mysql，最近3个月的数据，包括零

Question

考虑以下架构:

CREATE TABLE IF NOT EXISTS `numbers` (
  `number` int(20) UNSIGNED NOT NULL AUTO_INCREMENT,
  UNIQUE KEY `number` (`number`)
) ENGINE=MyISAM AUTO_INCREMENT=8 DEFAULT CHARSET=utf8;

INSERT INTO `numbers` (`number`) VALUES
(1),
(2),
(3),
(4),
(5),
(6),
(7);

CREATE VIEW `calendar_3_months`  AS  select date_format((curdate() + interval (-(`numbers`.`number`) + 1) month),'%Y-%m') AS `date` from `numbers` limit 3 ;

CREATE TABLE IF NOT EXISTS `investments` (
  `id` int(11) UNSIGNED NOT NULL AUTO_INCREMENT,
  `uuid` char(36) NOT NULL,
  `name` varchar(64) NOT NULL,
  PRIMARY KEY (`id`),
  UNIQUE KEY `uuid` (`uuid`)
) ENGINE=InnoDB DEFAULT CHARSET=utf8;

INSERT INTO `investments` (`id`, `uuid`, `name`) VALUES
(1, 'f684a5a0-0c42-11e9-b39f-00163e2daa7c', 'Dom'),
(2, 'f200018c-0c43-11e9-b39f-00163e2daa7c', 'Dom2');

CREATE TABLE IF NOT EXISTS `schedules` (
  `id` int(10) UNSIGNED NOT NULL AUTO_INCREMENT,
  `uuid` char(36) NOT NULL,
  `id_investment` int(10) UNSIGNED NOT NULL,
  PRIMARY KEY (`id`),
  UNIQUE KEY `uuid` (`uuid`),
  KEY `id_investment` (`id_investment`)
) ENGINE=InnoDB DEFAULT CHARSET=utf8;



INSERT INTO `schedules` (`id`, `uuid`, `id_investment`) VALUES
(1, 'f684c2d2-0c42-11e9-b39f-00163e2daa7c', 1),
(2, 'f2018aeb-0c43-11e9-b39f-00163e2daa7c', 2);

CREATE TABLE IF NOT EXISTS `schedule_items` (
  `id` bigint(20) UNSIGNED NOT NULL AUTO_INCREMENT,
  `id_schedule` int(10) UNSIGNED NOT NULL,
  `uuid` char(36) NOT NULL,
  `name` varchar(128) CHARACTER SET utf8mb4 NOT NULL,
  PRIMARY KEY (`id`),
  UNIQUE KEY `uuid` (`uuid`),
  KEY `id_schedule` (`id_schedule`)
) ENGINE=InnoDB DEFAULT CHARSET=utf8;

INSERT INTO `schedule_items` (`id`, `id_schedule`, `uuid`, `name`) VALUES
(1, 1, 'f684e067-0c42-11e9-b39f-00163e2daa7c', ''),
(2, 1, 'f684ed90-0c42-11e9-b39f-00163e2daa7c', 'I. Stan zerowy'),
(3, 1, 'f685bc4f-0c42-11e9-b39f-00163e2daa7c', 'Roboty ziemne'),
(4, 1, 'f685d614-0c42-11e9-b39f-00163e2daa7c', 'Fundamenty'),
(5, 1, 'f6868511-0c42-11e9-b39f-00163e2daa7c', 'Izolacje'),
(6, 2, 'f201a5a7-0c43-11e9-b39f-00163e2daa7c', ''),
(7, 2, 'f203bfe5-0c43-11e9-b39f-00163e2daa7c', 'II. Stan surowy zamknięty'),
(8, 2, 'f2048a51-0c43-11e9-b39f-00163e2daa7c', 'Ściany konstrukcyjne'),
(9, 2, 'f204a3fa-0c43-11e9-b39f-00163e2daa7c', 'Stropy i kominy'),
(10, 2, 'f204c0ff-0c43-11e9-b39f-00163e2daa7c', 'Schody'),
(11, 2, 'f204ebd5-0c43-11e9-b39f-00163e2daa7c', 'Ściany działowe');



CREATE TABLE IF NOT EXISTS `expenses` (
  `id` bigint(20) UNSIGNED NOT NULL AUTO_INCREMENT,
  `uuid` char(36) NOT NULL,
  `price` decimal(12,2) NOT NULL,
  `name` varchar(128) CHARACTER SET utf8mb4 NOT NULL,
  `creation_time` datetime NOT NULL,
  PRIMARY KEY (`id`),
  UNIQUE KEY `uuid` (`uuid`)
) ENGINE=InnoDB DEFAULT CHARSET=utf8;


INSERT INTO `expenses` (`id`, `uuid`, `price`, `name`, `creation_time`) VALUES
(1, 'b0701faf-0ffd-11e9-b39f-00163e2daa7c', '122.00', 'dokumentacja', '2019-11-04 09:49:47'),
(2, '480fcc4b-11bd-11e9-befe-00163e2daa7c', '2000.00', 'koparka', '2019-07-06 15:13:46'),
(3, '081a5d1f-129a-11e9-befe-00163e2daa7c', '3500.00', 'beton', '2019-12-07 17:33:58'),
(4, 'f7e93abd-1501-11e9-befe-00163e2daa7c', '7430.00', 'pustaki', '2020-01-10 19:03:00'),
(5, '38c310a1-1900-11e9-befe-00163e2daa7c', '1750.00', 'piasek', '2020-01-15 21:00:35'),
(6, '09e39ee1-1b1a-11e9-b1e6-00163e2daa7c', '11600.00', 'dachowka', '2019-10-18 13:10:26'),
(7, '1c196498-5c77-11e9-92a6-00163e2daa7c', '22036.94', 'cegly', '2019-09-11 18:30:25'),
(8, 'ee8255a9-5c77-11e9-92a6-00163e2daa7c', '5300.00', 'beton b20', '2019-08-11 18:36:18'),
(9, '3141fd72-5c8a-11e9-92a6-00163e2daa7c', '10509.30', 'podsypka', '2019-11-11 20:47:01'),
(10, '510f9ee1-5c8a-11e9-92a6-00163e2daa7c', '1083.96', 'pcv', '2019-12-11 20:47:54');



CREATE TABLE IF NOT EXISTS `schedule_expenses` (
  `id_expense` bigint(20) UNSIGNED NOT NULL,
  `id_item` bigint(20) UNSIGNED NOT NULL,
  KEY `id_expense` (`id_expense`),
  KEY `id_schedule` (`id_item`)
) ENGINE=InnoDB DEFAULT CHARSET=utf8;


INSERT INTO `schedule_expenses` (`id_expense`, `id_item`) VALUES
(1, 1),
(2, 2),
(3, 2),
(4, 1),
(5, 8),
(7, 3),
(7, 3),
(7, 1),
(7, 9),
(10, 9);

我需要提出一个 SQL 查询，它将产生类似这样的结果

 date   | investment ID | total
---------------------------------
2020-01 | 1             | 0
2019-12 | 1             | 1000.00
2019-11 | 1             | 0
2020-01 | 2             | 750.00
2019-12 | 2             | 14345.67
2019-11 | 2             | 0
2020-01 | ...           | ...
2019-12 | ...           | ...
2019-11 | ...           | ...

日期必须包括最近3个月，并且应该是动态的(当查询在2020年1月执行时，它包括2020年1月、2019年12月和2019年11月；如果在2020年8月，它应该包括2020年6月、7月和2020年8月)。一般来说，查询应该是最近3个月内每项投资的所有费用总和的报告，包括没有任何费用的月份。

涉及的表说明:投资 - 存储有关投资的信息，例如正在 build 的房屋、公寓；时间表 - 每项投资都遵循特定的时间表Schedule_Items - 时间表元素费用 - 存储有关费用的信息Schedule_expenses - 费用与日程项目关系表

到目前为止我所做的:我创建了一个名为“calendar_3_months”的 View ，其中包含动态值，始终为最近 3 个月，格式如下:“%Y-%m”。我想出了一个查询，它产生几乎预期的结果(不正确的“总计”值)。我尝试了不同的关节，但它们没有给我想要的结果。我想出的最好的办法是:

SELECT
    `date` as `calendar_date`,
    `investments`.`id`,
    COALESCE(sum(`expenses`.`price`), 0) as `total`
FROM
    (`investments`, `calendar_3_months`)
inner join `schedules` on `schedules`.`id_investment` = `investments`.`id`
inner join `schedule_items` on `schedule_items`.`id_schedule` = `schedules`.`id`
left join `schedule_expenses` ON `schedule_expenses`.`id_item` = `schedule_items`.`id`
left join `expenses` ON `schedule_expenses`.`id_expense` = `expenses`.`id`
GROUP BY
    `calendar_date`, `investments`.`id`
ORDER BY
    `investments`.`id`

http://sqlfiddle.com/#!9/5e2f3f/1

请让我知道完成此任务的更改方向？

Answer 1

如果您的问题是缺少 0 行，那么这包括它:

SELECT c.date as calendar_date,
       i.id,
       COALESCE(sum(e.price), 0) as `total`
FROM investments i CROSS JOIN
     calendar_3_months c LEFT JOIN
     schedules s
     ON s.id_investment = i.id LEFT JOIN
     schedule_items si
     ON si.id_schedule = s.id LEFT JOIN
     schedule_expenses se
     ON se.id_item = si.id LEFT JOIN
     expenses e
     ON se.id_expense = e.id AND
        DATE_FORMAT(e.creation_time, '%Y-%m') = c.date        
GROUP BY calendar_date, i.id
ORDER BY i.id, calendar_date;

Here是一个 SQL Fiddle。