Question

我正在尝试按如下方式加入两个表：

tours                                 tours_schedules
---------------------                 ---------------------------------
tour_id | title     |                 schedule_id | tour_id | startdate
---------------------                 ---------------------------------
1       | Day Trip 1                  1           | 1       | 2014-05-23
2       | Day Trip 2                  2           | 1       | 2014-05-24
                                      3           | 2       | 2014-06-01
                                      4           | 1       | 2014-05-17

我想选择能告诉我tours.tour_id，tours.title的结果，tour_schedules中的tours记录，其中tours_schedules.tour_id与tours.tour_id匹配，以及最早匹配的tours_schedules.startdate的日期。

结果应如下所示

results
---------------------------------------------------------------
tour_id  | title         | schedule_count | earliest_startdate 
---------------------------------------------------------------
1        | Day Trip 1    | 3              | 2014-05-17
2        | Day Trip 2    | 1              | 2014-06-01

我做不到，有人能把我从痛苦中解救出来吗？！以下是我尝试过的众多内容。计算tour_schedules似乎会在选择的游览中停止多行。

SELECT DISTINCT  t.title, a.startdate, COUNT (DISTINCT a.startdate) AS schedule_count
FROM tours t
INNER JOIN tours_schedules a 
ON t.tour_id = a.tour_id
WHERE t.category_id = 1
ORDER BY a.startdate ASC


SELECT DISTINCT  t.title
FROM tours t
RIGHT JOIN tours_schedules a
ON t.tour_id = a.tour_id
ORDER BY a.startdate


SELECT a.schedule_id, COUNT(a.schedule_id) AS schedule_count, t.title
FROM tours_schedules a
JOIN tours t
ON a.tour_id = t.tour_id
ORDER BY a.startdate

SELECT title, tour_id FROM tours
UNION ALL
SELECT startdate FROM tours_schedules

SELECT t.title, t.tour_id, (SELECT a.startdate)
FROM tours t
JOIN tours_schedules a
WHERE t.tour_id = a.tour_id 

SELECT t.title, t.tour_id, (SELECT a.startdate LIMIT 1), (SELECT COUNT(a.startdate) WHERE t.tour_id = a.tour_id)
FROM tours t
JOIN tours_schedules a
WHERE t.tour_id = a.tour_id

Answer 1

SELECT a.schedule_id, 
       COUNT(a.schedule_id) AS schedule_count, 
       MIN(a.startdate) AS earliest_startdate , 
       t.title
FROM tours_schedules a
JOIN tours t
ON a.tour_id = t.tour_id
GROUP BY t.tour_id

MySQL SELECT用join，count，order by

1 个答案: