php - MySQL 按天分组，然后显示分组结果中的日期

Question

所以我有一大堆选票进入投票系统。我想显示我在任何一天有多少票。但我也想然后，显示每天的票数并吐出他们在哪一天投票，即 24k 票在 05/06/12，27k 票在 06/06/12

SELECT count(*) AS count
FROM results 
GROUP BY DAY(datesubmitted), YEAR(datesubmitted), MONTH(datesubmitted) 
ORDER  BY DAY(datesubmitted) DESC, YEAR(datesubmitted) DESC, MONTH(datesubmitted) DESC

是我的查询，我尝试添加类似

DAY(FROM_UNIXTIME(datesubmitted)) 作为 order_day

但这只是抛出一个我觉得有趣的空值，因为我希望查询失败，因为没有任何外部。

Answer 1

你为什么不干脆GROUP BY datesubmitted DESC？此外，如果它遵循与 GROUP BY 相同的标准，则无需 ORDER BY。

Answer 2

特定日期的投票数：

SELECT COUNT(*) AS total
FROM results 
WHERE datesubmitted BETWEEN @dateMin AND @dateMax

每一天的票数：

SELECT COUNT(*) AS total, DATE(datesubmitted) AS day
FROM results 
GROUP BY DATE(datesubmitted)
ORDER BY DATE(datesubmitted) DESC

再次更新

Answer 3

所以刚刚看到一个新的答案，但对我来说，我是如何让它工作的：

SELECT count(*) AS count, DAY(datesubmitted) AS newday, 
YEAR(datesubmitted) as newyear ,MONTH(datesubmitted) as newmonth
FROM results 
GROUP BY DAY(datesubmitted), YEAR(datesubmitted), MONTH(datesubmitted) 
ORDER  BY  YEAR(datesubmitted) DESC, MONTH(datesubmitted) DESC, DAY(datesubmitted) DESC

这样我就可以正确地获得年、月和日的正确排序，并且它还显示日期。我可以连接它们，但那是另一天。