8

我有一个 SQL 查询，它返回有关单个团队的下表：

date         gameid     pointsfor     pointsagainst

2011-03-20   15         1             10
2011-03-27   17         7             3
2011-04-03   23         6             5
2011-04-10   30         5             4
2011-04-17   35         4             8
2011-05-01   38         8             1
2011-05-08   43         3             7
2011-05-15   48         6             2
2011-05-22   56         10            2
2011-05-29   59         4             5
2011-06-05   65         2             3
2011-06-19   71         5             6
2011-06-19   74         12            2
2011-06-19   77         5             2
2011-06-19   80         5             4

从这张表中，谁能帮我计算一下最长的连胜和连败是多少？

我在这里查看了其他一些示例，但由于它们与我的不太一样，因此很难遵循它们。任何帮助将不胜感激。谢谢！

4

7 回答 7

2

你必须实现一些 MySQL 变量来帮助处理这个比多个查询/加入/分组更有效。这一次通过所有记录，然后再次获得每种类型的赢/输（或平局）的最大值。我假设您提供的数据是单次的，日期是明显的比赛顺序……积分是您感兴趣的球队，而反对的对手是谁。也就是说，我的“别名”名称将是“YourResultSingleTeam”。

内部查询将预先确定游戏的状态为“W”in 或“L”oss。然后，查看该值是否与团队的前一个实例相同。如果是这样，将 1 添加到现有的赢/输计数器。如果不是，则将计数器设置回 1。然后，将当前游戏的状态保留回“LastStatus”值，以便与下一个游戏进行比较。

完成后，它是一个简单的游戏结果，max() 按游戏结果状态分组

select
      StreakSet.GameResult,
      MAX( StreakSet.WinLossStreak ) as MaxStreak
   from
      ( select YR.Date,
               @CurStatus := if( YR.PointsFor > YR.PointsAgainst, 'W', 'L' ) as GameResult,
               @WinLossSeq := if( @CurStatus = @LastStatus, @WinLossSeq +1, 1 ) as WinLossStreak,
               @LastStatus := @CurStatus as carryOverForNextRecord
            from 
               YourResultSingleTeam YR,
               ( select @CurStatus := '',
                        @LastStatus := '',
                        @WinLossSeq := 0 ) sqlvars
            order by
               YR.Date ) StreakSet
   group by
      StreakSet.GameResult

正如 Nikola 所提供的，如果您想考虑“平局”游戏，我们可以通过将 @CurStatus 更改为 case/when 条件来进行调整

@CurStatus := case when YR.PointsFor > YR.PointsAgainst then 'W'
                   when YR.PointsFor < YR.PointsAgainst then 'L'
                   else 'T' end as GameResult,
于 2012-05-07T12:53:27.187 回答
1

有一个解决方案，但我认为您不会喜欢它，因为它需要自联接，而且您的表不是表而是查询。

内部查询会将日期转换为范围 - 对于表中的每个日期，它会找到具有不同结果的第一个日期，或者，如果是上一场比赛，则是本场比赛的日期。该数据将按不同条纹的第一个日期汇总，以平整和计数条纹；外部查询然后根据结果查找极端情况。

select case Outcome 
            when -1 then 'Losses'
            when 1 then 'Wins'
            else 'Undecided'
        end Title
      , max(Streak) Streak
from
(
  select min(date) date, date_to, Outcome, count(*) Streak
  from
  (
    select t1.date, 
           sign (t1.pointsfor - t1.pointsagainst) Outcome, 
           ifnull (min(t2.date), t1.date) date_to
     from table1 t1
     left join table1 t2
       on t1.date < t2.date
      and sign (t1.pointsfor - t1.pointsagainst) 
       <> sign (t2.pointsfor - t2.pointsagainst)
    group by t1.date, sign (t1.pointsfor - t1.pointsagainst)
  ) a
  group by date_to, Outcome
) a
group by Outcome

为了避免需要替换table1- 可能 - 繁琐的查询，您可能会使用临时表，或者在辅助表中已经有适当格式的数据。在Sql fiddle进行了实时测试，以及另一个可能表现更好的子查询驱动版本 - 您应该同时尝试它们。

