有一个表stop_times.txt
,其格式(GTFS)类似于:
+------------------+---------------+
| trip_id | stop_sequence |
+------------------+---------------+
| 4503599630773892 | 0 |
| 4503599630773892 | 1 |
| ... | ... |
| 4503599630773892 | 27 |
| 4503599630810392 | 0 |
| 4503599630810392 | 1 |
| ... | ... |
| 4503599630810392 | 17 |
| 4503599631507892 | 0 |
| 4503599631507892 | 1 |
| ... | ... |
| 4503599631507892 | 29 |
| ... | ... |
+------------------+---------------+
我期待的结果是:
+------------------+------------+-----------+
| trip_id | first_stop | last_stop |
+------------------+------------+-----------+
| 4503599630773892 | 0 | 27 |
| 4503599630810392 | 0 | 17 |
| 4503599631507892 | 0 | 19 |
| ... | ... | ... |
+------------------+------------+-----------+
PS:标题可能不准确。请改进它。
还有一个问题:如何将与stop_name
对应的stop_sequence
添加到此表中?
以下是不正确的代码,因为first_stop
和last_stop
的停止名称应与不同的stop_id
对应:
(SELECT routes.route_short_name, MIN(stop_times.stop_sequence) AS first_stop, stops.stop_name, MAX(stop_times.stop_sequence) AS last_stop, stops.stop_name
FROM stop_times
JOIN stops ON stops.stop_id=stop_times.stop_id
JOIN trips ON stop_times.trip_id=trips.trip_id
JOIN routes ON routes.route_id=trips.route_id
GROUP BY stop_times.trip_id);
编辑:我在几个小时的工作后完成了。这是密钥源代码:
SELECT T1.trip_id, T1.stop_sequence, T1.stop_id, T2.stop_sequence, T2.stop_id
FROM
-- create a new table T1: trip_id, stop_sequence=0, stop_id (first stop)
(SELECT st_first1.trip_id, st_first1.stop_sequence, st_first1.stop_id
FROM stop_times st_first1
INNER JOIN
-- filter out the first stop: trip_id, stop_sequence=0
(SELECT stop_times.trip_id, MIN(CAST(stop_times.stop_sequence AS UNSIGNED)) AS first_stop
FROM stop_times
GROUP BY stop_times.trip_id
) st_first2
ON st_first1.trip_id=st_first2.trip_id AND st_first1.stop_sequence=st_first2.first_stop
) T1
LEFT JOIN -- combine T1 and T2
-- create a new table T2: trip_id, stop_sequence=MAX, stop_id (last stop)
(SELECT st_last1.trip_id, st_last1.stop_sequence, st_last1.stop_id
FROM stop_times st_last1
INNER JOIN
-- filter out the last stop: trip_id, stop_sequence=MAX
(SELECT stop_times.trip_id, MAX(CAST(stop_times.stop_sequence AS UNSIGNED)) AS last_stop
FROM stop_times
GROUP BY stop_times.trip_id
) st_last2
ON st_last1.trip_id=st_last2.trip_id AND st_last1.stop_sequence=st_last2.last_stop
) T2
ON T1.trip_id=T2.trip_id
答案 0 :(得分:2)
您可以GROUP BY
trip_id
然后获取MIN
和MAX
stop_sequence
值,分别获得第一站和最后一站。
SELECT DISTINCT st.trip_id, s.stop_name, t.first_stop, t.last_stop
FROM stop_times st INNER JOIN stops s
ON st.stop_id = s.stop_id
RIGHT JOIN
(
SELECT trip_id, MIN(stop_sequence) AS first_stop, MAX(stop_sequence) AS last_stop
FROM stop_times
GROUP BY trip_id
) t
ON t.trip_id = st.trip_id
答案 1 :(得分:1)
如果您想创建一个新表,那么您可以使用以下查询:
create table new_table as
SELECT trip_id, MIN(stop_sequence) AS first_stop,MAX(stop_sequence) AS last_stop
FROM stop_times
GROUP BY trip_id