Mysql SQL分组依据：获取“max”记录的值

我有一个分数表：

现在，我可以很容易地用

SELECT user, score, date FROM scores ORDER BY score DESC

然而，我只想包括每个用户的一个分数，即他的最高分数。我将从以下内容开始

SELECT user, MAX(score) AS score FROM scores GROUP BY user ORDER BY score DESC

然而，现在我已经失去了记录最高分数的日期。如何获得它？

您可以再次加入到表中：

SELECT s1.user, max(s1.dt), s2.mxscore as score
FROM scores s1
inner join 
(
    select user, max(score) mxscore
    from scores
    GROUP BY user 
) s2
    on s1.user = s2.user
    and s1.score = s2.mxscore
GROUP BY s1.username, s2.mxscore
ORDER BY score DESC

---

请参见

您需要将结果与原始表格相关联：

select a.user, a.maxScore, b.maxDate
from (
    select user, max(score) as maxScore 
    from scores group by user ) as a
inner join (
    select user, score, max(date) as maxDate 
    from scores group by user, score) as b on a.user = b.user and a.maxScore=b.score
order by
    a.maxScore desc

此查询将返回每个用户的最高分数，以及该最高分数被打分的最后日期，但为true

SELECT   a.*
FROM     scores a
JOIN     (
         SELECT   MAX(a.id) AS id
         FROM     scores a
         JOIN     (
                  SELECT   user, MAX(score) AS score
                  FROM     scores
                  GROUP BY user
                  ) b ON a.user  = b.user
                     AND a.score = b.score
         GROUP BY a.user, 
                  a.score
         ) b ON a.id = b.id
ORDER BY a.score DESC

这将解释每个用户有多个相同最高分数的情况。在这种情况下，它只需要最大id。

事实上，您根本不需要一个GroupBy

问题是：

SELECT scores.id, scores.user, scores.score, scores.date
FROM scores
WHERE NOT EXISTS (
  SELECT * 
  FROM scores AS _scores
  WHERE _scores.user = scores.user
  AND (
    _scores.score > scores.score
    OR
    _scores.score = scores.score AND _scores.id < scores.id) 
)

并看到它的工作

---

请注意，此查询正确地处理了用户多次获得最大分数的情况，它返回了第一个最大分数的记录。

如果有多个日期具有相同的分数，会发生什么情况？id字段不太可能是用户id，而是一个自动递增的字段，作为每个分数实例的主键。

SELECT scores.id, scores.user, scores.score, scores.date
FROM scores
WHERE NOT EXISTS (
  SELECT * 
  FROM scores AS _scores
  WHERE _scores.user = scores.user
  AND (
    _scores.score > scores.score
    OR
    _scores.score = scores.score AND _scores.id < scores.id) 
)