我有如下表格:

id     homeTeam    awayTeam   homeScore    awayScore
1         t1          t2          3            2
2         t3          t7          0            1
.          .           .          .            .
.          .           .          .            .
.          .           .          .            .

这是本地联赛的足球比赛结果。 我想获得“最长连胜”、“最长连败”和......只需要一个查询。我找了一圈，找到了oracle版本，但是我找不到如何做？ PS:我有mysql数据库。 提前致谢

最佳答案

这是一种方法，但我觉得您不会喜欢它...

考虑以下数据 DDL...

CREATE TABLE results
(id     INT NOT NULL AUTO_INCREMENT PRIMARY KEY
,homeTeam    INT NOT NULL
,awayTeam    INT NOT NULL
,homeScore    INT NOT NULL
,awayScore INT NOT NULL
);

INSERT INTO results VALUES
(1,1,2,3,2),
(2,3,4,0,1),
(3,2,1,2,0),
(4,4,3,1,0),
(5,3,2,1,2),
(6,2,3,0,2),
(7,1,4,4,1),
(8,4,1,1,2),
(9,1,3,3,0),
(10,3,1,1,0),
(11,4,2,1,0),
(12,2,4,1,2);

从这里，我们可以得到如下的中间结果......

SELECT x.*, COUNT(*) rank
  FROM
     ( SELECT id,hometeam team, CASE WHEN homescore > awayscore THEN 'w' ELSE 'l' END result FROM results 
       UNION
       SELECT id,awayteam, CASE WHEN awayscore > homescore THEN 'w' ELSE 'l' END result FROM results
     ) x
  JOIN 
     ( SELECT id,hometeam team, CASE WHEN homescore > awayscore THEN 'w' ELSE 'l' END result FROM results 
       UNION
       SELECT id,awayteam, CASE WHEN awayscore > homescore THEN 'w' ELSE 'l' END result FROM results
     ) y
    ON y.team = x.team
   AND y.id <= x.id
 GROUP
    BY x.id
     , x.team
 ORDER
    BY team, rank;

+----+------+--------+------+
| id | team | result | rank |
+----+------+--------+------+
|  1 |    1 | w      |    1 |
|  3 |    1 | l      |    2 |
|  7 |    1 | w      |    3 |
|  8 |    1 | w      |    4 |
|  9 |    1 | w      |    5 |
| 10 |    1 | l      |    6 |
|  1 |    2 | l      |    1 |
|  3 |    2 | w      |    2 |
|  5 |    2 | w      |    3 |
|  6 |    2 | l      |    4 |
| 11 |    2 | l      |    5 |
| 12 |    2 | l      |    6 |
|  2 |    3 | l      |    1 |
|  4 |    3 | l      |    2 |
|  5 |    3 | l      |    3 |
|  6 |    3 | w      |    4 |
|  9 |    3 | l      |    5 |
| 10 |    3 | w      |    6 |
|  2 |    4 | w      |    1 |
|  4 |    4 | w      |    2 |
|  7 |    4 | l      |    3 |
|  8 |    4 | l      |    4 |
| 11 |    4 | w      |    5 |
| 12 |    4 | w      |    6 |
+----+------+--------+------+

通过观察，我们可以看到团队 1 的连胜时间最长(连续 3 次“w”)。你可以设置几个@vars 来跟踪这个，或者，如果你有点自虐(像我一样)，你可以做一些更慢、更长、更复杂的事情......

SELECT a.team
     , MIN(c.rank) - a.rank + 1 streak
  FROM (SELECT x.*, COUNT(*) rank
  FROM
     ( SELECT id,hometeam team, CASE WHEN homescore > awayscore THEN 'w' ELSE 'l' END result FROM results 
       UNION
       SELECT id,awayteam, CASE WHEN awayscore > homescore THEN 'w' ELSE 'l' END result FROM results
     ) x
  JOIN 
     ( SELECT id,hometeam team, CASE WHEN homescore > awayscore THEN 'w' ELSE 'l' END result FROM results 
       UNION
       SELECT id,awayteam, CASE WHEN awayscore > homescore THEN 'w' ELSE 'l' END result FROM results
     ) y
    ON y.team = x.team
   AND y.id <= x.id
 GROUP
    BY x.id
     , x.team
     ) a
  LEFT 
  JOIN (SELECT x.*, COUNT(*) rank
  FROM
     ( SELECT id,hometeam team, CASE WHEN homescore > awayscore THEN 'w' ELSE 'l' END result FROM results 
       UNION
       SELECT id,awayteam, CASE WHEN awayscore > homescore THEN 'w' ELSE 'l' END result FROM results
     ) x
  JOIN 
     ( SELECT id,hometeam team, CASE WHEN homescore > awayscore THEN 'w' ELSE 'l' END result FROM results 
       UNION
       SELECT id,awayteam, CASE WHEN awayscore > homescore THEN 'w' ELSE 'l' END result FROM results
     ) y
    ON y.team = x.team
   AND y.id <= x.id
 GROUP
    BY x.id
     , x.team
     ) b 
    ON b.team = a.team
   AND b.rank = a.rank - 1 
   AND b.result = a.result
  LEFT 
  JOIN (SELECT x.*, COUNT(*) rank
  FROM
     ( SELECT id,hometeam team, CASE WHEN homescore > awayscore THEN 'w' ELSE 'l' END result FROM results 
       UNION
       SELECT id,awayteam, CASE WHEN awayscore > homescore THEN 'w' ELSE 'l' END result FROM results
     ) x
  JOIN 
     ( SELECT id,hometeam team, CASE WHEN homescore > awayscore THEN 'w' ELSE 'l' END result FROM results 
       UNION
       SELECT id,awayteam, CASE WHEN awayscore > homescore THEN 'w' ELSE 'l' END result FROM results
     ) y
    ON y.team = x.team
   AND y.id <= x.id
 GROUP
    BY x.id
     , x.team
     ) c 
    ON c.team = a.team
   AND c.rank >= a.rank 
   AND c.result = a.result
  LEFT 
  JOIN (SELECT x.*, COUNT(*) rank
  FROM
     ( SELECT id,hometeam team, CASE WHEN homescore > awayscore THEN 'w' ELSE 'l' END result FROM results 
       UNION
       SELECT id,awayteam, CASE WHEN awayscore > homescore THEN 'w' ELSE 'l' END result FROM results
     ) x
  JOIN 
     ( SELECT id,hometeam team, CASE WHEN homescore > awayscore THEN 'w' ELSE 'l' END result FROM results 
       UNION
       SELECT id,awayteam, CASE WHEN awayscore > homescore THEN 'w' ELSE 'l' END result FROM results
     ) y
    ON y.team = x.team
   AND y.id <= x.id
 GROUP
    BY x.id
     , x.team
     ) d 
    ON d.team = a.team
   AND d.rank = c.rank + 1 
   AND d.result = a.result
 WHERE a.result = 'w'
   AND b.id IS NULL
   AND c.id IS NOT NULL
   AND d.id IS NULL
 GROUP 
    BY a.team
     , a.rank
 ORDER 
    BY streak DESC 
 LIMIT 1; 

 +------+--------+
 | team | streak |
 +------+--------+
 |    1 |      3 |
 +------+--------+

请注意，这不考虑个人比赛关系(对重复子查询的适度更改)，也不考虑两支球队是否拥有最长的相等长度的连胜纪录(需要将此处的所有内容重新加入自身!)。

关于mysql - 查询最长连胜纪录，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/20028832/