显示正确的"访问最多"日期

Question

我有一些问题,以显示访问最多的日期.以下是数据库中的样子:

date_firstvisit
2012-11-25 15:49:16
2012-11-21 17:42:40
2012-11-21 15:36:04
2012-11-20 07:39:11
2012-11-19 09:19:36
2012-11-15 23:03:24
2012-11-15 22:57:47
2012-11-15 03:10:20
2012-11-15 03:10:14

此SQL查询应该打印2012-11-15但只打印0000-00-00:

SELECT date_firstvisit FROM visitors
GROUP BY DATE(date_firstvisit)
ORDER BY COUNT(date_firstvisit) DESC
LIMIT 1

如果我更换SELECT date_firstvisit FROM visitors到SELECT * FROM visitors它打印2012-11-28.这当然是错的!

我不知道我的SQL查询有什么问题,所以我现在问你,我错过了什么吗？

编辑

此代码显示了我如何将数据打印到我的网站.根据njk这个SQL是正确的,但它只是继续打印0000-00-00.

$get_mostactive_date = $sql->query("SELECT DATE(date_firstvisit), COUNT(date_firstvisit) FROM visitors GROUP BY DATE(date_firstvisit) ORDER BY COUNT(date_firstvisit) DESC LIMIT 1");
$mostactive_date = $get_mostactive_date->fetch(PDO::FETCH_ASSOC);

echo $mostactive_date['date_lastactive'];

提前致谢.

Answer 1

这将显示具有相同最高总访问次数的日期

SELECT DATE(date_firstvisit) Dates, COUNT(*) totalCount
FROM visitors
GROUP BY DATE(date_firstvisit)
HAVING COUNT(*) = 
(
  SELECT MAX(s)
  FROM(
  SELECT COUNT(*) s
  FROM visitors
  GROUP BY DATE(date_firstvisit))s
)

• SQLFiddle演示
• SQLFiddle演示(带有重复项)