Question

所以我有一大堆投票进入投票系统。我想显示我在任何一天有多少票。但我当时也希望显示每天的投票数量并吐出他们被投票的那天，即2012年6月5日的24k票，2012年6月6日的27k票。

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，我试图添加类似

的内容

DAY（FROM_UNIXTIME（datesubmitted））作为order_day

但是这只会抛出一个null，我发现它很有趣，因为我希望查询失败，因为没有任何outers。

Answer 1

为什么不简单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

每个单独日的投票数：

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

所以刚刚看到了一个新答案，但对我而言，我的工作方式是：

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

通过这种方式，我可以正确地获得正确的年，月，日订购，并显示日期。我可以把它们连在一起，但那是另一天。

MySQL按天分组然后显示分组结果中的日期

3 个答案: