获取每个ID的最后一条记录

Question

我有这个问题：

SELECT  Truck.Name AS name,
        timestamp,
        oil,
        diesel,
        cargo,
        Truck.notes AS Remarks

        FROM trip
        INNER JOIN Truck USING (idTruck)
        WHERE idTruck IN('2','4','5','6','7','8','9','11','12','13','14','15','16')
        ORDER BY name, timestamp DESC

其中 returns 个 ID 的所有记录我正在考虑将 CURDATE 用作

AND DATE(timestamp) = CURDATE()

为了获得最后的记录，但是当一天没有修改记录时（他们通常会这样做，但是有时情况并非如此）我正在丢失记录。我将如何修改查询以获取每个 idTruck 的最后一个条目，而不考虑单个查询中的时间戳？

Answer 1

建议使用sql的MAX函数获取最后一条记录(MAX(Truck.Name)):

SELECT  MAX(Truck.Name) AS name,
        timestamp,
        oil,
        diesel,
        cargo,
        Truck.notes AS Remarks

FROM trip
INNER JOIN Truck USING (idTruck)
WHERE idTruck IN('2','4','5','6','7','8','9','11','12','13','14','15','16')
ORDER BY name, timestamp DESC

获取每个ID的最后一条记录：

SELECT  MAX(Truck.Name) AS name,
        timestamp,
        oil,
        diesel,
        cargo,
        Truck.notes AS Remarks

FROM trip
INNER JOIN Truck USING (idTruck)
WHERE idTruck IN('2','4','5','6','7','8','9','11','12','13','14','15','16')
GROUP BY idTruck 
ORDER BY name, timestamp DESC

因为根据 sql.sh(https://sql.sh/fonctions/agregation/max):

the aggregation function MAX() allows to return the maximum value of a column in a record set. The function can be applied to numeric or alphanumeric data. For example, it is possible to search for the most expensive product in a table in an online shop.

获取每个ID的最后一条记录

Get last record of each ID

php

mysql

mariadb

groupwise-maximum