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我是使用高级 SQL 查询的新手，我正在为一个查询而苦苦挣扎。

我在 php 中创建了预订系统，它使用 4 个表：

  • site_days
  • site_timeslots
  • 网站预订
  • 站点团队
  • 每个 site_team 都与 site_booking 有关
  • 每个 site_booking 都与 site_timeslot 相关
  • 每个 site_timeslot 都与 site_days 相关

一个站点_day 可以有更多 site_timeslots 一个站点_timeslot 可以有更多 site_bookings 一个站点_bookings 可以有更多 site_teams

您可以使用此 sql 创建测试表：

-- Adminer 3.6.3 MySQL dump

SET NAMES utf8;
SET foreign_key_checks = 0;
SET time_zone = 'SYSTEM';
SET sql_mode = 'NO_AUTO_VALUE_ON_ZERO';

DROP TABLE IF EXISTS `site_bookings`;
CREATE TABLE `site_bookings` (
  `id` int(11) NOT NULL auto_increment,
  `timeslot_id` int(11) NOT NULL,
  PRIMARY KEY  (`id`)
) ENGINE=InnoDB AUTO_INCREMENT=2 DEFAULT CHARSET=utf8 COLLATE=utf8_bin;

INSERT INTO `site_bookings` (`id`, `timeslot_id`) VALUES
(1, 6443);

DROP TABLE IF EXISTS `site_days`;
CREATE TABLE `site_days` (
  `id` int(11) NOT NULL auto_increment,
  `date` date NOT NULL,
  PRIMARY KEY  (`id`)
) ENGINE=InnoDB AUTO_INCREMENT=93 DEFAULT CHARSET=utf8 COLLATE=utf8_bin;

INSERT INTO `site_days` (`id`, `date`) VALUES
(85,    '2013-04-01'),
(92,    '2013-04-02');

DROP TABLE IF EXISTS `site_teams`;
CREATE TABLE `site_teams` (
  `id` int(11) NOT NULL auto_increment,
  `booking_id` int(11) NOT NULL,
  `name` varchar(100) collate utf8_bin NOT NULL,
  PRIMARY KEY  (`id`)
) ENGINE=InnoDB AUTO_INCREMENT=3 DEFAULT CHARSET=utf8 COLLATE=utf8_bin;

INSERT INTO `site_teams` (`id`, `booking_id`, `name`) VALUES
(1, 1,  'Avengers'),
(2, 1,  'Big Five');

DROP TABLE IF EXISTS `site_timeslots`;
CREATE TABLE `site_timeslots` (
  `id` int(11) NOT NULL auto_increment,
  `day_id` int(11) NOT NULL,
  `date` date NOT NULL,
  `starts` time NOT NULL,
  PRIMARY KEY  (`id`)
) ENGINE=InnoDB AUTO_INCREMENT=7152 DEFAULT CHARSET=utf8 COLLATE=utf8_bin;

INSERT INTO `site_timeslots` (`id`, `day_id`, `date`, `starts`) VALUES
(6443,  85, '2013-04-01',   '08:00:00'),
(6444,  85, '2013-04-01',   '08:10:00'),
(7098,  92, '2013-04-02',   '08:00:00'),
(7099,  92, '2013-04-02',   '08:10:00');

因此，我想获取表 site_timeslots 的所有时间段，并提供一些附加信息： - 对于每个 site_timeslot，我想知道该时间段的所有相关预订中的 site_teams 总数（例如，如果该 site_timeslot 有 2 个 site_bookings 并且每个有 2 个 site_teams，那么总计数应该是 4) 以及相关预订的计数。

我试过这个sql：

SELECT `site_teams`.`id` AS site_teams_id, `site_teams`.`name` AS site_teams_name, `site_teams`.`booking_id` AS site_teams_booking_id, `site_days`.`id` AS site_days_id, `site_days`.`date` AS site_days_date, `site_timeslots`.`id` AS site_timeslots_id, `site_timeslots`.`starts` AS site_timeslots_starts, `site_bookings`.`id` AS site_bookings_id, `site_bookings`.`timeslot_id` AS site_bookings_timeslot_id
FROM (`site_days`)
LEFT JOIN `site_timeslots` ON `site_timeslots`.`day_id` = `site_days`.`id`
LEFT JOIN `site_bookings` ON `site_bookings`.`timeslot_id` = `site_timeslots`.`id`
LEFT JOIN `site_teams` ON `site_teams`.`booking_id` = `site_bookings`.`id`
GROUP BY `site_teams`.`booking_id`

-> 但我不会得到没有任何 site_bookings 的时间段，请问我应该如何更改这个 sql 查询以获得结果：

  1. 每行的 site_timeslot
  2. 新列“count_of_site_bookings”中与该 site_timeslot 相关的 site_booking 计数
  3. 与新列“count_of_site_teams”中与该站点时间段相关的所有站点预订相关的站点团队计数
4

1 回答 1

1

您可以通过从 site_timeslots 开始的 LEFT JOINing 来做到这一点，然后在 2 个相关字段上使用 COUNT 来获得您所追求的总数

SELECT

  sti.*,
  COUNT(DISTINCT sb.id) AS count_of_site_bookings,
  COUNT(DISTINCT ste.id) AS count_of_site_teams

FROM site_timeslots sti

INNER JOIN site_days sd
  ON sd.id = sti.day_id


LEFT JOIN site_bookings sb
ON sb.timeslot_id = sti.id

LEFT JOIN site_teams ste
ON ste.booking_id = sb.id

GROUP BY sti.id

您可以在 SQL Fiddle http://sqlfiddle.com/#!2/1a253/2上找到它

由于我的假设不正确，我还做了一个使用子查询的先前版本，如果你想看看它以供参考，它也可以在http://sqlfiddle.com/#!2/9ccf2/ 10

于 2013-04-04T11:54:06.013 回答