1

我在mysql中有一个这样的表：

+------------+------------+------------+------------+ 
|     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   | 
+------------+------------+------------+------------+

我尝试提取我在这里找到的用户的时间范围和示例，但我无法在小时内完成这项工作

我试过查询： 

 $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);";

我需要创建一个从 08:00 到 21:00 的范围，其中约会之间的空闲块如下所示：

free time
08:15:00 to 14:00:00
14:30:00 to 17:00:00
17:45:00 to 21:00:00
4

4 回答 4

2

试试这个查询

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

已编辑

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     |
+----------------------+---------------------------------+

从这个查询中学习的要点是

  1. Timediff 函数用法。timediff('结束时间','开始时间')
  2. 加入大号
  3. 避免连接中的第一条记录具有较长的偏移量，并且限制从 1 开始而不是 0
  4. IFNULL 用法 ifnull('if here come null','select this then')
于 2012-06-29T13:53:04.447 回答
1

试试这个查询我希望它对你有用.....

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);
于 2012-06-29T15:24:33.297 回答
0

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对接其他插槽的开始或结束。所以 10-11 11-12 将给出 10 开始，12 结束 = 空闲时间是 08-10 和 12-21

所以空闲时间是 08:00（除非它是开始时间）到第一个“开始”时间，然后匹配对 END->START 到最后一个 END->21:00（除非 21:00 是最后一个结束）

注意您还需要添加日期。

于 2012-06-29T13:20:14.447 回答
0

也许这已经结束了，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
于 2012-06-29T13:50:19.687 回答