Question

我需要为我的查询总结一个组。类似的东西：

SELECT customer_id, SUM(weekly) FROM earnings GROUP BY customer_id

上述工作，但我需要更复杂的东西，我的查询是这样的：

SELECT
    (SELECT SUM(weekly) FROM earnings) + 
    (SELECT SUM(day_earnings) FROM earnings) / .75

以上是所有收入的总和，每周和每日，但我需要在第一个示例中按customer_id like对它们进行分组。我如何实现这一目标？

Answer 1

根据您在上面尝试的操作顺序，只有SUM(day_earnings)除以.75。如果它应该是每周＆amp; day_earnings汇总在一起，然后将括号括在(SUM(weekly) + SUM(day_earnings)) / 0.75 AS earnings。

SELECT 
  customer_id,
  customer_name,
  SUM(weekly) + SUM(day_earnings) / 0.75 AS earnings
FROM earnings JOIN customers ON earnings.customer_id = customers.customer_id
GROUP BY customer_id, customer_name

Answer 2

这有用吗？

SELECT customer_id, SUM(weekly + (day_earnings / .75)) 
FROM earnings 
GROUP BY customer_id

Answer 3

SELECT 
  customer_id, 
  SUM(weekly) weekly, 
  SUM(day_earnings) daily, 
  ((SUM(weekly) + SUM(day_earnings)) / .75) calc
FROM 
  earnings 
GROUP BY 
  customer_id

Answer 4

你不能这样做：

SELECT customer_id, (SUM(weekly) + SUM(day_earnings) / .75) AS earnSum FROM earnings GROUP BY customer_id

如何在mySQL中使用SUM？

4 个答案: