MySQL 销售数据组（按月平均）答案

【问题标题】：MySQL sales data group by monthly averageMySQL 销售数据组（按月平均）
【发布时间】：2016-10-14 17:59:29
【问题描述】：

我有一个包含以下列的销售表：

mysql> select * from sales;
+-------------+--------+------------+
| customer_id | amount | date       |
+-------------+--------+------------+
|           1 |     12 | 2015-01-01 |
|           1 |      1 | 2015-01-02 |
|           1 |    663 | 2015-02-12 |
|           2 |     22 | 2015-01-03 |
|           2 |     21 | 2015-02-12 |
|           2 |     11 | 2015-02-12 |
|           2 |      9 | 2015-04-12 |
+-------------+--------+------------+

我可以使用这个查询来接近我想要的：

SELECT
  customer_id,
  sum(if(month(date) = 1, amount, 0))  AS Jan,
  sum(if(month(date) = 2, amount, 0))  AS Feb,
  sum(if(month(date) = 3, amount, 0))  AS Mar,
  sum(if(month(date) = 4, amount, 0))  AS Apr,
  sum(if(month(date) = 5, amount, 0))  AS May,
  sum(if(month(date) = 6, amount, 0))  AS Jun,
  sum(if(month(date) = 7, amount, 0))  AS Jul,
  sum(if(month(date) = 8, amount, 0))  AS Aug,
  sum(if(month(date) = 9, amount, 0))  AS Sep,
  sum(if(month(date) = 10, amount, 0)) AS Oct,
  sum(if(month(date) = 11, amount, 0)) AS Nov,
  sum(if(month(date) = 12, amount, 0)) AS `Dec`
FROM sales
GROUP BY customer_id;

所需的输出格式：

+-------------+------+------+------+------+------+------+------+------+------+------+------+------+
| customer_id | Jan  | Feb  | Mar  | Apr  | May  | Jun  | Jul  | Aug  | Sep  | Oct  | Nov  | Dec  |
+-------------+------+------+------+------+------+------+------+------+------+------+------+------+
|           1 |   13 |  663 |    0 |    0 |    0 |    0 |    0 |    0 |    0 |    0 |    0 |    0 |
|           2 |   22 |   32 |    0 |    9 |    0 |    0 |    0 |    0 |    0 |    0 |    0 |    0 |
+-------------+------+------+------+------+------+------+------+------+------+------+------+------+

但是我想要每月的平均值，当我将 sum 更改为 avg 时，我没有得到平均值。

SELECT
  customer_id,
  avg(if(month(date) = 1, amount, 0))  AS Jan,
  avg(if(month(date) = 2, amount, 0))  AS Feb,
  avg(if(month(date) = 3, amount, 0))  AS Mar,
  avg(if(month(date) = 4, amount, 0))  AS Apr,
  avg(if(month(date) = 5, amount, 0))  AS May,
  avg(if(month(date) = 6, amount, 0))  AS Jun,
  avg(if(month(date) = 7, amount, 0))  AS Jul,
  avg(if(month(date) = 8, amount, 0))  AS Aug,
  avg(if(month(date) = 9, amount, 0))  AS Sep,
  avg(if(month(date) = 10, amount, 0)) AS Oct,
  avg(if(month(date) = 11, amount, 0)) AS Nov,
  avg(if(month(date) = 12, amount, 0)) AS `Dec`
FROM sales
GROUP BY customer_id;

产出（不是每月平均）：

+-------------+--------+----------+--------+--------+--------+--------+--------+--------+--------+--------+--------+--------+
| customer_id | Jan    | Feb      | Mar    | Apr    | May    | Jun    | Jul    | Aug    | Sep    | Oct    | Nov    | Dec    |
+-------------+--------+----------+--------+--------+--------+--------+--------+--------+--------+--------+--------+--------+
|           1 | 4.3333 | 221.0000 | 0.0000 | 0.0000 | 0.0000 | 0.0000 | 0.0000 | 0.0000 | 0.0000 | 0.0000 | 0.0000 | 0.0000 |
|           2 | 5.5000 |   8.0000 | 0.0000 | 2.2500 | 0.0000 | 0.0000 | 0.0000 | 0.0000 | 0.0000 | 0.0000 | 0.0000 | 0.0000 |
+-------------+--------+----------+--------+--------+--------+--------+--------+--------+--------+--------+--------+--------+

我有点想弄清楚发生了什么，有人提供一些建议吗？谢谢

CREATE TABLE `sales` (
  `customer_id` int(11) NOT NULL,
  `amount` int(11) DEFAULT NULL,
  `date` date DEFAULT NULL
) 

INSERT INTO `sales` VALUES (1,12,'2015-01-01'),(1,1,'2015-01-02'),(1,663,'2015-02-12'),(2,22,'2015-01-03'),(2,21,'2015-02-12'),(2,11,'2015-02-12'),(2,9,'2015-04-12');

【问题讨论】：

尝试在 avg 中将 0 更改为 NULL：avg(if(month(date) = 1, amount, NULL))
根据其他现有记录在一个月内插入任意多个0，您会偏向您的平均值。也许更好group by customer_id, MONTH(date)？
考虑处理表示层/应用程序级代码中的数据显示问题（如果可用）

