php - 离开加入以获得最小日期

Question

我想在我的数据库中获取每个预订的第一个日期。

预订表是这样的:

id | id_client_booking | the_date

表 CLIENT_BOOKINGS 是

id | etc | etc

关于 BOOKINGS每个id_client_booking对应于一个id的 CLIENT_BOOKINGS , 并且可以有很多日期。

我想获取每个客户预订的第一个日期。这个选择只给我所有客户预订的第一个日期。我错过了什么？

SELECT
min(the_date) AS start_date, id_client_booking
FROM `day_bookings`AS db
LEFT JOIN `client_bookings`AS cb ON db.id_client_booking = cb.id

也试过:

SELECT
min(the_date) AS db.start_date, db.id_client_booking
                 ^^             ^^
FROM `day_bookings`AS db
LEFT JOIN `client_bookings`AS cb ON db.id_client_booking = cb.id

但这给出了错误:You have an error in your SQL syntax;

Answer 1

你缺少一个 GROUP BY，试试这个:

SELECT MIN(the_date) AS start_date, id_client_booking
FROM `day_bookings`AS db
LEFT JOIN `client_bookings`AS cb 
ON db.id_client_booking = cb.id
GROUP BY id_client_booking