mysql - 如何从日历数据库中获取可用时间段列表

Question

我有一个事件/日历 MySQL 表，其中每个用户全天都有多个约会/事件。如果一个用户不能进行那个约会/事件“因为他/她在其他约会上落后了”，我需要能够将这个约会重新分配给另一个可用的用户。因此，我需要显示在预定时间范围内可用且可以接受此约会的前 5 名用户的建议，经理将能够将此约会重新分配给建议的用户之一。

我的事件表看起来像这样

CREATE TABLE `calendar_events` (
   `event_id` int(11) unsigned NOT NULL AUTO_INCREMENT,
   `start_on` datetime NOT NULL,
   `end_on` datetime NOT NULL,
   `subject` varchar(255) NOT NULL,
   `event_type` enum('Phone Call','Meeting','Event','Appointment','Other') CHARACTER SET latin1 COLLATE latin1_general_ci NOT NULL DEFAULT 'Phone Call',
   `all_day_event` tinyint(1) DEFAULT '0' COMMENT '1 = all day event, 0 = no',
   `phone_call_id` int(11) unsigned DEFAULT NULL,
   `account_id` int(11) unsigned DEFAULT NULL,
   `client_id` int(11) unsigned DEFAULT NULL,
   `owner_id` int(11) unsigned NOT NULL,
   `created_by` int(11) unsigned NOT NULL,
   `created_on` datetime NOT NULL DEFAULT CURRENT_TIMESTAMP,
   `modified_by` int(11) unsigned DEFAULT NULL,
   `modified_on` datetime DEFAULT NULL,
   `event_location` varchar(255) DEFAULT NULL,
   `event_notes` varchar(10000) DEFAULT NULL,
   `status` tinyint(1) NOT NULL DEFAULT '1' COMMENT '0 = purged, 1 = active, 2=pass, 3 = cancled, 5 = waiting for auditor to be enabled',
   PRIMARY KEY (`event_id`),
   UNIQUE KEY `phone_call_id` (`phone_call_id`,`account_id`,`client_id`),
   KEY `client_id` (`client_id`),
   KEY `account_id` (`account_id`)
 ) ENGINE=InnoDB AUTO_INCREMENT=1 DEFAULT CHARSET=utf8;

所以让 event_id = 100 分配给 user_id = 2 并安排到 start_on = '2014-09-21 10:00:00' 和 end_on '2014-09-21 10:00:00'

and user_id = 5 有约会 start_on '2014-09-21 11:45:00' 和 end_on '2014-09-21 12:30:00'

并且 user_id = 2 无法完成他安排在“2014-09-21 10:00:00”的约会，因此他们的系统将建议 user_id = 5，因为他将在接下来的 105 分钟内。

最终的数据集需要是

event_id org_owner  suggested_owner   available_for
100      2          5                 105

如果用户安排了一个事件(一个用户可以有多个记录)，下面的查询将从 users 表中为我提供所有可用用户的列表以及 start_on end_on 值。此查询中的 start_on 为 null，这意味着此用户没有任何事件，否则它将返回每个事件的开始。

所以如果用户 ID 出现在上面的查询中并且在 start_on 列中有一个 NULL 值，这意味着这个用户全天都有空所以这个用户应该是推荐的 5 个用户中的一个，因为它是最高的可用性。但是如果一个用户在数据集中有一行/多行在start on then中有一个非空值，我们需要查看距离事件最近的start_on，然后推荐具有最大可用性的前5个值(value)。

SELECT user_id, start_on, end_on, subject
FROM view_users AS su
LEFT JOIN calendar_events AS c ON c.owner_id = su.user_id AND c.start_on NOT BETWEEN '2014-09-30 00:00:00' AND '2014-09-30 23:59:59' AND c.status = 1
WHERE su.is_available_today = 1

我怎样才能提取这个数据集？

Answer 1

感谢您的帮助编辑了第一个提案，只需要照顾没有任何事件的用户(可以通过在“t”子查询中进行左连接来实现)。这可以改进很多，但现在我有点累了:)

SELECT
c.event_id,                     -- Event id
c.owner_id AS org_owner,        -- Original owner of event
t.owner_id AS suggested_owner,  -- Suggested new user
c.start_on,                     -- Event start
t.free_from,                    -- Owner free slot start
t.free_to,                  -- Owner free slot end
TIME_TO_SEC( TIMEDIFF( t.free_to, c.start_on ) ) /60 AS available_for   -- Availibility of minutes (diff between event start and free slot end)

FROM calendar_events AS c

-- Join with free slots
LEFT JOIN (
    -- Add a slot for beginning, 1999-01-01 to first event start
    SELECT * FROM (
        SELECT owner_id, '1900-01-01' AS free_from, MIN( start_on ) AS free_to
        FROM calendar_events c3
        GROUP BY owner_id
    ) AS deb

    UNION

    -- select free slots by taking the event end and the following event start
    SELECT owner_id, `end_on` AS free_from, (
        SELECT start_on
        FROM calendar_events c2
        WHERE c2.owner_id = c1.owner_id
        AND c2.start_on > c1.end_on
        ORDER BY c2.start_on
        LIMIT 0 , 1
    ) AS free_to

    FROM calendar_events c1

    UNION 

    -- Add a slot for end, last event end to 2100-01-01
    SELECT * FROM (
        SELECT owner_id, MAX( end_on ) AS free_from, '2100-01-01' AS free_to
        FROM calendar_events c3
        GROUP BY owner_id
    ) AS end
) AS t ON t.owner_id <> c.owner_id
-- Join avoid using same user and ensure free slot matches event dates
AND t.free_from <= c.start_on AND t.free_to >= c.end_on
WHERE c.status = 1
AND c.event_id =52
GROUP BY t.owner_id     -- To avoid multiple free slots by user
ORDER BY available_for DESC -- Sort to list biggest slots first
LIMIT 0, 5              -- Only five first matching users

祝你好运:)