标签： mysql sql

【解决方案1】：

问题是0s。不会影响sum()，但会影响平均值。您可以将它们更改为NULL：

SELECT customer_id,
       avg(if(month(date) = 1, amount, NULL))  AS Jan,
       avg(if(month(date) = 2, amount, NULL))  AS Feb,
       avg(if(month(date) = 3, amount, NULL))  AS Mar,
       avg(if(month(date) = 4, amount, NULL))  AS Apr,
       avg(if(month(date) = 5, amount, NULL))  AS May,
       avg(if(month(date) = 6, amount, NULL))  AS Jun,
       avg(if(month(date) = 7, amount, NULL))  AS Jul,
       avg(if(month(date) = 8, amount, NULL))  AS Aug,
       avg(if(month(date) = 9, amount, NULL))  AS Sep,
       avg(if(month(date) = 10, amount, NULL)) AS Oct,
       avg(if(month(date) = 11, amount, NULL)) AS Nov,
       avg(if(month(date) = 12, amount, NULL)) AS `Dec`
FROM sales
GROUP BY customer_id;

作为说明：我更喜欢case，因为它是 ANSI 标准，所以我会这样写：

       avg(case when month(date) = 1 then amount end) as Jan,

【讨论】：

谢谢，Mike 也发现了这一点并添加了评论。答案有效，但也解释了我在平均值中包含的 0 周围缺少什么。
3,3,0 的平均值为 (3+3+0)/3，但 3,3,NULL 的平均值为 (3+3)/2。它在 avg 计算中不考虑 NULL，但会考虑 0。

【解决方案2】：

试试这样的：

SELECT
    customer_id,
  if(month(hlp_tab2.date) = 1, hlp_tab2.amount, 0)  AS Jan,
  if(month(hlp_tab2.date) = 2, hlp_tab2.amount, 0)  AS Feb,
  if(month(hlp_tab2.date) = 3, hlp_tab2.amount, 0)  AS Mar,
  if(month(hlp_tab2.date) = 4, hlp_tab2.amount, 0)  AS Apr,
  if(month(hlp_tab2.date) = 5, hlp_tab2.amount, 0)  AS May,
  if(month(hlp_tab2.date) = 6, hlp_tab2.amount, 0)  AS Jun,
  if(month(hlp_tab2.date) = 7, hlp_tab2.amount, 0)  AS Jul,
  if(month(hlp_tab2.date) = 8, hlp_tab2.amount, 0)  AS Aug,
  if(month(hlp_tab2.date) = 9, hlp_tab2.amount, 0)  AS Sep,
  if(month(hlp_tab2.date) = 10, hlp_tab2.amount, 0) AS Oct,
  if(month(hlp_tab2.date) = 11, hlp_tab2.amount, 0) AS Nov,
  if(month(hlp_tab2.date) = 12, hlp_tab2.amount, 0) AS `Dec`
FROM (
    SELECT
        hlp_tab.customer_id,
        avg(hlp_tab.amount) 'amount',
        hlp_tab.date
    FROM (
        SELECT
          customer_id,
          amount,
          month(date) 'date'
        FROM sales
    ) hlp_tab
    group by
        hlp_tab.customer_id,
        hlp_tab.date
) hlp_tab2

【讨论】：

【解决方案3】：

SELECT 
customer_id, 
SUM(Jan) AS Jun,
SUM(Feb) AS Feb,
SUM(Mar) AS Mar,
SUM(Apr) AS Apr,
SUM(May) AS May,
SUM(Jun) AS Jun,
SUM(Jul) AS Jul,
SUM(Aug) AS Aug,
SUM(Sep) AS Sep,
SUM(Oct) AS Oct,
SUM(Nov) AS Nov,
SUM(`Dec`) AS `Dec`
FROM (
SELECT
    customer_id,
    avg(if(month(date) = 1, amount, 0))  AS Jan,
    avg(if(month(date) = 2, amount, 0))  AS Feb,
    avg(if(month(date) = 3, amount, 0))  AS Mar,
    avg(if(month(date) = 4, amount, 0))  AS Apr,
    avg(if(month(date) = 5, amount, 0))  AS May,
    avg(if(month(date) = 6, amount, 0))  AS Jun,
    avg(if(month(date) = 7, amount, 0))  AS Jul,
    avg(if(month(date) = 8, amount, 0))  AS Aug,
    avg(if(month(date) = 9, amount, 0))  AS Sep,
    avg(if(month(date) = 10, amount, 0)) AS Oct,
    avg(if(month(date) = 11, amount, 0)) AS Nov,
    avg(if(month(date) = 12, amount, 0)) AS `Dec`
  FROM sales
  GROUP BY customer_id, month(date)
) AS a GROUP BY customer_id

【讨论】：

【解决方案4】：

一个快速的解决方法是：

    SELECT 
  customer_id,
  MAX(if(`month` = 1, amount, 0))  AS Jan,
  MAX(if(`month` = 2, amount, 0))  AS Feb,
  MAX(if(`month` = 3, amount, 0))  AS Mar,
  MAX(if(`month` = 4, amount, 0))  AS Apr,
  MAX(if(`month` = 5, amount, 0))  AS May,
  MAX(if(`month` = 6, amount, 0))  AS Jun,
  MAX(if(`month` = 7, amount, 0))  AS Jul,
  MAX(if(`month` = 8, amount, 0))  AS Aug,
  MAX(if(`month` = 9, amount, 0))  AS Sep,
  MAX(if(`month` = 10, amount, 0)) AS Oct,
  MAX(if(`month` = 11, amount, 0)) AS Nov,
  MAX(if(`month` = 12, amount, 0)) AS `Dec`
FROM
(
  SELECT
    customer_id,
    (month(date)) `month`,
    AVG(amount) amount
  FROM sales
  GROUP BY customer_id,(month(date))
) AS T
GROUP BY customer_id;

SQLFiddle

【讨论】：