MySQL - how to retrieve columns in same row as the values returned by min/mx - sql

Question

I couldn't frame the question's title properly. Suppose a table of weekly movie earnings as below:
MovieName <Varchar(450)>
MovieGross <Decimal(18)>
WeekofYear <Integer>
Year <Integer>
So how do I get the names of top grossers for each week of this year, if I do:
select MovieName , Max(MovieGross) , WeekofYear
from earnings where year = 2010 group by WeekofYear;
Then obviously the query wont run, while
select Max(MovieName) , Max(MovieGross) , WeekofYear
from earnings where year = 2010 group by WeekofYear;
would just give movies starting with lowest alphabet. Is using group_concat() and then substring_index() the only option here?
select
substring_index(group_concat(MovieName order by MovieGross desc),',',1),
Max(MovieGross) , WeekofYear from earnings where year = 2010
group by WeekofYear ;
Seems clumsy. Is there any better way of achieving this?

Answer 1

It's the ever-recurring max-per-group problem. You solve it by selecting the defining properties of your group and then joining your "real" data against that.
select
e.MovieName,
e.MovieGross,
e.WeekofYear
from
earnings e
inner join (
select Max(MovieGross) MovieGross, Year, WeekofYear
from earnings
group by Year, WeekofYear
) max on max.Year = e.Year
and max.WeekofYear = e.WeekofYear
and max.MovieGross = e.MovieGross
where
e.year = 2010
The defining properties of your group are Year, WeekofYear and MAX(MovieGross). There will be one row with different values for each group range.
An INNER JOIN against your data table elimitates all rows that do not fulfill the defining properties of your group. This also means that it lets through all rows that do - you could end up with two movies that made the same amount of money in any particular week. Group the "outer" query again to eliminate the duplicate rows in favor of a single movie.

Answer 2

You need to determine the max weekly gross and then select the movie name based on that criterion. Something like this:
SELECT e.MovieName, m.Gross, m.WeekofYear
FROM earnings e JOIN
(SELECT MAX(MovieGross) Gross, WeekofYear
FROM earnings WHERE `year` = 2010 GROUP BY WeekofYear) m
ON e.MovieGross=m.Gross AND e.WeekofYear=m.WeekofYear

Answer 3

This is pretty fast query, that does the job:
SELECT e.WeekofYear as WeekofYear
, max(MovieGross) as MovieGross
, (SELECT MovieName FROM earnings
WHERE WeekofYear=e.WeekofYear ORDER BY MovieGross DESC LIMIT 1
) as MovieName
FROM earnings AS e
WHERE year='2010'
GROUP BY WeekofYear
ORDER BY WeekofYear;
Happy to help you :)
P.S. and thanks for ratings ;)

Answer 4

Ok, trying again with the having clause, I cannot help myself.
Hopefully this will help you get started. First, create a list of the weeks of the year, then for the inner query, find the one that has the max for that week.
Select MovieName, max(MovieGross) as max_gross, WeekofYear
from earnings
where year = 2010
order by MovieGross desc
Having MovieGross=max_gross
group by WeekofYear
This should return the top grossing movie for each week. This should also return multiple entries for a week in the event of a tie.