首先，我想说我已经发现MySQL不支持SQL的FULL OUTER JOIN。但是，我需要以这种方式加入数据...希望我说“我需要全部加入”是正确的。如果我错了，请纠正我。
我的情况是什么？
一个表（users）描述了注册玩游戏的用户。表格results1描述了播放 game1 的用户的结果，表results2描述了播放 game2 的用户的结果，依此类推。
现在我想编写一个查询，在一段时间内获取所有游戏中的用户和用户点列表。积分必须加总。查询必须按user_id和日期（每月）对结果进行分组
最大的问题是没有一个表有完整的月份（所以我不能只做LEFT或RIGHT连接）。我考虑过制作某种临时日历表（在一段时间内只有几年和几个月），然后将表与点（results1，results2，results3等连接起来。 。）到那个日历表。但是这种解决方案似乎也很复杂。还有其他想法吗？
我的情况（MySQL转储）：
-- --------------------------------------------------------
-- Host: 192.168.0.60
-- Server version: 5.5.40-cll-lve - MySQL Community Server (GPL) by Atomicorp
-- Server OS: Linux
-- HeidiSQL Version: 9.1.0.4867
-- --------------------------------------------------------
/*!40101 SET @OLD_CHARACTER_SET_CLIENT=@@CHARACTER_SET_CLIENT */;
/*!40101 SET NAMES utf8mb4 */;
/*!40014 SET @OLD_FOREIGN_KEY_CHECKS=@@FOREIGN_KEY_CHECKS, FOREIGN_KEY_CHECKS=0 */;
/*!40101 SET @OLD_SQL_MODE=@@SQL_MODE, SQL_MODE='NO_AUTO_VALUE_ON_ZERO' */;
-- Dumping database structure for example
CREATE DATABASE IF NOT EXISTS `example` /*!40100 DEFAULT CHARACTER SET utf8 COLLATE utf8_unicode_ci */;
USE `example`;
-- Dumping structure for table example.results1
CREATE TABLE IF NOT EXISTS `results1` (
`id` int(11) unsigned NOT NULL AUTO_INCREMENT,
`user_id` int(11) DEFAULT NULL,
`points` int(11) DEFAULT NULL,
`date` date DEFAULT NULL,
PRIMARY KEY (`id`)
) ENGINE=InnoDB AUTO_INCREMENT=9 DEFAULT CHARSET=utf8 COLLATE=utf8_unicode_ci;
-- Dumping data for table example.results1: ~8 rows (approximately)
/*!40000 ALTER TABLE `results1` DISABLE KEYS */;
INSERT INTO `results1` (`id`, `user_id`, `points`, `date`) VALUES
(1, 1, 5, '2014-01-17'),
(2, 1, 5, '2014-01-18'),
(3, 2, 10, '2014-02-17'),
(4, 9, 8, '2014-03-17'),
(5, 1, 15, '2014-07-17'),
(6, 3, 9, '2014-10-17'),
(7, 1, 20, '2015-02-17'),
(8, 5, 10, '2014-06-17');
/*!40000 ALTER TABLE `results1` ENABLE KEYS */;
-- Dumping structure for table example.results2
CREATE TABLE IF NOT EXISTS `results2` (
`id` int(11) unsigned NOT NULL AUTO_INCREMENT,
`user_id` int(11) DEFAULT NULL,
`points` int(11) DEFAULT NULL,
`date` date DEFAULT NULL,
PRIMARY KEY (`id`)
) ENGINE=InnoDB AUTO_INCREMENT=9 DEFAULT CHARSET=utf8 COLLATE=utf8_unicode_ci ROW_FORMAT=COMPACT;
-- Dumping data for table example.results2: ~8 rows (approximately)
/*!40000 ALTER TABLE `results2` DISABLE KEYS */;
INSERT INTO `results2` (`id`, `user_id`, `points`, `date`) VALUES
(1, 1, 50, '2014-01-01'),
(2, 2, 35, '2014-01-02'),
(3, 3, 14, '2014-01-03'),
(4, 4, 18, '2014-06-01'),
(5, 5, 16, '2014-06-01'),
(6, 5, 16, '2014-06-02'),
(7, 6, 4, '2014-10-29'),
(8, 1, 20, '2014-01-16');
/*!40000 ALTER TABLE `results2` ENABLE KEYS */;
-- Dumping structure for table example.results3
CREATE TABLE IF NOT EXISTS `results3` (
`id` int(11) unsigned NOT NULL AUTO_INCREMENT,
`user_id` int(11) DEFAULT NULL,
`points` int(11) DEFAULT NULL,
`date` date DEFAULT NULL,
PRIMARY KEY (`id`)
) ENGINE=InnoDB AUTO_INCREMENT=4 DEFAULT CHARSET=utf8 COLLATE=utf8_unicode_ci ROW_FORMAT=COMPACT;
-- Dumping data for table example.results3: ~3 rows (approximately)
/*!40000 ALTER TABLE `results3` DISABLE KEYS */;
