SQL：我用select max（）搜索好查询，然后选择count（）

Question

I have a table named arrived (this table is simple, and it just saves the users that arrive at school at such day and at such hour) with the following data :

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

Now, I search the good query that find for each user : the most frequent day and the most frequent hour, that is say, the result of the query must return this lines :

Alice   Monday      11
Céline  Wednesday   14
Maud    Saturday    15

=> because :

• Alice frequently arrives at Monday, and frequently at 11h
• Céline frequently arrives at Wednesday, and frequently at 14h
• Maud frequently arrives at Saturday, and frequently at 15h

My query is below, but it doesn't give me the good result :

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 

Thank you, cordially.

Answer 1

Try the greatest(day, hour) when selecting columns. You can also try greatest(max(day), max(hour))

Cheers!

Answer 2

Not sure is this the right way to do it but should work!

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) 

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