于 2012-05-07T12:12:17.893 回答
1

MySQL 没有 CTE 也没有窗口函数（例如 SUM OVER、ROW_NUMBER OVER 等）。但它有一个赎回因素。变量！

用这个：

select 
   min(date) as start_date,
   max(date) as end_date,
   count(date) as streak,
   group_concat(gameid) as gameid_list
from
( 
  select *,      
    IF(
        pointsfor > pointsagainst 
        and 
        @pointsfor > @pointsagainst, 
           @gn, @gn := @gn + 1)                
    as group_number,

    @date as old_date, @gameid as old_gameid, 
    @pointsfor as old_pointsfor,
    @pointsagainst as old_pointsagainst,

    @date := date, @gameid := gameid, 
    @pointsfor := pointsfor, @pointsagainst := pointsagainst      
  from tbl
  cross join 
  (
    select 
      @date := CAST(null as date) as xa,
      @gameid := null + 0 as xb, -- why CAST(NULL AS INT) doesn't work?
      @pointsfor := null + 0 as xc, @pointsagainst := null + 0 as xd, @gn := 0
  ) x
  order by date
) as y
group by group_number
order by streak desc;

输出：

START_DATE                    END_DATE                      STREAK  GAMEID_LIST
March, 27 2011 08:00:00-0700  April, 10 2011 08:00:00-0700  3       17,23,30
June, 19 2011 08:00:00-0700   June, 19 2011 08:00:00-0700   3       74,77,80
May, 15 2011 08:00:00-0700    May, 22 2011 08:00:00-0700    2       48,56
March, 20 2011 08:00:00-0700  March, 20 2011 08:00:00-0700  1       15
April, 17 2011 08:00:00-0700  April, 17 2011 08:00:00-0700  1       35
May, 01 2011 08:00:00-0700    May, 01 2011 08:00:00-0700    1       38
May, 08 2011 08:00:00-0700    May, 08 2011 08:00:00-0700    1       43
May, 29 2011 08:00:00-0700    May, 29 2011 08:00:00-0700    1       59
June, 05 2011 08:00:00-0700   June, 05 2011 08:00:00-0700   1       65
June, 19 2011 08:00:00-0700   June, 19 2011 08:00:00-0700   1       71

现场测试：http ://www.sqlfiddle.com/#!2/bbe78/8

请注意我在 sqlfiddle 上的解决方案，它有两个查询。1. 模拟在上面。2.下面的最终查询

于 2012-05-07T13:54:09.803 回答
1

最新版本的 MySQL 具有 CTE 并且具有窗口功能。

这是一个解决方案。

第一步，通过分配他们自己的 streak_group 编号来分组输赢：

with t as 
(
    select
        *,      
        pointsfor - pointsagainst > 0 is_winner,
        case when pointsfor - pointsagainst > 0 
            and lag(pointsfor) over(order by date, pointsfor - pointsagainst desc) 
                - lag(pointsagainst) over(order by date, pointsfor - pointsagainst desc) > 0 
        then
            0
        else
            1
        end as is_new_group
    from tbl
)
select *, sum(is_new_group) over(order by date, pointsfor - pointsagainst desc) as streak_group
from t

输出：

date                |gameid |pointsfor |pointsagainst |is_winner |is_new_group |streak_group |
--------------------|-------|----------|--------------|----------|-------------|-------------|
2011-03-20 15:00:00 |15     |1         |10            |0         |1            |1            |
2011-03-27 15:00:00 |17     |7         |3             |1         |1            |2            |
2011-04-03 15:00:00 |23     |6         |5             |1         |0            |2            |
2011-04-10 15:00:00 |30     |5         |4             |1         |0            |2            |
2011-04-17 15:00:00 |35     |4         |8             |0         |1            |3            |
2011-05-01 15:00:00 |38     |8         |1             |1         |1            |4            |
2011-05-08 15:00:00 |43     |3         |7             |0         |1            |5            |
2011-05-15 15:00:00 |48     |6         |2             |1         |1            |6            |
2011-05-22 15:00:00 |56     |10        |2             |1         |0            |6            |
2011-05-29 15:00:00 |59     |4         |5             |0         |1            |7            |
2011-06-05 15:00:00 |65     |2         |3             |0         |1            |8            |
2011-06-19 15:00:00 |74     |12        |2             |1         |1            |9            |
2011-06-19 15:00:00 |77     |5         |2             |1         |0            |9            |
2011-06-19 15:00:00 |80     |5         |4             |1         |0            |9            |
2011-06-19 15:00:00 |71     |5         |6             |0         |1            |10           |

最终查询。计算连胜次数：

with t as 
(
    select
        *,      
        pointsfor - pointsagainst > 0 is_winner,
        case when pointsfor - pointsagainst > 0 
            and lag(pointsfor) over(order by date, pointsfor - pointsagainst desc) 
                - lag(pointsagainst) over(order by date, pointsfor - pointsagainst desc) > 0 
        then
            0
        else
            1
        end as is_new_group
    from tbl
)
, streak_grouping as
(
    select
        *, sum(is_new_group) over(order by date, pointsfor - pointsagainst desc) as streak_group
    from t
)
select 
    min(date) as start_date,
    max(date) as end_date,
    count(*) as streak,
    group_concat(gameid order by gameid) as gameid_list
from streak_grouping
group by streak_group
order by streak desc, start_date

输出：

start_date          |end_date            |streak |gameid_list |
--------------------|--------------------|-------|------------|
2011-03-27 15:00:00 |2011-04-10 15:00:00 |3      |17,23,30    |
2011-06-19 15:00:00 |2011-06-19 15:00:00 |3      |74,77,80    |
2011-05-15 15:00:00 |2011-05-22 15:00:00 |2      |48,56       |
2011-03-20 15:00:00 |2011-03-20 15:00:00 |1      |15          |
2011-04-17 15:00:00 |2011-04-17 15:00:00 |1      |35          |
2011-05-01 15:00:00 |2011-05-01 15:00:00 |1      |38          |
2011-05-08 15:00:00 |2011-05-08 15:00:00 |1      |43          |
2011-05-29 15:00:00 |2011-05-29 15:00:00 |1      |59          |
2011-06-05 15:00:00 |2011-06-05 15:00:00 |1      |65          |
2011-06-19 15:00:00 |2011-06-19 15:00:00 |1      |71          |
于 2019-03-12T12:24:54.540 回答
0

您在这里处理的是跟踪获胜和失败的趋势，这需要使用某种带有运行计数器的循环来计算，而不是在 SQL 中。SQL 查询处理单个行、分组、排序等；您正在尝试使用一种并非旨在解决此类问题的语言。

于 2012-05-07T08:24:32.737 回答
0

你必须创建一个游标，读取所有行，计算数据......每次你想要获得最长的条纹......

我建议一种解决方法，这将使事情变得更容易。您在表“streakFor”中添加一列。每次插入一行时：

//pseudo code
if pointsFor > pointsAgainst
    if last_streakFor > 0 
        then streakFor++ 
        else streakFor = 1

else
    if last_streakFor > 0 
        then streakFor = -1 
        else streakFor--

last_streakFor 在最后插入的行中是 streakFor
然后你插入带有列 streakFor 的行

现在你可以

  • select max(streakFor) from yourTable where yourConditions这将为您提供“pointsFor”的最长连胜和“pointsAgainst”的最长连败
  • select min(streakFor) from yourTable where yourConditions将为您提供“pointsAgainst”的最长连胜和“pointsFor”的最长连败
于 2012-05-07T08:26:03.957 回答
0

感谢所有的帮助家伙。我最终按照建议使用 php 循环。如果有人想知道，这是我的代码：

$streakSQL = "SELECT date, gameid, pointsfor, pointsagainst FROM result WHERE teamid = ".$_GET['teamid']." AND bye = 0 AND COMPLETED = 1 AND seasonid > 7 AND roundwd = 0 AND (pointsfor != 0 OR pointsagainst != 0)";
            $streak = mysql_query($streakSQL);

            $winstreak = 0;
            $maxwinstreak = 0;
            $losestreak = 0;
            $maxlosestreak = 0;
            while($streakRow = mysql_fetch_array($streak))
            {
                //calculate winning streak
                if($streakRow['pointsfor'] > $streakRow['pointsagainst'])
                { 
                    $winstreak++; 
                    if($winstreak > $maxwinstreak)
                    {
                        $maxwinstreak = $winstreak;
                    }
                }
                else{ $winstreak = 0; }
                //calculate losing streak
                if($streakRow['pointsfor'] < $streakRow['pointsagainst'])
                { 
                    $losestreak++; 
                    if($losestreak > $maxlosestreak)
                    {
                        $maxlosestreak = $losestreak;
                    }
                }
                else{ $losestreak = 0; }
            }
            echo "Biggest Winning Streak: ".$maxwinstreak;
            echo "<br />Biggest Losing Streak: ".$maxlosestreak;
于 2012-05-07T15:32:00.130 回答