Question

我有三顿饭，费用，用户！我想从这些表中计算出每个用户的总费用，总费用！在一个查询中！我在计算总费用时遇到问题！

totalcost = meal rate *每位用户的总餐食

（餐费=所有用户费用之和/所有用户膳食之和）！

当我朗读下面的查询时，它仅返回一行！我尝试了group by，但是它给了我错误的输出结果：(有人可以帮我吗？

 {
    SELECT u.name , 
           ((SUM(tot_expenseamount)/ SUM(tot_noofmeal))*tot_noofmeal) as total_Cost,
           IFNULL(t.tot_expenseamount,0) AS tot_expense_amount ,
           IFNULL(n.tot_noofmeal,0) AS tot_no_of_meal 
    FROM users u
    LEFT JOIN (
        SELECT e.user_id , SUM(e.expenseamount) AS tot_expenseamount 
        FROM expenses e 
        WHERE e.expensedate >= '2019-04-01' + INTERVAL 0 MONTH AND e.expensedate < '2019-04-01' + INTERVAL 1 MONTH
        GROUP BY e.user_id ) t ON t.user_id = u.id
    LEFT JOIN (
        SELECT m.user_id , SUM(m.noofmeal) AS tot_noofmeal 
        FROM meals m 
        WHERE m.mealdate >= '2019-04-01' + INTERVAL 0 MONTH AND m.mealdate < '2019-04-01' + INTERVAL 1 MONTH
        GROUP BY m.user_id ) n ON n.user_id = u.id 
    ORDER BY u.name

Answer 1

我替你改了。

如果您使用的是MySQL 8.0.2 +：

SELECT u.name 
     , (expense.TotalExpenseAmount / SUM(MealCount) OVER ()) * MealsPerUser.MealCount
  FROM users u
 CROSS 
  JOIN (SELECT SUM(e.expenseamount) AS TotalExpenseAmount 
          FROM expenses e 
         WHERE e.expensedate >= '2019-04-01' + INTERVAL 0 MONTH AND e.expensedate < '2019-04-01' + INTERVAL 1 MONTH
     ) expense
  LEFT 
  JOIN (SELECT m.user_id, SUM(m.noofmeal) AS MealCount 
          FROM meals m 
         WHERE m.mealdate >= '2019-04-01' + INTERVAL 0 MONTH AND m.mealdate < '2019-04-01' + INTERVAL 1 MONTH
         GROUP BY m.user_id
     ) MealsPerUser
    ON MealsPerUser.user_id = u.id 
 ORDER BY u.name

如果没有，它将变得更加复杂：

SELECT u.name 
     , (expense.TotalExpenseAmount / meals.TotalMealCount) * MealsPerUser.MealCount
  FROM users u
 CROSS 
  JOIN (SELECT SUM(e.expenseamount) AS TotalExpenseAmount 
          FROM expenses e 
         WHERE e.expensedate >= '2019-04-01' + INTERVAL 0 MONTH AND e.expensedate < '2019-04-01' + INTERVAL 1 MONTH
     ) expense
 CROSS
  JOIN (SELECT SUM(m.noofmeal) AS TotalMealCount
          FROM meals m 
         WHERE m.mealdate >= '2019-04-01' + INTERVAL 0 MONTH AND m.mealdate < '2019-04-01' + INTERVAL 1 MONTH
     ) meals
  LEFT 
  JOIN (SELECT m.user_id, SUM(m.noofmeal) AS MealCount 
          FROM meals m 
         WHERE m.mealdate >= '2019-04-01' + INTERVAL 0 MONTH AND m.mealdate < '2019-04-01' + INTERVAL 1 MONTH
         GROUP BY m.user_id
     ) MealsPerUser
    ON MealsPerUser.user_id = u.id 
 ORDER BY u.name

仅返回一行！但所有用户的预期输出

1 个答案: