Question

假设我有一张人名（1-8）和人物角色（1-4）表：

CREATE TABLE personRole (
PersonId int NOT NULL,
RoleId int NOT NULL
);

INSERT INTO personRole
VALUES
(1, 1),
(1, 2),
(2, 1),
(2, 3),
(3, 3),
(4, 3),
(1, 4),
(5, 2),
(6, 1),
(7, 1),
(7, 4),
(8, 1),
(8, 2),
(8, 4)
;

我的目标是选择拥有3个或更多角色的人员ID，角色特别是1,2和4.这是我的第一个解决方案：

SELECT PersonId FROM personRole
WHERE RoleID in (1,2,4)
GROUP BY PersonId
HAVING count(*) >= 3

但后来我被告知没有GROUP BY这样做，因为它很慢，所以我提出了这个解决方案：

SELECT distinct PersonId
FROM 
(
  SELECT PersonId, count(*) over(partition by PersonId) AS pcount
  FROM (SELECT * FROM personRole WHERE RoleID in (1,2,4)) AS A
) AS S
WHERE pcount >= 3

我已经将这些包含在内，以举例说明我想要实现的目标。但是现在我被告知要不加指望地尝试它。我目前能够找到所有具有重复/重复人ID的行：

 SELECT personId
 FROM personRole AS a
 WHERE EXISTS (
   SELECT 1
   FROM   personRole AS a2
   WHERE  a2.PersonId = a.PersonId
   AND    a2.RoleID <> a.RoleID
 );

但是如果他们重复3次或更多次，我就会试图找出如何只选择它们。如果我能够，那么我怀疑我可以用它来交换：

SELECT PersonId FROM personRole
WHERE RoleID in (1,2,4)

获得我的完整解决方案。到目前为止，我是否正确地解决了这个问题，还是我走向了错误的方向？

Answer 1

通过“不计数”，独裁者是否意味着没有聚合函数？你总是可以求和（1）而不是计数（*）。

否则，请尝试自我加入。

select a.PersonId, 
   a.RoleId, 
   b.RoleId, 
   c.RoleId,
   d.RoleId
from personRole a
    left join personRole b
    on a.PersonId = b.PersonId
    and a.RoleId <> b.RoleId
left join personRole c
    on a.PersonId = c.PersonId
    and a.RoleId <> c.RoleId
    and b.RoleId <> c.RoleId
left join personRole d
    on a.PersonId = d.PersonId
    and a.RoleId <> d.RoleId
    and b.RoleId <> d.RoleId
    and c.RoleId <> d.RoleId
order by a.PersonId, a.RoleId
;

+----------+--------+--------+--------+--------+
| PersonId | RoleId | RoleId | RoleId | RoleId |
+----------+--------+--------+--------+--------+
|        1 |      1 |      4 |      2 |   NULL |
|        1 |      1 |      2 |      4 |   NULL |
|        1 |      2 |      4 |      1 |   NULL |
|        1 |      2 |      1 |      4 |   NULL |
|        1 |      4 |      2 |      1 |   NULL |
|        1 |      4 |      1 |      2 |   NULL |
|        2 |      1 |      3 |   NULL |   NULL |
|        2 |      3 |      1 |   NULL |   NULL |
|        3 |      3 |   NULL |   NULL |   NULL |
|        4 |      3 |   NULL |   NULL |   NULL |
|        5 |      2 |   NULL |   NULL |   NULL |
|        6 |      1 |   NULL |   NULL |   NULL |
|        7 |      1 |      4 |   NULL |   NULL |
|        7 |      4 |      1 |   NULL |   NULL |
|        8 |      1 |      2 |      4 |   NULL |
|        8 |      1 |      4 |      2 |   NULL |
|        8 |      2 |      1 |      4 |   NULL |
|        8 |      2 |      4 |      1 |   NULL |
|        8 |      4 |      2 |      1 |   NULL |
|        8 |      4 |      1 |      2 |   NULL |
+----------+--------+--------+--------+--------+
20 rows in set (0.00 sec)

使用在c.RoleId中查找值的where子句限制它 - 并使用你的幻数来剔除笛卡尔积，如下所示：

select a.PersonId, 
       a.RoleId, 
       b.RoleId, 
       c.RoleId
from personRole a
left join personRole b
    on a.PersonId = b.PersonId
left join personRole c
    on a.PersonId = c.PersonId
where 
    b.RoleId <> a.RoleId
    and b.RoleId <> c.RoleId
    and c.RoleId <> a.RoleId
    and c.RoleId <> b.RoleId
    and a.RoleId = 1
    and b.RoleId = 2
    and c.RoleId = 4
order by a.PersonId, a.RoleId
;

+----------+--------+--------+--------+
| PersonId | RoleId | RoleId | RoleId |
+----------+--------+--------+--------+
|        1 |      1 |      2 |      4 |
|        8 |      1 |      2 |      4 |
+----------+--------+--------+--------+
2 rows in set (0.00 sec)

如果你想要它更紧凑，并且你只是在搜索这一个案例，你可以一起取消左连接和值比较

mysql> select a.PersonId, 
    ->        a.RoleId, 
    ->        b.RoleId, 
    ->        c.RoleId
    -> from personRole a,
    ->      personRole b,
    ->      personRole c
    -> where 
    ->     a.PersonId = b.PersonId
    ->     and a.PersonId = c.PersonId
    ->     and a.RoleId = 1
    ->     and b.RoleId = 2
    ->     and c.RoleId = 4
    -> order by a.PersonId, a.RoleId
    -> ;
+----------+--------+--------+--------+
| PersonId | RoleId | RoleId | RoleId |
+----------+--------+--------+--------+
|        1 |      1 |      2 |      4 |
|        8 |      1 |      2 |      4 |
+----------+--------+--------+--------+
2 rows in set (0.00 sec)

Answer 2

你可以进行自我加入，但我不知道它会比你的其他解决方案更有效率。它会消除任何聚合函数，因为你似乎被限制使用它们。

select a.PersonId
from personRole a
    join personRole b on a.PersonId = b.PersonId
        and b.RoleId = 2
    join personRole c on a.PersonId = c.PersonId
        and c.RoleId = 4
where a.RoleId = 1

选择重复/重复值，计数N不包含COUNT或GROUP BY

2 个答案: