查找用户的最低单圈时间排名

Question

我有很多赛车手，他们在不同的赛道上比赛。我想获得他们在每个特定赛道上的最佳单圈时间，并将其与其他用户在该特定赛道上的最佳单圈时间进行排名。我想为所有轨道上的所有用户执行此操作。所以我想返回一个大排行榜。

基本上，如果我有 10 位赛车手，每人在 1 号赛道完成 25 圈，我只想确定每位赛车手的最佳单圈时间，并在每条赛道上对这些时间进行排名。

我想我几乎拥有它，请参阅下面的查询和 SQLFiddle 链接：http://sqlfiddle.com/#!9/64d6c/2

为确保您拥有所有作品，还有一个“tracks.verified”列。这确定轨道是否已被验证。你真的不需要担心这个参数，如果你愿意，可以把它省掉。我指的是这一行：

AND s1.track_id IN (SELECT id FROM tracks WHERE verified=1)

http://sqlfiddle.com/#!9/64d6c/2

SELECT track_id, user_id, duration, @rank := @rank + 1 AS rank FROM
(
  SELECT DISTINCT s1.track_id, s1.user_id, s1.duration
  FROM session_lap_times s1
  LEFT JOIN session_lap_times s2
    ON s1.user_id=s2.user_id
   AND s1.track_id = s2.track_id
   AND s1.duration > s2.duration
  WHERE s2.duration IS NULL
    AND s1.track_id IN (SELECT id FROM tracks WHERE verified=1)
  ORDER BY s1.duration ASC
) zz, (SELECT @rank := 0) z;

这个输出：

+----------+---------+------------------+------+
| track_id | user_id | duration         | rank |
+----------+---------+------------------+------+
|       15 |       2 | 71001.5129995350 |    1 |
|       15 |       1 | 75001.5129995350 |    2 |
|       17 |       1 | 90258.1180334090 |    3 |
|       17 |       2 | 90897.0659971240 |    4 |
+----------+---------+------------------+------+

问题在于它的排名都是相互对立的，而不是每条赛道上的所有排名。这些结果应如下所示：

+----------+---------+------------------+------+
| track_id | user_id | duration         | rank |
+----------+---------+------------------+------+
|       15 |       2 | 71001.5129995350 |    1 |
|       15 |       1 | 75001.5129995350 |    2 |
|       17 |       1 | 90258.1180334090 |    1 |
|       17 |       2 | 90897.0659971240 |    2 |
+----------+---------+------------------+------+

我认为这行是问题所在：

AND s1.track_id IN (SELECT id FROM tracks WHERE verified=1)

因为如果我设置 s1.track_id = 15（例如），这对于单个轨道可以正常工作。只是不是全部。

感谢任何帮助！

编辑：

我正在尝试按 user_id 过滤结果。所以我把你的整个查询放到了一个子查询中。有没有办法做到这一点而不将其放入子查询中？谢谢

select track_id, user_id, duration, rank from (SELECT track_id, user_id,duration, 
  CASE WHEN track_id!=@track THEN @rank:=0 END as reset,
  @rank := @rank + 1 AS rank,
  @track:=track_id
FROM
(
  SELECT DISTINCT s1.track_id, s1.user_id, s1.duration
  FROM session_lap_times s1
  LEFT JOIN session_lap_times s2
    ON s1.user_id=s2.user_id
   AND s1.track_id = s2.track_id
   AND s1.duration > s2.duration
  WHERE s2.duration IS NULL
    AND s1.track_id IN (SELECT id FROM tracks WHERE verified=1)
  ORDER BY s1.duration ASC
) zz, (SELECT @rank := 0, @track := null) z) as x where user_id=1

至于计算我得到的驱动程序查询总数：

SELECT track_id, COUNT(user_id) FROM (SELECT track_id,user_id FROM session_lap_times GROUP BY track_id,user_id) as z GROUP BY track_id

同样，没有子查询有没有更好的方法呢？谢谢！ :)

Answer 1

SELECT track_id, user_id, duration, 
  CASE WHEN track_id!=@track THEN @rank:=0 END as reset,
  @rank := @rank + 1 AS rank,
  @track:=track_id
FROM
(
  SELECT DISTINCT s1.track_id, s1.user_id, s1.duration
  FROM session_lap_times s1
  LEFT JOIN session_lap_times s2
    ON s1.user_id=s2.user_id
   AND s1.track_id = s2.track_id
   AND s1.duration > s2.duration
  WHERE s2.duration IS NULL
    AND s1.track_id IN (SELECT id FROM tracks WHERE verified=1)
  ORDER BY s1.duration ASC
) zz, (SELECT @rank := 0, @track := null) z;

用你的小提琴测试过。

引入一个附加变量并在 track_id 更改时重置 rank

Answer 2

我发现您的 SQL 令人困惑。我会使用 GROUP BY。

SELECT track_id, user_id, MIN(duration) as best_time
FROM session_lap_times
GROUP BY track_id, user_id
ORDER BY track_id, best_time;