLException {
    Staff staff;
    int id = rs.getInt("id");
    String staffID = rs.getString("staffID");
    String staffName = rs.getString("staffName");
    String staffPassword = rs.getString("staffPassword");
    String staffNRIC = rs.getString("staffNRIC");
    String staffGender = rs.getString("staffGender");
    int staffContactNo = rs.getInt("staffContactNo");
    String staffEmail = rs.getString("staffEmail");
    String dob = rs.getString("dob");
    String department = rs.getString("department");
    staff = new Staff(id, staffID, staffName, staffNRIC, staffGender, staffContactNo, staffEmail, dob, department, staffPassword);

    return staff;
}

public static Staff retrieveNameByStaffID(String staffID) {

    ResultSet rs = null;
    DBController db = new DBController();
    String dbQuery;
    PreparedStatement pstmt;
    Staff staffName = null;

    // step 1 - connect to database
    db.getConnection();

    // step 2 - declare the SQL statement
    dbQuery = "SELECT * FROM staff WHERE staffID = ?";
    pstmt = db.getPreparedStatement(dbQuery);

    // step 3 - execute query
    try {

        pstmt.setString(1, staffID);
        rs = pstmt.executeQuery();
        if (rs.next()) { // first record found
            staffName = convertToStaff(rs);
        }
    } catch (Exception e) {
        e.printStackTrace();
    }
    db.terminate();
    return staffName;
}

private void actionPerformedStaffLogin() {
    String staffID = txtID.getText();
    String staffPassword = passwordField.getText();
    String adminID = txtID.getText();
    String adminPassword = passwordField.getText();
    Staff staffName = StaffDA.retrieveNameByStaffID(staffID);
    try {
                    //StaffDA.logIn(staffID, staffPassword);
        //StaffController.logIn(staffID, staffPassword);

        if (StaffController.logIn(staffID, staffPassword) == 1) {

            JOptionPane.showMessageDialog(null, "Welcome " + staffName);
            System.out.println(staffName);
            JPanel contentPane = new StaffMenu(HMSFrame);
            HMSFrame.setContentPane(contentPane);
            HMSFrame.setVisible(true);
        } else if (StaffController.adminlogIn(adminID, adminPassword) == 1) {
            JOptionPane.showMessageDialog(null, "Welcome " + adminID);
            JPanel contentPane = new AdminMenu(HMSFrame);
            HMSFrame.setContentPane(contentPane);
            HMSFrame.setVisible(true);
        } else if (StaffController.logIn(staffID, staffPassword) > 1) {

            JOptionPane.showMessageDialog(null, "duplicate");

        } else {
            JOptionPane.showMessageDialog(null, "ID and Password mismatch");
        }
    } catch (Exception ex) {
        JOptionPane.showMessageDialog(null, ex);
    }
}

Java jbdc从mysql中检索特定数据

2 个答案: