5

我一直在尝试为公共交通系统编写查询,当我输入起点和终点站号时,该系统将输出路线列表。

这是我的 MySQL 表:

mysql> desc route_timings;
+----------------+---------+------+-----+---------+----------------+
| Field          | Type    | Null | Key | Default | Extra          |
+----------------+---------+------+-----+---------+----------------+
| ID             | int(11) | NO   | PRI | NULL    | auto_increment |
| route_number   | int(11) | NO   |     | NULL    |                |
| stop_number    | int(11) | NO   |     | NULL    |                |
| arrival_time   | time    | YES  |     | NULL    |                |
| departure_time | time    | YES  |     | NULL    |                |
+----------------+---------+------+-----+---------+----------------+
5 rows in set (0.00 sec)

以下是示例值:

mysql> select * from route_timings;
+----+--------------+-------------+--------------+----------------+
| ID | route_number | stop_number | arrival_time | departure_time |
+----+--------------+-------------+--------------+----------------+
|  1 |           54 |           1 | 10:00:00     | 10:05:00       |
|  2 |           54 |           2 | 11:00:00     | 11:05:00       |
|  3 |           54 |           3 | 12:00:00     | 12:05:00       |
|  4 |           55 |           3 | 13:00:00     | 13:05:00       |
|  5 |           55 |           4 | 14:00:00     | 14:05:00       |
|  6 |           55 |           5 | 15:00:00     | 15:05:00       |
|  7 |           60 |           3 | 10:00:00     | 10:05:00       |
|  8 |           60 |           2 | 11:00:00     | 11:05:00       |
|  9 |           60 |           1 | 12:00:00     | 12:05:00       |
+----+--------------+-------------+--------------+----------------+
9 rows in set (0.01 sec)

我的问题是:如果我想列出包含 stop_number 1 和 stop_number 3 的 route_number,我会写一个类似这样的查询:

SELECT DISTINCT `route_number` FROM `route_timings` WHERE `route_number` IN (
    SELECT `route_number` FROM `route_timings` WHERE `stop_number`=1
) AND `route_number` IN (
    SELECT `route_number` FROM `route_timings` WHERE `stop_number`=3
);

但是,上述查询将仅返回包含两个站点的 route_numbers,而不是源站点 (1) 在目标站点 (3) 之前到达的路线。

该查询将返回以下内容:

+--------------+
| route_number |
+--------------+
|           54 |
|           60 |
+--------------+

route_number 60 不是从 1 开始到 3,而是从 3 开始到 1。有人可以帮我将该位添加到查询中,以便查询仅输出 stop_number 1 的到达时间较少的 route_numbers比 stop_number 3 的到达时间。

提前致谢。-谢恩

4

3 回答 3

1
SELECT DISTINCT `route_number` R1 FROM R1.`route_timings` WHERE R1.`stop_number` =3 AND R1.`route_number` IN (
SELECT `route_number` R2 FROM `route_timings` WHERE R2.`stop_number`=1 AND R2.`arrival_time` > R1.`arrival_time`
);

我现在不能尝试,但我希望语法正确。我所做的是选择了经过 3 的路线,然后我检查了它们是否稍后在 1 处停止。我通常使用允许重命名表以组合查询的 SQLite,就像我在 SELECT route_numberR2中所做的那样

于 2012-03-13T09:13:09.973 回答
1

怎么样:

select fromStop.RouteNumber  
from routeTimings fromStop  
  inner join routeTimings toStop on toStop.RouteNumber = fromStop.RouteNumber and toStop.StopNumber = 3 and toStop.ArrivalTime > fromStop.DepartureTime  
where fromStop.StopNumber = 1

请注意,这是使用 SQL server 语法和名称调整为我习惯的约定,但原则应该很清楚

于 2012-03-13T09:25:16.673 回答
0

采用 :

SELECT DISTINCT `route_number` FROM `route_timings` WHERE `stop_number` IN (1,3)

注意:为什么离开时间大于到达时间?

于 2012-03-13T09:09:13.480 回答