INSERT INTO `results3` (`id`, `user_id`, `points`, `date`) VALUES
(1, 9, 6, '2014-12-17'),
(2, 1, 10, '2014-01-01'),
(3, 1, 2, '2014-10-17'),
(4, 1, 8, '2014-01-03');
/*!40000 ALTER TABLE `results3` ENABLE KEYS */;
-- Dumping structure for table example.results4
CREATE TABLE IF NOT EXISTS `results4` (
`id` int(11) unsigned NOT NULL AUTO_INCREMENT,
`user_id` int(11) DEFAULT NULL,
`points` int(11) DEFAULT NULL,
`date` date DEFAULT NULL,
PRIMARY KEY (`id`)
) ENGINE=InnoDB AUTO_INCREMENT=3 DEFAULT CHARSET=utf8 COLLATE=utf8_unicode_ci ROW_FORMAT=COMPACT;
-- Dumping data for table example.results4: ~2 rows (approximately)
/*!40000 ALTER TABLE `results4` DISABLE KEYS */;
INSERT INTO `results4` (`id`, `user_id`, `points`, `date`) VALUES
(1, 4, 41, '2015-03-17'),
(2, 1, 2, '2014-12-17');
/*!40000 ALTER TABLE `results4` ENABLE KEYS */;
-- Dumping structure for table example.users
CREATE TABLE IF NOT EXISTS `users` (
`id` int(11) NOT NULL AUTO_INCREMENT,
`name` varchar(1000) COLLATE utf8_unicode_ci DEFAULT NULL,
PRIMARY KEY (`id`)
) ENGINE=InnoDB AUTO_INCREMENT=11 DEFAULT CHARSET=utf8 COLLATE=utf8_unicode_ci;
-- Dumping data for table example.users: ~10 rows (approximately)
/*!40000 ALTER TABLE `users` DISABLE KEYS */;
INSERT INTO `users` (`id`, `name`) VALUES
(1, 'Sophie'),
(2, 'Joshua'),
(3, 'Isabelle'),
(4, 'Jack'),
(5, 'Emily'),
(6, 'Harry'),
(7, 'Olivia'),
(8, 'Oliver'),
(9, 'Lily'),
(10, 'Charlie');
/*!40000 ALTER TABLE `users` ENABLE KEYS */;
/*!40101 SET SQL_MODE=IFNULL(@OLD_SQL_MODE, '') */;
/*!40014 SET FOREIGN_KEY_CHECKS=IF(@OLD_FOREIGN_KEY_CHECKS IS NULL, 1, @OLD_FOREIGN_KEY_CHECKS) */;
/*!40101 SET CHARACTER_SET_CLIENT=@OLD_CHARACTER_SET_CLIENT */;
我加入两个表的查询：
SELECT
r1_r2.user_id, r1_r2.points, r1_r2.`date`
FROM (
SELECT
IFNULL(rr1.user_id, rr2.user_id) as user_id, (IFNULL(rr1.points,0) + IFNULL(rr2.points,0)) as points, IFNULL(rr1.`date`, rr2.`date`) as `date`
FROM (
SELECT
r1.user_id, SUM(r1.points) as points, DATE_FORMAT(r1.`date`, '%Y-%m') as `date`
FROM results1 as r1
GROUP BY r1.user_id, DATE_FORMAT(r1.`date`, '%Y-%m')
) as rr1
LEFT JOIN (
SELECT
r2.user_id, SUM(r2.points) as points, DATE_FORMAT(r2.`date`, '%Y-%m') as `date`
FROM results2 as r2
GROUP BY r2.user_id, DATE_FORMAT(r2.`date`, '%Y-%m')
) as rr2 ON (rr1.user_id = rr2.user_id AND rr1.`date` = rr2.`date`)
UNION
SELECT
IFNULL(rl1.user_id, rl2.user_id) as user_id, (IFNULL(rl1.points,0) + IFNULL(rl2.points,0)) as points, IFNULL(rl1.`date`, rl2.`date`) as `date`
FROM (
SELECT
r1.user_id, SUM(r1.points) as points, DATE_FORMAT(r1.`date`, '%Y-%m') as `date`
FROM results1 as r1
GROUP BY r1.user_id, DATE_FORMAT(r1.`date`, '%Y-%m')
) as rl1
RIGHT JOIN (
SELECT
r2.user_id, SUM(r2.points) as points, DATE_FORMAT(r2.`date`, '%Y-%m') as `date`
FROM results2 as r2
GROUP BY r2.user_id, DATE_FORMAT(r2.`date`, '%Y-%m')
) as rl2 ON (rl1.user_id = rl2.user_id AND rl1.`date` = rl2.`date`)
) as r1_r2
HAVING
r1_r2.`date` BETWEEN '2014-01' AND '2014-12'
ORDER BY
r1_r2.user_id ASC, r1_r2.`date` ASC
它适用于两个表（results1和results2），但问题是我需要以相同的方式连接两个以上的表...
我有一些解决方案（比如一次又一次地嵌套表格），但问题是我的解决方案变得非常复杂（篇幅很长，阅读和理解也很复杂）。此外，在不久的将来，还有机会出现一些额外的表格。添加3或5个表后，查询将如何添加？如果我将继续执行相同类型的嵌套连接，整个查询将变得越来越复杂，阅读，理解，修改......
以下是已加入的3个表的查询（results1，results2，results3）：
SELECT
r1_r2_r3.user_id, r1_r2_r3.points, r1_r2_r3.`date`
FROM (
SELECT
IFNULL(r1_r2_l.user_id, r3_l.user_id) as user_id, (IFNULL(r1_r2_l.points,0) + IFNULL(r3_l.points,0)) as points, IFNULL(r1_r2_l.`date`, r3_l.`date`) as `date`
FROM (
# BEGIN. RESULT FROM BEFORE
SELECT
r1_r2.user_id, r1_r2.points, r1_r2.`date`
FROM (
SELECT
IFNULL(rr1.user_id, rr2.user_id) as user_id, (IFNULL(rr1.points,0) + IFNULL(rr2.points,0)) as points, IFNULL(rr1.`date`, rr2.`date`) as `date`
FROM (
SELECT
r1.user_id, SUM(r1.points) as points, DATE_FORMAT(r1.`date`, '%Y-%m') as `date`
FROM results1 as r1
GROUP BY r1.user_id, DATE_FORMAT(r1.`date`, '%Y-%m')
) as rr1
LEFT JOIN (
SELECT
r2.user_id, SUM(r2.points) as points, DATE_FORMAT(r2.`date`, '%Y-%m') as `date`
FROM results2 as r2
GROUP BY r2.user_id, DATE_FORMAT(r2.`date`, '%Y-%m')
) as rr2 ON (rr1.user_id = rr2.user_id AND rr1.`date` = rr2.`date`)
UNION
SELECT
IFNULL(rl1.user_id, rl2.user_id) as user_id, (IFNULL(rl1.points,0) + IFNULL(rl2.points,0)) as points, IFNULL(rl1.`date`, rl2.`date`) as `date`
FROM (
SELECT
r1.user_id, SUM(r1.points) as points, DATE_FORMAT(r1.`date`, '%Y-%m') as `date`
FROM results1 as r1
GROUP BY r1.user_id, DATE_FORMAT(r1.`date`, '%Y-%m')
) as rl1
RIGHT JOIN (
SELECT
r2.user_id, SUM(r2.points) as points, DATE_FORMAT(r2.`date`, '%Y-%m') as `date`
FROM results2 as r2
GROUP BY r2.user_id, DATE_FORMAT(r2.`date`, '%Y-%m')
) as rl2 ON (rl1.user_id = rl2.user_id AND rl1.`date` = rl2.`date`)
) as r1_r2
# END. RESULT FROM BEFORE
) as r1_r2_l
LEFT JOIN (
SELECT
r3.user_id, SUM(r3.points) as points, DATE_FORMAT(r3.`date`, '%Y-%m') as `date`
FROM results3 as r3
GROUP BY r3.user_id, DATE_FORMAT(r3.`date`, '%Y-%m')
) as r3_l ON (r1_r2_l.user_id = r3_l.user_id AND r1_r2_l.`date` = r3_l.`date`)
UNION
SELECT
IFNULL(r1_r2_r.user_id, r3_r.user_id) as user_id, (IFNULL(r1_r2_r.points,0) + IFNULL(r3_r.points,0)) as points, IFNULL(r1_r2_r.`date`, r3_r.`date`) as `date`
FROM (
# BEGIN. RESULT FROM BEFORE
SELECT
r1_r2.user_id, r1_r2.points, r1_r2.`date`
FROM (
SELECT
