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。如果一个用户无法进行该约会/事件“因为他/她在其他约会上落后”，我需要能够将此约会重新分配给另一个可用用户。因此，我需要显示在预定时间范围内可用并且可以接受此约会的前 5 位用户的建议，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;

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

并且 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 无法进行安排在 '2014-09-21 10:00:00' 的约会，因此系统会建议 user_id = 5，因为他将在接下来的 105 分钟内进行。

最终的数据集需要是

event_id org_owner  suggested_owner   available_for
100      2          5                 105

以下查询将为我提供users 表中所有可用用户的列表以及 start_on end_on 值，如果用户安排了一个事件（一个用户可以有多个记录）。如果 start_on 在这个查询意味着这个用户没有任何事件，否则它将返回每个事件的开始。

因此，如果用户 ID 出现在上面的查询中并且在 start_on 列中具有 NULL 值，这意味着该用户全天都可用，因此该用户应该是要推荐的 5 个用户中的 1 个，因为它是最高的用户之一可用性。但是如果一个用户在数据集中有一个/多行并且start on then中的值是非空的，我们需要查看最接近事件的start_on，然后推荐具有最大可用性的前5个价值。

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

如何提取此数据集？

Answer 1

在您的帮助下编辑了第一个提案，只需要照顾没有任何事件的用户（可以通过“t”子查询中的左连接来实现）。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:)

Answer 2

这个怎么样：

SELECT event_id, owner_id, start_on INTO @eventid, @user, @start_on 
FROM calender_events WHERE event_id = 100;

SELECT @event_id event_id,
    @user org_owner,
    c.owner_id suggested_owner,
    TIMESTAMPDIFF(MINUTE, $start_on, COALESCE(c.min_start, DATE(@start_on) + INTERVAL 18 HOUR)) available_for
FROM
    users u
LEFT JOIN
    (SELECT
          owner_id,
          MIN(start_on) 
     FROM
          calender_events
     WHERE
          (start_on BETWEEN @start_on AND DATE(@start_on) + INTERVAL 18 HOUR) 
          OR
          (start_on BETWEEN DATE(@start_on) AND DATE(@start_on) + INTERVAL 18 HOUR AND all_day_event = 1)
     GROUP BY owner_id
    ) c
    ON u.user_id = c.owner_id
WHERE u.user_id <> @user
ORDER BY available_for DESC
LIMIT 5

也许你必须调整 INTERVAL，我只是假设当天下午 6 点结束

Answer 3

试试这个：

SELECT
    co.event_id,
    co.owner_id org_owner,
    su.user_id suggested_owner,
    ifnull(min((to_seconds(c.start_on) - to_seconds(co)) / 60), 999) available
FROM calendar_events co
CROSS JOIN view_users su
LEFT JOIN calendar_events c ON c.owner_id = su.user_id
    AND c.start_on BETWEEN co.start_on AND date(adddate(co.start_on, 1))
    AND c.status = 1
WHERE co.event_id = 100
AND su.is_available_today = 1
GROUP BY 1, 2, 3
ORDER BY 4 DESC
LIMIT 5

在目标事件后的第二天没有约会的用户被分配了“999”的可用值，将他们放在列表的顶部。

每个用户的下一个事件使用min() 在时间间隔内找到，所有用户都按时间间隔最大排序，limit 为您提供前 5 个。