验证组中特定且唯一的值组

Question

我有一个具有以下结构的MySQL表：

我想要一个查询，它会收到一组uid s（或单个uid），然后检查它们是否存在于特定中间的封闭组中。如果它们存在，则查询应返回它们所在的mid。例如，在上表中：

('chuks.obima', 'crackhead') should return '2
('vweetah','crackhead') should return '1'
('vweetah','crackhead','chuks.obima') should return 3
('crackhead') should return an empty result

Answer 1

我认为你需要这样的东西：

SELECT mid
FROM your_table
WHERE uid in ('favour','crackhead','charisma')
GROUP BY mid
HAVING COUNT(*)=3

编辑：根据您的第二个示例，这就是您要寻找的内容：

SELECT mid
FROM your_table
WHERE uid in ('vweetah', 'crackhead')
GROUP BY mid
HAVING
  COUNT(distinct uid)=
  (select count(*)
   from (select 'vweetah' union select 'crackhead') s)

或者您可以使用您要查找的元素数替换最后一个子查询，例如HAVING COUNT(distinct uid) = 2

EDIT2 ：现在我完全理解你在寻找什么。这应该会给你正确的结果：

SELECT your_table.mid, s.tot_count, count(distinct uid)
FROM
  your_table inner join
  (select mid, seq, count(distinct uid) tot_count from your_table group by mid, seq) s
  on your_table.mid = s.mid and your_table.seq=s.seq
WHERE your_table.uid in ('crackhead')
GROUP BY your_table.mid
HAVING COUNT(distinct uid)=s.tot_count AND COUNT(distinct uid)=1

其中最后一次计数等于您要查找的元素数。这可以简化为：

SELECT your_table.mid
FROM your_table
GROUP BY your_table.mid
HAVING
  count(distinct uid)=
  count(distinct case when your_table.uid in ('vweetah','crackhead','chuks.obima') then your_table.uid end)
  and count(distinct uid)=3

如果所有uid都在同一个seq下，系统会被视为已关闭，则您还必须使用以下代码修改分组：group by your_table.mid, your_table.seq，并选择SELECT distinct your_table.mid

Answer 2

要验证它是否为封闭组，您可以获取该COUNT()组的所有成员的汇总mid，并将其与列表中的人数进行比较。如果它们相等，它就会被关闭。

如果所有3人都在群组中，则以下将返回1，并且该群组中的总人数也为3。

SELECT
  (((SELECT COUNT(*) FROM yourtable WHERE `uid` IN ('favour','crackhead','charisma') AND `mid` = 2) 
  =
  (SELECT COUNT(*) FROM yourtable WHERE `mid` = 2)) 
  AND (SELECT COUNT(*) FROM yourtable WHERE `mid` = 2) = 3) AS group_is_closed

将其包装在子查询中以避免计算mid两次。

SELECT
  /* 3 is the number of uid you are looking for */
  (mid_count = 3 AND mid_count = member_count) AS group_is_closed
FROM (
  SELECT
   /* Find how many of your uids are in the `mid` */
   (SELECT COUNT(*) FROM yourtable WHERE `uid` IN ('favour','crackhead','charisma') AND `mid` = 2) AS member_count,
   /* Find the total number of uids in the `mid` */
   (SELECT COUNT(*) FROM yourtable WHERE `mid` = 2) AS mid_count
) subq

SQLFiddle演示（又名哇，实际上可行）：

• Positive result（只有3个被选中的mid，返回1）
• Negative result（不在3中的用户也在mid中，返回0）
• Negative result 2（3个用户中的一个<{1>}中不，返回0）

Answer 3

试试这个：

SELECT mid
FROM your_table
WHERE uid in ('favour','crackhead','charisma')
GROUP BY mid
HAVING COUNT(DISTINCT uid) = 3