Question

我有两个不同的查询。一个用于＆＃34;加＆＃34;和＃34;减去＆＃34;。我想找到每个玩家的计数值差异。

我尝试了所有联盟，得到了一些非常奇怪的数字。

以下是每个查询的结果，我发现它们的区别在于：

fk_plus_player1_id  cnt
90                  71
65                  68
79                  66
45                  59
64                  57
27                  56
55                  56
93                  55
37                  55
40                  44
 1                  36
84                  33
20                  31
24                  28
 8                  23


fk_minus_player1_id cnt
93                  44
64                  42
79                  40
37                  35
90                  33
20                  31
84                  31
27                  30
65                  30
40                  26
 1                  26
24                  25
45                  25
55                  22
 8                  10

我将如何做到这一点？在此先感谢您的帮助。我是个菜鸟......

UGH ...尝试进行连接方法。有问题，没有结果，只有4个空列。这就是我想要的

SELECT * FROM（此处为plus结果的select语句）AS tp

JOIN

（在此选择减去结果的陈述）AS tm

ON tp.fk_plus_player1_id = tm.fk_minus_player1_id

GROUP BY fk_plus_player1_id

建议??

Answer 1

你有两张桌子。你想要每个球员，计数的差异。

所以：

SELECT t1.fk_minus_player1_id AS player, ABS(t1.cnt - t2.cnt) AS difference
FROM table1 t1, table2 t2
WHERE t1.fk_minus_player1_id = t2.fk_plus_player1_id
GROUP BY t1.fk_minus_player1_id;

也许这就是你要找的东西？

WITH query1 AS 
(SELECT t1.fk_minus_player1_id AS player, (t1.cnt - IFNULL(t2.cnt,0)) AS difference
FROM table1 t1 LEFT OUTER JOIN table2 t2 ON t1.fk_minus_player1_id = t2.fk_plus_player1_id
GROUP BY t1.fk_minus_player1_id), 
query2 AS (SELECT t2.fk_plus_player1_id AS player, (IFNULL(t1.cnt,0) - t2.cnt) AS difference
FROM table2 t2 LEFT OUTER JOIN table1 t1 ON t1.fk_minus_player1_id = t2.fk_plus_player1_id
GROUP BY t2.fk_plus_player1_id)

(SELECT player, difference
FROM query1)

UNION

(SELECT player, difference 
FROM query2 WHERE player NOT IN (SELECT player FROM query1))

Answer 2

您冒着两个名单中不存在相同玩家的风险。解决方案为union all group by：

select player1id, sum(pluscnt) as pluscnt, sum(minuscnt) as minuscnt,
       (sum(pluscnt) - sum(minuscnt)) as diff
from ((select player1id, cnt as pluscnt, 0 as minuscnt
       from plustable
      ) union all
      (select player1id, 0, cnt
       from minustable
      )
     ) t
group by player1id;

MySQL计算两个查询之间的计数列的差异

2 个答案: