SQL ：我用 select max() 和 select count() 搜索好的查询

Question

我有一个table叫arrived（这个table很简单，就是把当天到校的用户保存起来这样的小时）具有以下数据：

id  name    day         hour
1   Alice   Monday      11
2   Alice   Monday      13
3   Alice   Tuesday     11
4   Céline  Wednesday   14
5   Céline  Wednesday   13
6   Céline  Thursday    14
7   Maud    Friday      15
8   Maud    Saturday    15
9   Maud    Saturday    16

现在，我搜索为每个用户找到的好查询：最频繁的一天和最频繁的小时，也就是说，查询的结果必须return这一行：

Alice   Monday      11
Céline  Wednesday   14
Maud    Saturday    15

=> 因为：

Alice 经常在星期一到达，也经常在 11 点到达
Céline 经常在星期三到达，而且经常在 14 点到达
Maud 经常在周六到达，而且经常在 15 点到达

我的查询如下，但没有给我好的结果:

SELECT NAME,
       day,
       Max(count_hour)
FROM   (SELECT NAME,
               day,
               Count(hour) AS count_hour
        FROM   arrived
        GROUP  BY NAME,
                  day) AS alias_table
GROUP  BY NAME

衷心感谢。

Answer 1

选择列时尝试greatest(day, hour)。您也可以尝试 greatest(max(day), max(hour))

干杯！

Answer 2

不确定这样做是否正确，但应该可行！

SELECT *
FROM   arrived A
WHERE  hour = (SELECT hour
               FROM   arrived B
               WHERE  a.NAME = b.NAME
               GROUP  BY b.hour,
                         b.NAME
               ORDER  BY Count(b.hour) DESC Limit 1)
       AND day = (SELECT day
                  FROM   arrived B
                  WHERE  a.NAME = b.NAME
                  GROUP  BY b.day,
                            b.NAME
                  ORDER  BY Count(b.day) DESC Limit 1)

SQLFIDDLE DEMO

SQL ：我用 select max() 和 select count() 搜索好的查询

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