Mysql SQL从一个表中选择在另一个表中出现

我有两张桌子：

Event
+----------+---------+--------------+
| event_id | name    | date         |
+----------+------+-----------------+
| 1        | Event 1 | 26/03/2012   |
+----------+------+-----------------+
| 2        | Event 2 | 27/03/2012   |
+----------+------+-----------------+

Reservation
+----------------+------------+--------------+
| reservation_id | date       | themed_id    |
+----------------+------------+--------------+
| 1              | 26/03/2012 | 1            |
+----------------+------------+--------------+
| 2              | 26/03/2012 | 1            |
+----------------+------------+--------------+
| 3              | 27/03/2012 | 2            |
+----------------+------------+--------------+
| 4              | 26/03/2012 | 1            |
+----------------+------------+--------------+

如何显示预订表中出现少于3次的所有事件

输出将是：

+----------+---------+--------------+
| event_id | name    | date         |
+----------+------+-----------------+
| 2        | Event 2 | 27/03/2012   |
+----------+------+-----------------+

因为事件二只在预定中出现过一次

谢谢

我猜是Event.Event\u id=Reservation.themed\u id？如果是：

编辑：更改为左联接，以包含保留为0的事件

SELECT
    Event.event_id,
    Event.name,
    Event.date
FROM
    Event
LEFT JOIN Reservation ON Event.event_id = Reservation.themed_id
GROUP BY
    Event.event_id
HAVING
    COUNT(DISTINCT Reservation.reservation_id) < 3

---

我还没有测试过，但基本想法是这样。

你自己先试过吗？我们是来帮助你的，不是给你一个完整的书面解决方案。

SELECT *
FROM Event
WHERE event_id IN (
SELECT themed_id
FROM Reservation
GROUP BY themed_id
HAVING COUNT(*) < 3)