sql - 如何在 SQL 中找到多行匹配项？

Question

有下表

CREATE TABLE T1 (
    A varchar(2),
    B varchar(2)
);

INSERT INTO T1 VALUES 
    ('aa', 'm'), ('aa', 'n'),
    ('bb', 'n'), ('bb', 'o'),
    ('cc', 'n'), ('cc', 'o'),
    ('dd', 'c'), ('dd', 'a'), ('dd', 'r'),
    ('ee', 'a'), ('ee', 'c'), ('ee', 'r')

A   | B
----+----
aa  | m
aa  | n
bb  | n
bb  | o
cc  | n
cc  | o
dd  | c
dd  | a
dd  | r
ee  | a
ee  | c
ee  | r

我如何选择 A 中匹配 B 中所有相应值的值并将其分组。例如，bb 和 cc 组成一个组，因为它们都包含“n”和“o”。

那么结果就是

Group | A
----------
1     | bb
1     | cc
2     | dd
2     | ee

Answer 1

这是一种方法:它首先计算匹配的“集合”，其中一个集合是一组两个匹配的 A。然后它计算“头”，或同一组中集合的最低 A。使用 dense_rank，您可以对头进行编号，然后重新加入集合列表以创建所有集合成员的列表。

查询SE Data .

; with  groups  as
        (
        select  distinct A
        from    @t
        )
,       vals as
        (
        select  distinct B
        from    @t
        )
,       sets as
        (
        select  g1.A as g1
        ,       g2.A as g2
        from    groups g1
        join    groups g2
        on      g1.A < g2.A
        cross join
                vals v
        left join
                @t v1
        on      g1.A = v1.A
                and v.B = v1.B
        left join
                @t v2
        on      g2.A = v2.A
                and v.B = v2.B
        group by
                g1.A
        ,       g2.A
        having  count(case when isnull(v1.B,'') <> isnull(v2.B,'') then 1 end) = 0
        )
,       heads as
        (
        select  s1.g1
        ,       s1.g2
        ,       head.head
        from    sets s1
        cross apply
                (
                select  min(g1) as head
                from    sets s2
                where   s1.g2 = s2.g2
                ) head
        )
select  distinct dense_rank() over (order by h.head)
,       g.g
from    (
        select  distinct head
        from    heads
        ) h
left join
        (
        select  g1 as g
        ,       head
        from    heads
        union all
        select  g2
        ,       head
        from    heads
        ) g
on      h.head = g.head