IFNULL(rr1.user_id, rr2.user_id) as user_id, (IFNULL(rr1.points,0) + IFNULL(rr2.points,0)) as points, IFNULL(rr1.`date`, rr2.`date`) as `date`
FROM (
SELECT
r1.user_id, SUM(r1.points) as points, DATE_FORMAT(r1.`date`, '%Y-%m') as `date`
FROM results1 as r1
GROUP BY r1.user_id, DATE_FORMAT(r1.`date`, '%Y-%m')
) as rr1
LEFT JOIN (
SELECT
r2.user_id, SUM(r2.points) as points, DATE_FORMAT(r2.`date`, '%Y-%m') as `date`
FROM results2 as r2
GROUP BY r2.user_id, DATE_FORMAT(r2.`date`, '%Y-%m')
) as rr2 ON (rr1.user_id = rr2.user_id AND rr1.`date` = rr2.`date`)
UNION
SELECT
IFNULL(rl1.user_id, rl2.user_id) as user_id, (IFNULL(rl1.points,0) + IFNULL(rl2.points,0)) as points, IFNULL(rl1.`date`, rl2.`date`) as `date`
FROM (
SELECT
r1.user_id, SUM(r1.points) as points, DATE_FORMAT(r1.`date`, '%Y-%m') as `date`
FROM results1 as r1
GROUP BY r1.user_id, DATE_FORMAT(r1.`date`, '%Y-%m')
) as rl1
RIGHT JOIN (
SELECT
r2.user_id, SUM(r2.points) as points, DATE_FORMAT(r2.`date`, '%Y-%m') as `date`
FROM results2 as r2
GROUP BY r2.user_id, DATE_FORMAT(r2.`date`, '%Y-%m')
) as rl2 ON (rl1.user_id = rl2.user_id AND rl1.`date` = rl2.`date`)
) as r1_r2
# END. RESULT FROM BEFORE
) as r1_r2_r
RIGHT JOIN (
SELECT
r3.user_id, SUM(r3.points) as points, DATE_FORMAT(r3.`date`, '%Y-%m') as `date`
FROM results3 as r3
GROUP BY r3.user_id, DATE_FORMAT(r3.`date`, '%Y-%m')
) as r3_r ON (r1_r2_r.user_id = r3_r.user_id AND r1_r2_r.`date` = r3_r.`date`)
) as r1_r2_r3
HAVING
r1_r2_r3.`date` BETWEEN '2014-01' AND '2014-12'
ORDER BY
r1_r2_r3.user_id ASC, r1_r2_r3.`date` ASC
...我认为如果我们继续使用更多以相同方式加入的表格，那么查询变得非常复杂，我认为你可以得到我的意思。
顺便说一下，这只是一个真实情况的简化版本。实际上，results1，results2，results3和results4是已经通过加入其他表而获得的表，计算两者之间的值...所以最后的查询我必须工作on比上面的例子中提到的要复杂得多。
我的问题是： 我可以将查询加入两个以上的表格更短，更容易理解吗？
答案 0 :(得分：0)
我认为你可以使用union all和聚合做你想做的事。我认为以下是您想要的：
select user_id, year(date), month(date), sum(points1) as point31,
sum(points2) as points2, sum(points3) as points3
from users u left join
((select r1.user_id, r1.date, r1.points as points1, NULL as points2, NULL as points3
from results1 r1
) union all
(select r2.user_id, r2.date, NULL as points1, r2.points as points2, NULL as points3
from results2 r2
) union all
(select r3.user_id, r3.date, NULL as points1, NULL as points2, r3.points as points3
from results3 r3
)
) r
on u.id = r.user_id
group by user_id, year(date), month(date);