集合包含SQL中的其他集合

有人知道如何找到包含所有其他集合的集合吗

桌子

我想得到这个：

id 2 3 在这种情况下，ID2包含一个，C-ID1的所有集合 而id 3包含F-所有id 4的集合

在我的查询中，我需要获取包含所有set-all元素（至少一个id）的all-id。

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我将非常感激。谢谢。

这一次我没有太多东西可供参考，但它很感兴趣，所以我猜。我正在使用SQL Server

我提交了包含嵌套选择的答案，然后我按照一步一步的逻辑来澄清发生了什么：

建立一个快速表

CREATE TABLE a (id int, t varchar(10))
INSERT INTO a (id, t)
VALUES (1,'A'),(1,'B'),(1,'B'),(1,'B'),(2,'A')
,(2,'B'),(2,'C'),(3,'C'),(3,'D'),(4,'A'),(5,'P');

以下是解决方案：

select distinct p_id supersets from 
(
    select e.p_id, e.c_id, count(*) matches from (
        select distinct c.id p_id, c.t p_t, d.id c_id, d.t c_t from a c
        inner join a d on c.t = d.t and c.id <> d.id) e
    group by e.p_id, e.c_id) sup
inner join
    (select id, count(distinct t) reqs from a group by id) sub
on sub.id = sup.c_id and sup.matches = sub.reqs;

以下是逻辑步骤，有助于解释我为什么要这么做：

--Step1 
--Create a list of matches between (distinct) values where IDs are not the same
select distinct c.id p_id, c.t p_t, d.id c_id, d.t c_t from a c
inner join a d on c.t = d.t and c.id <> d.id;

--Step2
--Create a unique list of parent IDs and their child IDs and the # of distinct matches
--For example 2 has 2 matches with 1 and vice versa
select e.p_id, e.c_id, count(*) matches from (
        select distinct c.id p_id, c.t p_t, d.id c_id, d.t c_t from a c
        inner join a d on c.t = d.t and c.id <> d.id) e
group by e.p_id, e.c_id;

--Step2a
--Create a sub query to see how many distinct values are in each "Set"
select id, count(distinct t) reqs from a group by id;

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现在我们将其全部放在上面的联接中，首先要确保父级到子级的匹配总数占子级值的100%是一个超集

我认为以下内容符合您的要求：

select t1.id
from t t1 join
     (select t.*, count(*) over (partition by id) as cnt
      from t
     ) t2
     on t1.elem = t2.elem and t1.id <> t2.id
group by t1.id, t2.id, t2.cnt
having count(*) = cnt;

这会根据元素将每个id与另一个id相匹配。如果匹配的数量等于第二个集合中的计数，则所有匹配-并且您有一个超集

我注意到您有重复的元素。让我们用CTE来处理这个问题：

with t as (
      select distinct id, elem
      from t
     )
select t1.id
from t t1 join
     (select t.*, count(*) over (partition by id) as cnt
      from t
     ) t2
     on t1.elem = t2.elem and t1.id <> t2.id
group by t1.id, t2.id, t2.cnt
having count(*) = cnt;

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您当然可以尝试一下？您使用的是哪种sql？什么是rdbms ie mysql、oracle等谢谢，请尽可能详细易懂

with t as (
      select distinct id, elem
      from t
     )
select t1.id
from t t1 join
     (select t.*, count(*) over (partition by id) as cnt
      from t
     ) t2
     on t1.elem = t2.elem and t1.id <> t2.id
group by t1.id, t2.id, t2.cnt
having count(*) = cnt;