Question

我希望在分组之前看到总计数。

我有这张桌子。

+----------+------------+------------+---------+---------+--------+--------+---------+
| match_id | date       | tournament | playerA | playerB | scoreA | scoreB | offline |
+----------+------------+------------+---------+---------+--------+--------+---------+
|        1 | 2012-12-04 |        799 |       4 |      55 |      1 |      3 |       0 |
|        2 | 2012-12-03 |      11921 |       2 |      41 |      2 |      0 |       0 |
|        3 | 2012-12-03 |      11921 |      21 |      41 |      0 |      2 |       0 |
|        4 | 2012-12-03 |      11921 |       3 |       2 |      2 |      1 |       0 |
|        5 | 2012-12-03 |      11921 |      41 |       2 |      1 |      2 |       0 |
|        6 | 2012-12-03 |      11921 |      21 |       3 |      1 |      2 |       0 |
|        7 | 2012-12-03 |      11924 |       1 |       8 |      2 |      1 |       1 |
|        8 | 2012-12-03 |      11924 |       1 |       8 |      2 |      3 |       1 |
|        9 | 2012-12-03 |      11924 |       8 |      19 |      3 |      2 |       1 |
|       10 | 2012-12-03 |      11924 |      19 |      12 |      2 |      1 |       1 |
+----------+------------+------------+---------+---------+--------+--------+---------+

Sqlfiddle相同：http://sqlfiddle.com/#!9/09d661/1

这就是我尝试过的。

SELECT *, count(*), count(scoreA > scoreB) as SA, count(scoreA < scoreB) as SB
from matches
    Group By playerA, scoreA > scoreB, scoreA < scoreB;

然后我认为有一个子查询可能会起作用。

SELECT count(playerA)
from matches
Group By playerA
(
SELECT *, count(*), count(scoreA > scoreB) as SA, count(scoreA < scoreB) as SB
from matches
    Group By playerA, scoreA > scoreB, scoreA < scoreB);

这两种方法都不适用于我。

预期结果

+----------+------------+------------+---------+---------+--------+--------+---------+----------+-----+-----+
| match_id | date       | tournament | playerA | playerB | scoreA | scoreB | offline | count(*) | SA  | SB  |
+----------+------------+------------+---------+---------+--------+--------+---------+----------+-----+-----+
|      823 | 2012-11-04 |       3480 |       1 |       7 |      2 |      2 |       1 |        195 |   1 |   1 |
|        8 | 2012-12-03 |      11924 |       1 |       8 |      2 |      3 |       1 |       195 |  131 |  61 |
|        7 | 2012-12-03 |      11924 |       1 |       8 |      2 |      1 |       1 |      195 | 133 | 61 |

如果你看一下预期结果表中的最后几列，它是1,61和133.我真的不明白为什么SA和SB总是相同的。

1 + 61 + 133 = 195

Answer 1

您的预期结果看起来不太明显，但您可能需要尝试这样的事情，

select m.playerA, count(SA.*), count(SB.*)
from matches m
inner join 
    (select *
    from matches 
    where scoreA > scoreB) SA
on m.match_id = SA.match_id
inner join
    (select *
    from matches 
    where scoreA < scoreB) SB
on m.match_id = SB.match_id
group by m.playerA

Answer 2

根据你发布的sqlfiddle，我认为这是你正在尝试的：

SELECT player
      ,team
      ,start_date
      ,end_date
      ,(SELECT count(player) 
          FROM members M2
         WHERE M2.player = M.player
         group by player) AS count
  FROM members M
  group by player, team

这是一个经过编辑的sqlfiddle： http://sqlfiddle.com/#!9/09d661/4/1

sql在执行group by之前显示总计数

2 个答案: