Question

我在mysql中有一个像这样的表：

+------------+------------+------------+------------+ 
|     date   |   user_id  | start_hour | end_hour   | 
+------------+------------+------------+------------+ 
| 2010-12-15 |         20 | 08:00:00   | 08:15:00   | 
| 2010-12-15 |         20 | 14:00:00   | 14:30:00   | 
| 2010-12-15 |         20 | 17:00:00   | 17:45:00   | 
+------------+------------+------------+------------+

我尝试提取用户时间范围我找到了例子here，但我无法在数小时内完成这项工作

我尝试了查询：

 $sql="
SELECT a.end_hour AS 'Available From', Min(b.start_hour) AS 'To' 

FROM ( 
SELECT  0 as date, '08:00:00' as start_hour,'08:00:00' as end_hour 
UNION SELECT date, start_hour, end_hour FROM table
) 

AS a JOIN 
( SELECT  date, start_hour, end_hour  FROM table 
UNION SELECT 0,  '21:00:00' as start_hour, '22:00:00' as end_hour
) AS b ON

a.date=b.date AND  a.user_id=b.user_id AND a.end_hour < b.start_hour WHERE  
a.date='$date'  AND a.user_id='$user_id' GROUP BY a.end_hour 
HAVING a.end_hour < Min(b.start_hour);";

我需要在08:00到21:00之间创建一个范围，并在约会之间设置免费区块像这样：

free time
08:15:00 to 14:00:00
14:30:00 to 17:00:00
17:45:00 to 21:00:00

Answer 1

尝试此查询

SELECT
  a.id,
  a.start_hour,
  a.end_hour,
  TIMEDIFF(la.start_hour, a.end_hour) as `Free Time`
FROM appointment as a
  LEFT JOIN(SELECT * FROM appointment LIMIT 1,18446744073709551615) AS la
    ON la.id = a.id + 1
  LEFT JOIN (SELECT * FROM appointment) AS ra ON a.id = ra.id

这将显示这些结果

+---------------------------------------------+
¦ id ¦ start_hour BY ¦ end_hour   | Free Time |
¦----+---------------¦------------------------|
¦  1 ¦   08:00:00    ¦  08:15:00  | 05:45:00  |
¦  2 ¦   14:00:00    ¦  14:30:00  | 02:30:00  |
¦  3 ¦   17:00:00    ¦  17:45:00  | 03:15:00  |
¦  4 ¦   21:00:00    ¦  21:00:00  | (NULL)    |
+--------------------+------------------------+

此外，您必须在表格中使用21:00:00，否则您将无法获得最后的时差。我在表格中输入21:00:00作为开始和结束日期。

EDITED

这是修改后的查询

SELECT
  a.id,
  a.end_hour AS `Free time Start`,
  IFNULL(la.start_hour,a.end_hour) AS `Free Time End`,
  IFNULL(TIMEDIFF(la.start_hour, a.end_hour),'00:00:00') AS `Total Free Time`
FROM appointment AS a
  LEFT JOIN (SELECT * FROM appointment LIMIT 1,18446744073709551615) AS la
    ON la.id = (SELECT MIN(id) FROM appointment where id > a.id LIMIT 1)

结果是

+--------------------------------------------------------+
¦ id ¦ Free time Start ¦ Free Time End | Total Free Time |
¦----+-----------------¦---------------------------------|
¦  1 ¦   08:15:00      ¦ 14:00:00      |    05:45:00     |
¦  2 ¦   14:30:00      ¦ 17:00:00      |    02:30:00     |
¦  3 ¦   17:45:00      ¦ 21:00:00      |    03:15:00     |
¦  4 ¦   21:00:00      ¦ 21:00:00      |    00:00:00     |
+----------------------+---------------------------------+

要从此查询中学习的要点是

Timediff功能用法。 timediff（'结束时间'，'开始时间'）
加入号码
避免使用长偏移量加入第一条记录，并从1开始而不是从零开始限制
IFNULL使用ifnull（'如果这里为空'，'选择此然后'）

Answer 2

尝试此查询我希望它能为您工作.....

