2

I have tables & data like this:

venues table contains : id

+----+---------+
| id | name    |
+----+---------+
|  1 | venue 1 |
|  2 | venue 2 |
---------------

event_dates : id, event_id, event_from_datetime, event_to_datetime, venue_id
+----+----------+---------------------+---------------------+----------+
| id | event_id | event_from_datetime | event_to_datetime   | venue_id |
+----+----------+---------------------+---------------------+----------+
|  1 |        1 | 2009-12-05 00:00:00 | 2009-12-07 00:00:00 |        1 |
|  2 |        1 | 2009-12-09 00:00:00 | 2009-12-12 00:00:00 |        1 |
|  3 |        1 | 2009-12-15 00:00:00 | 2009-12-20 00:00:00 |        2 |
+----+----------+---------------------+---------------------+----------+

This is my requirement: I want venues that will be free on 2009-12-06 00:00:00 i.e.

I should get

|venue_id|

|2 |

Currently I'm having the following query, 

select ven.id , evtdt.event_from_datetime, evtdt.event_to_datetime 
from venues ven 
left join event_dates evtdt 
on (ven.id=evtdt.venue_id) 
where evtdt.venue_id is null 
or not ('2009-12-06 00:00:00' between evtdt.event_from_datetime 
                                  and evtdt.event_to_datetime);
+----+---------------------+---------------------+
| id | event_from_datetime | event_to_datetime   |
+----+---------------------+---------------------+
|  1 | 2009-12-09 00:00:00 | 2009-12-12 00:00:00 |
|  2 | 2009-12-15 00:00:00 | 2009-12-20 00:00:00 |
|  3 | NULL                | NULL                |
|  5 | NULL                | NULL                |
+----+---------------------+---------------------+

If you note the results, its not including venue id 1 where date is in between 2009-12-06 00:00:00 but showing other bookings. Please help me correct this query.

Thanks in advance.

4

3 回答 3

2 
SELECT  *
FROM    venue v
WHERE   NOT EXISTS
        (
        SELECT  NULL
        FROM    event_dates ed
        WHERE   ed.venue_id = v.id
                AND '2009-12-06 00:00:00' BETWEEN ed.event_from_datetime AND ed.event_to_datetime
        )
于 2010-01-12T15:43:15.933 回答
1 
or not ('2009-12-06 00:00:00' between evtdt.event_from_datetime 
                              and evtdt.event_to_datetime);

2009年 12 月 6 日介于 12/5/09 和 12/7/09之间...这就是为什么将venue_id 1 排除在外的原因...您要从数据中准确提取什么？

您构建的连接查询说，获取场所表，并为其中具有匹配场所_id 的每一行制作场所表行的副本并附加匹配的行。因此，如果您刚刚这样做：

select * 
  from venues ven 
  left join event_dates evtdt 
  on (ven.id=evtdt.venue_id);

它会产生：

+----+---------+------+----------+---------------------+---------------------+----------+
| id | name    | id   | event_id | event_from_datetime | event_to_datetime   | venue_id |
+----+---------+------+----------+---------------------+---------------------+----------+
|  1 | venue 1 |    1 |        1 | 2009-12-05 00:00:00 | 2009-12-07 00:00:00 |        1 |
|  1 | venue 1 |    2 |        1 | 2009-12-09 00:00:00 | 2009-12-12 00:00:00 |        1 |
|  2 | venue 2 |    3 |        1 | 2009-12-15 00:00:00 | 2009-12-20 00:00:00 |        2 |
+----+---------+------+----------+---------------------+---------------------+----------+

如果您随后添加了条件，即感兴趣的日期不在事件的开始日期和结束日期之间，则查询如下所示：

select * 
  from venues ven 
  left join event_dates evtdt 
  on (ven.id=evtdt.venue_id) 
  where not ('2009-12-06' between evtdt.event_from_datetime and evtdt.event_to_datetime)

这会产生以下结果：

+----+---------+------+----------+---------------------+---------------------+----------+
| id | name    | id   | event_id | event_from_datetime | event_to_datetime   | venue_id |
+----+---------+------+----------+---------------------+---------------------+----------+
|  1 | venue 1 |    2 |        1 | 2009-12-09 00:00:00 | 2009-12-12 00:00:00 |        1 |
|  2 | venue 2 |    3 |        1 | 2009-12-15 00:00:00 | 2009-12-20 00:00:00 |        2 |
+----+---------+------+----------+---------------------+---------------------+----------+

这些是我在 MySQL 中使用您的数据的实际实验结果。

如果您想在提议的日期获得免费的场地 ID，那么您可以编写如下内容：

  select ven.id, SUM('2009-12-06' between evtdt.event_from_datetime and evtdt.event_to_datetime) as num_intersects 
    from venues ven left join event_dates evtdt on (ven.id=evtdt.venue_id) 
    group by ven.id
    having num_intersects = 0;

产生：

+----+----------------+
| id | num_intersects |
+----+----------------+
|  2 |              0 |
+----+----------------+

如果您的场地在 event_date 表中没有活动，这也会得出正确的答案（无需修改）。

于 2010-01-12T15:43:08.973 回答
0

猜测一下，如果你删除not from

or not ('2009-12-06 00:00:00' between evtdt.event_from_datetime 
                                  and evtdt.event_to_datetime)

然后，这将从事件日期返回第 1 行，但不返回其他事件日期行。

我说“猜测”是因为您的 where 子句有点难以理解。也许你的意思是

select ven.id , evtdt.event_from_datetime, evtdt.event_to_datetime 
from venues ven 
left join event_dates evtdt 
on (ven.id=evtdt.venue_id) 
where '2009-12-06 00:00:00' between evtdt.event_from_datetime 
                                  and evtdt.event_to_datetime;
于 2010-01-12T15:45:41.233 回答