sql 生成桌上足球的比赛时间表（可变的 2 人团队和所有可能的比赛）

Question

好的，我想问你一些类似的问题：Cross Join without duplicate combinations 我有 8 名桌上足球运动员。我们想玩所有可能的组合游戏。因此，为了生成所有可能的球队（2 名球员），我可以使用这个解决方案（我认为 28 支球队是正确的）：

select distinct
        case when a.id<=b.id then a.id else b.id end as p1,
        case when a.id<=b.id then b.id else a.id end as p2   
from
        scores.players a join scores.players b on a.id!=b.id

但是我怎样才能在所有可能的团队之间生成所有可能的游戏而不重复呢？我不知道如何比较所有列。我尝试使用这个查询，结果是 420 个组合，但在我看来太多了：

select distinct
    t1.p1 as t1p1,
    t1.p2 as t1p2,
    t2.p1 as t2p1,
    t2.p2 as t2p2
from
    (select distinct
        case when a.id<=b.id then a.id else b.id end as p1,
        case when a.id<=b.id then b.id else a.id end as p2   
    from
        scores.players a join scores.players b on a.id!=b.id) t1
     join
            (select distinct
                case when a.id<=b.id then a.id else b.id end as p1,
                case when a.id<=b.id then b.id else a.id end as p2   
            from
                scores.players a
                join scores.players b on a.id!=b.id) t2 on (t1.p1!=t2.p1 and t1.p1!=t2.p2 and t1.p2!=t2.p1 and t1.p2!=t2.p2)

Answer 1

再试一次：

select t1.id as t1p1,
       t2.id as t1p2
from players t1
  join players t2 on t1.id > t2.id
  join (select t3.id as t2p1,
               t4.id as t2p2
        from players t3
          join players t4 on t3.id > t4.id)
    on  t2p1 not in (t1.id,t2.id)
    and t2p2 not in (t1.id,t2.id)
    and t1.id > t2p1

返回 210 行！

Answer 2

我检查过：

select t1.id as t1p1,
       t2.id as t1p2,
       t3.id as t2p1,
       t4.id as t2p2
from scores.players t1
  join scores.players t2 on t1.id > t2.id
  join scores.players t3 on t2.id > t3.id
  join scores.players t4 on t3.id > t4.id

结果（70 行）：

    t1p1    t1p2    t2p1    t2p2
------------------------------
    d   c   b   a
    e   c   b   a
    e   d   b   a
    e   d   c   a
    e   d   c   b
    f   c   b   a
    f   d   b   a
    f   d   c   a
    f   d   c   b
    f   e   b   a
    f   e   c   a
    f   e   c   b
    f   e   d   a
    f   e   d   b
    f   e   d   c
    g   c   b   a
    g   d   b   a
    g   d   c   a
    g   d   c   b
    g   e   b   a
    g   e   c   a
    g   e   c   b
    g   e   d   a
    g   e   d   b
    g   e   d   c
    g   f   b   a
    g   f   c   a
    g   f   c   b
    g   f   d   a
    g   f   d   b
    g   f   d   c
    g   f   e   a
    g   f   e   b
    g   f   e   c
    g   f   e   d
    h   c   b   a
    h   d   b   a
    h   d   c   a
    h   d   c   b
    h   e   b   a
    h   e   c   a
    h   e   c   b
    h   e   d   a
    h   e   d   b
    h   e   d   c
    h   f   b   a
    h   f   c   a
    h   f   c   b
    h   f   d   a
    h   f   d   b
    h   f   d   c
    h   f   e   a
    h   f   e   b
    h   f   e   c
    h   f   e   d
    h   g   b   a
    h   g   c   a
    h   g   c   b
    h   g   d   a
    h   g   d   b
    h   g   d   c
    h   g   e   a
    h   g   e   b
    h   g   e   c
    h   g   e   d
    h   g   f   a
    h   g   f   b
    h   g   f   c
    h   g   f   d
    h   g   f   e

所以我试图找到我们玩的第一个真正的游戏：

d e a f 

意思是：d & e VS a & f

这种组合存在于结果中（f e d a - f & e VS d & a），但又不是完全一样的游戏

我希望现在我能更清楚地解释我的问题。