SELECT
  a.id,
  a.end_hour    AS `Free time Start`,
  la.start_hour AS `Free Time End`,
  IFNULL(TIMEDIFF(la.start_hour, a.end_hour),'00:00:00') AS `Total Free Time`
FROM appointment AS a
  LEFT JOIN (SELECT *
             FROM appointment
             LIMIT 1,18446744073709551615) AS la
    ON la.id = (SELECT
                  id
                FROM appointment
                WHERE id NOT IN(a.id)
                ORDER BY (a.id > id)ASC
                LIMIT 1);

Answer 3

这将为您提供结束时间块

SELECT T_a.end_hour
FROM (table AS T_a) LEFT JOIN (table AS T_b)
ON (T_a.`date`=T_b.`date` AND T_b.start_hour<=T_a.end_hour AND T_b.end_hour>T_a.end_hour)
WHERE T_b.user_id IS NULL

这将为您提供开始时间块

SELECT T_a.start_hour
FROM (table AS T_a) LEFT JOIN (table AS T_b)
ON (T_a.`date`=T_b.`date` AND T_b.start_hour<T_a.start_hour AND T_b.end_hour>=T_a.start_hour)
WHERE T_b.user_id IS NULL

这些查询将查找未交叉或与其他插槽对接的开始或结束。所以10-11 11-12将给出10开始，12结束=自由时间是08-10和12-21

所以空闲时间是08:00（除非它是一个开始时间）直到第一个“开始”时间，然后匹配END-＆gt; START到最后END-＆gt; 21:00的对（除非是最后一个21:00）端）

N.B。您还需要添加日期。

Answer 4

也许这是过度杀戮但这适用于sql-server，不确定转换为mysql的努力

declare @AvailableTime table ( AvailableDate date,  StartTime time, EndTime time )

insert into @AvailableTime values 
('15 dec 2010', '08:00:00', '21:00:00')

declare @Booking table (BookingDate date, UserId int, StartTime Time, EndTime Time)

insert into @Booking values 
('15 dec 2010', 20, '08:00:00', '08:15:00'),
('15 dec 2010', 20, '14:00:00', '14:30:00'),
('15 dec 2010', 20, '17:00:00', '17:45:00')

select 
    *
from
(
    -- first SELECT get start boundry
    select t.StartTime s, b.StartTime e
    from @AvailableTime t, @Booking b
    where 
        t.AvailableDate = b.BookingDate and (t.StartTime <= b.EndTime and b.StartTime <= t.EndTime) and t.StartTime <> b.StartTime
    and
        not exists (select 1 from @Booking b2 where b2.BookingDate = b.BookingDate and b2.StartTime < b.StartTime)

    union

    -- second SELECT get spikes ie middle
    select b1.EndTime s, b2.StartTime e
    from @AvailableTime t, @Booking b1, @Booking b2
    where 
        t.AvailableDate = b1.BookingDate and (t.StartTime <= b1.EndTime and b1.StartTime <= t.EndTime) 
    and 
        t.AvailableDate = b2.BookingDate and (t.StartTime <= b2.EndTime and b2.StartTime <= t.EndTime) 
    and 
        b1.EndTime < b2.StartTime
    and
        not exists (select 1 from @Booking b3 where b3.BookingDate = t.AvailableDate and b3.StartTime > b1.EndTime and b3.EndTime < b2.StartTime)

    union

    -- third SELECT get end boundry
    select b.EndTime s, t.EndTime e
    from @AvailableTime t, @Booking b
    where 
        t.AvailableDate = b.BookingDate and (t.StartTime <= b.EndTime and b.StartTime <= t.EndTime) and t.EndTime <> b.EndTime
    and 
        not exists (select 1 from @Booking b2 where b2.BookingDate = b.BookingDate and b2.StartTime > b.StartTime)
) t1

在mysql和php中查找空闲时间块？

4 个答案:

EDITED