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;
最佳答案
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
关于php - 离开加入以获得最小日期,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/21983795/