Question

我有三个表，结构如下所示。

此表（称为contest_submissions）存储提交和竞赛的关系。

ContestID | SubmissionID

1           1000
1           1001
1           1002
1           1003

第二个表（称为submissions）存储提交的详细信息：

SubmissionID | ProblemID | User | Score | Time

1000           1000        A      100     1000
1001           1000        A      40      1250
1002           1001        A      50      1500
1003           1001        B      20      1750

另一个表（称为contest_contestants）包含：

ContestID | User

1           A
1           B

我写了以下SQL：

SELECT *, (
    SELECT SUM(score)
    FROM  contest_submissions cs
    NATURAL JOIN submissions
    WHERE user = cc.user
    AND SubmissionID = cs.SubmissionID
) AS TotalScore, (
    SELECT SUM(Time)
    FROM contest_submissions cs
    NATURAL JOIN submissions
    WHERE user = cc.user
    AND SubmissionID = cs.SubmissionID
) AS TotalTime
FROM contest_contestants cc
WHERE contestID = 1

我得到了以下结果（假设为ContestID = 1）：

contestID | User | Total Score | Total Time
1           A      190           3750
1           B      20            1750

其中190 = 100 + 40 + 50。

但是，我希望得到以下结果：

contestID | User | Total Score | Total Time
1           A      150           2500
1           B      20            1750

其中150 = MAX(100, 40) + 50，因为100和40来自同一问题（具有相同的ProblemID）。

我该怎么办？

BTW，我正在使用MySQL。

Answer 1

你可以尝试这样的事情：

select User, sum(MaxScore)
from
(
select User, ProblemID, max(Score) as MaxScore
from submissions
group by User, ProblemId
) as t
group by User

Answer 2

嗯。我认为有一种方法可以只使用一个group by：

select s.user, sum(s.score)
from submissions s
where s.submissionId = (select s2.submissionId
                        from submissions s2
                        where s2.user = s.user and s2.ProblemId = s.ProblemId
                        order by s2.score desc
                        limit 1
                       )
group by s.user;

我提供此解决方案是因为submissions(user, ProblemId, score, submissionId)上的索引应该比具有两个聚合的解决方案具有更好的性能。

Answer 3

您可以使用嵌套查询 - 内部查询来获取用户＆＃39;每个问题的最佳答案和外部问题的总结：

SELECT   user, SUM(score) AS total_score
FROM     (SELECT   user, problemid, MAX(score) AS score
          FROM     submission
          GROUP BY user, problemid) t
GROUP BY user

MySQL中不同值的总和

3 个答案: