php - 验证组中特定且唯一的一组值

Question

我有一个具有以下结构的 MySQL 表:

我想要一个查询，该查询将接收一组 uid(或单个 uid)，然后检查它们是否存在于特定 mid 下的封闭组中。如果它们存在，查询应该返回它们所在的 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

我想你需要这样的东西:

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，这就是您要查找的内容:

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)

或者您可以将最后一个子查询替换为您要查找的元素数量，例如有计数(distinct uid)= 2

EDIT2:现在我明白你在找什么了。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

其中最后一个计数等于您要查找的元素数。这可以像这样简化:

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 都在相同的 seq 下，则认为该组已关闭，您还必须修改组依据:group by your_table.mid, your_table.seq 以及您使用 SELECT distinct your_table.mid

进行的选择