如何修复SQL分组依据并获得预期结果？

Question

我在PostgreSQL中进行此查询时遇到了麻烦。我按分组进行，它给了我适当的结果，但是我找不到我真正想要的具体解决方案。已返回两列。但是我想要此结果的total_amount的总和并仅返回一列

SELECT pi.user_id, 
       Sum(total_amount) AS total_amount, 
       emp_id, 
       status, 
       max_pi_pending_amount, 
       first_pending_pi_warning_percentage, 
       second_pending_pi_warning_percentage, 
       emp_id, 
       Coalesce(pending_pi_amount, 0) AS pending_pi_amount 
FROM   proforma_invoice pi 
       JOIN users usr 
         ON pi.user_id = usr.user_id 
       JOIN employee emp 
         ON usr.employee_id = emp.emp_id 
       JOIN alarm_parameter ap 
         ON pi.user_id = ap.user_id 
WHERE  pi.user_id = 18 
       AND (status = 'Created' 
        OR status = 'Approved')
GROUP  BY pi.user_id, 
          status, 
          max_pi_pending_amount, 
          first_pending_pi_warning_percentage, 
          second_pending_pi_warning_percentage, 
          emp_id, 
          pending_pi_amount 

Answer 1

只需将状态列从查询中删除，因为如果将它们组合在一起，则该列无效

DetailView

Answer 2

在我看来您需要子查询或cte

   with cte as
    (
    SELECT pi.user_id, 
           Sum(total_amount) AS total_amount, 
           emp_id, 
           status, 
           max_pi_pending_amount, 
           first_pending_pi_warning_percentage, 
           second_pending_pi_warning_percentage, 
           emp_id, 
           Coalesce(pending_pi_amount, 0) AS pending_pi_amount 
    FROM   proforma_invoice pi 
           JOIN users usr 
             ON pi.user_id = usr.user_id 
           JOIN employee emp 
             ON usr.employee_id = emp.emp_id 
           JOIN alarm_parameter ap 
             ON pi.user_id = ap.user_id 
    WHERE  pi.user_id = 18 
           AND (status = 'Created' 
            OR status = 'Approved')
    GROUP  BY pi.user_id, 
              status, 
              max_pi_pending_amount, 
              first_pending_pi_warning_percentage, 
              second_pending_pi_warning_percentage, 
              emp_id, 
              pending_pi_amount
    ) select user_id,sum(total_amount) from cte group by user_id