Question

我需要从我的数据库中每小时选择第一个值。但我不知道如何在GROUP BY语句上颠倒顺序。如何重写我的查询（现在它以小时为单位选择最后值）？

SELECT HOUR(`time`) as hour, mytable.*
FROM mytable
WHERE DATE(`time`) ="2015-09-12" GROUP BY HOUR(`time`) ORDER BY `time` ASC;

Answer 1

这个查询给了我预期的结果：

SELECT HOUR(`time`) as hour, sortedTable.* FROM
(SELECT  electrolysis.* FROM electrolysis
    WHERE DATE(`time`)='2015-09-12' ORDER BY `time`) as sortedTable
GROUP BY HOUR(`time`);

Answer 2

您只需在子查询中选择MIN HOUR，尝试使用查询：

SELECT * from mytable WHERE `time` IN (
SELECT MIN(HOUR(`time`)) as `hour`
FROM mytable
WHERE DATE(`time`) ="2015-09-12" 
GROUP BY HOUR(`time`) ) ORDER BY `time` ASC;

Answer 3

您可以这样做： -

SELECT sub0.min_time, 
        mytable.*
FROM mytable
INNER JOIN
(
    SELECT MIN(`time`) AS min_time
    FROM mytable
    GROUP BY HOUR(`time`)
) sub0
ON mytable.`time` = sub0.min_time
WHERE DATE(`time`) ="2015-09-12" 
ORDER BY `time` ASC

这是使用子查询来获取每小时的最小时间。然后在最短的时间内将其与主表连接起来以获得此时的记录。

请注意，如果有多个记录与一小时的最小记录共享同一时间，则此处存在潜在问题。有很多方法可以解决这个问题，但这取决于你的数据（例如，如果你有一个唯一的id字段总是随时间升序那么你可以选择每小时的最小id并加入基于那）

Answer 4

您可以使用以下查询，这是更优化的，只需确保时间字段应编入索引。

SELECT HOUR(m.time), m.* 
FROM mytable AS m
JOIN
(
SELECT MIN(`time`) AS tm
FROM mytable 
WHERE `time` >= '2015-09-12 00:00:00' AND `time` <= '2015-09-12 23:59:59' 
GROUP BY HOUR(`time`)
) AS a ON m.time=a.tm 
GROUP BY HOUR(m.time)
ORDER BY m.time;

MySQL中GROUP BY的逆序

4 个答案: