Java，如果可以，否則不沒有錯誤消息

Question

我在啟動else語句時遇到麻煩。此代碼位於GUI中，在用戶名在文本字段中輸入后，該GUI會向用戶發送電子郵件。 如果在SQL數據庫中找到它，它將檢索電子郵件地址，一切正常。 如果輸入的用戶名不在數據庫中，則不會執行任何操作，我不會收到錯誤消息，它只會掛起。 這是提交按鈕中的代碼

jButtonSubmit.addActionListener(new ActionListener() {
        @Override
        public void actionPerformed(ActionEvent evt) 
        {
           try 
             {
             username = (jTextFieldUsername.getText());

             connection = DriverManager.getConnection(DATABASE_URL, USERNAME, PASSWORD);
             query = "SELECT User_userName, User_email FROM user_details WHERE " +
                     "User_userName = '"+username+"'";

             PreparedStatement stm = connection.prepareStatement(query);
             rs =stm.executeQuery();
             while (rs.next()) 
             {  
                String eMail = rs.getString ("User_email");

                if (username.equals(rs.getString("User_userName"))) 
                {  
                    JOptionPane.showMessageDialog(frame, "An email with your password "
                            + "has been sent \nto " + eMail); 
                }// end if

                else 
                {
                 JOptionPane.showMessageDialog(frame, "Incorrect username.  If "
                         + "you are unable to login\n"
                         + "with your current username, please contact us on\n"
                         + "info@somewhere.com.au."); 
                }//end else
            } //end while  
          close();
        }//end try  //catch statement follows

期待中……

謝謝大家的幫助。 現在它可以正常工作，您已經有效地解決了我的另一個問題。 這是最終代碼。 再次感謝。

int count = 0;

while（rs.next（））{

   eMail = rs.getString ("User_email");
   count++;}

        if (count != 0) 
        {        
         JOptionPane.showMessageDialog(frame, "An email with your password "
                 + "has been sent \nto " + eMail); 
        }// end if

        else 
        {
            JOptionPane.showMessageDialog(frame, "Incorrect username.  If "
                 + "you are unable to login\n"
                 + "with your current username, please contact us on\n"
                 + "info@somewhere.com.au."); 
        }//end else

           // } //end while  

Answer 1

我認為無需在while循環中檢查用戶名是否相等，因為您已經使用該名稱進行了查詢。

int count=0;
while (rs.next()) 
         {
     count++;
}

 if (count>0) 
            {  
                JOptionPane.showMessageDialog(frame, "An email with your password "
                        + "has been sent \nto " + eMail); 
            }// end if

            else 
            {
             JOptionPane.showMessageDialog(frame, "Incorrect username.  If "
                     + "you are unable to login\n"
                     + "with your current username, please contact us on\n"
                     + "info@somewhere.com.au."); 
            }//end else

Answer 2

不是完整的代碼...而是您想要的代碼！

query = "SELECT User_userName, User_email FROM user_details WHERE User_userName = ?";

                 PreparedStatement stm = connection.prepareStatement(query);
                 stm.setString(1, username);
                 rs =stm.executeQuery();
..
..
//And to check if rs is null
if(rs!=null){
   while(rs.next()){
     ....
    }
}
else{
 JOptionPane.showMessageDialog(frame, "Incorrect username.  If "
                             + "you are unable to login\n"
                             + "with your current username, please contact us on\n"
                             + "info@somewhere.com.au."); 
}