MySql 旋转 Table 计数

Question

我有以下 table 持有客户订单。

order_id  |  customer_id  |  fulfilled |  order_date
---------------------------------------------------
123       |  1234         |  1         |  2022/01/12
126       |  1235         |  1         |  2022/01/18
127       |  1235         |  0         |  2022/01/19
128       |  1236         |  1         |  2022/01/01
129       |  1236         |  0         |  2022/01/03
130       |  1236         |  1         |  2022/01/04
131       |  1237         |  1         |  2022/01/01
132       |  1237         |  1         |  2022/01/03
133       |  1237         |  1         |  2022/01/04
134       |  1238         |  1         |  2022/02/12
135       |  1239         |  1         |  2022/02/18
136       |  1239         |  0         |  2022/02/19
137       |  1239         |  1         |  2022/02/20
138       |  1239         |  1         |  2022/02/21
139       |  1240         |  1         |  2022/02/01
140       |  1240         |  1         |  2022/02/03
141       |  1240         |  1         |  2022/02/04
142       |  1241         |  1         |  2022/02/01
143       |  1241         |  0         |  2022/02/03
144       |  1241         |  1         |  2022/02/04
145       |  1241         |  1         |  2022/02/04
146       |  1241         |  1         |  2022/02/04

SQL以下

CREATE TABLE `orders` (
  `order_id` int(11) NOT NULL,
  `customer_id` int(11) NOT NULL,
  `fulfilled` tinyint(1) NOT NULL DEFAULT 0,
  `order_date` date NOT NULL
) ENGINE=InnoDB DEFAULT CHARSET=latin1;

INSERT INTO `orders` (`order_id`, `customer_id`, `fulfilled`, `order_date`) VALUES
(123, 1234, 1, '2022-01-12'),
(126, 1235, 1, '2022-01-18'),
(127, 1235, 0, '2022-01-19'),
(128, 1236, 1, '2022-01-01'),
(129, 1236, 0, '2022-01-03'),
(130, 1236, 1, '2022-01-04'),
(131, 1237, 1, '2022-01-01'),
(132, 1237, 1, '2022-01-03'),
(133, 1237, 1, '2022-01-04'),
(134, 1238, 1, '2022-02-12'),
(135, 1239, 1, '2022-02-18'),
(136, 1239, 0, '2022-02-19'),
(137, 1239, 1, '2022-02-20'),
(138, 1239, 1, '2022-02-21'),
(139, 1240, 1, '2022-02-01'),
(140, 1240, 1, '2022-02-03'),
(141, 1240, 1, '2022-02-04'),
(142, 1241, 1, '2022-02-01'),
(143, 1241, 0, '2022-02-03'),
(144, 1241, 1, '2022-02-04'),
(145, 1241, 1, '2022-02-04'),
(146, 1241, 1, '2022-02-04');

ALTER TABLE `orders` ADD PRIMARY KEY (`order_id`);

我想创建一个 MySQL 查询来将此数据排列到一个数据透视表 table 中，显示有多少客户有 1、2、3 或 4 个 已完成订单[=每个月 24=]（已填满的列 = 1）。每个客户的订单永远不会超过 4 个。

Customers Customers Customers Customers Month/Year | with 1 Order | with 2 Orders | with 3 Orders | with 4 Orders ------------------------------------------------------------------------------------------- Jan 22 | 2 | 2 | 1 | 0 Feb 22 | 1 | 0 | 2 | 1

我在网上广泛查看，但在这项任务中惨遭失败。如有任何帮助，我们将不胜感激。

Answer 1

我们可以进行两步聚合。首先，按客户和月份汇总以获得每个月每个客户的各种计数。然后按月和数据透视汇总得到最终结果。

SELECT month_year,
       SUM(total = 1) AS one_order,
       SUM(total = 2) AS two_order,
       SUM(total = 3) AS three_order,
       SUM(total = 4) AS four_order,
FROM
(
    SELECT DATE_FORMAT(order_date, '%b %y') AS month_year, customer_id,
           SUM(fulfilled) AS total
    FROM orders
    GROUP BY 1, 2
) t
GROUP BY month_year;

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