Verify a specific and unique group of values in a group - php

Question

I have a MySQL table with the following structure:
I want a query that would receive a group of uids (or a single uid) and then check for their existence in a closed group under a specific mid. If they exist, the query should return the mid under which they exist. For example in the table above:
('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

I think you need something like this:
SELECT mid
FROM your_table
WHERE uid in ('favour','crackhead','charisma')
GROUP BY mid
HAVING COUNT(*)=3
EDIT: based on your second example, this is what you are looking for:
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)
or you can just substitute last subquery with the number of elements you are looking for, e.g. HAVING COUNT(distinct uid) = 2
EDIT2: now i understand exactly what you are looking for. This should give you the correct results:
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
where the last count is equal to the number of elements you are looking for. This could be simplified like this:
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
If the group is to considered closed if all uid are under the same seq, you also have to modify group by with: group by your_table.mid, your_table.seq and your select with SELECT distinct your_table.mid

Answer 2

To verify that it is a closed group, you can get the aggregate COUNT() of the total members of that mid group and compare it to the number of people in your list. If they are equal, it is closed.
The following would return a 1 if all 3 are in the group, and the total number of people in the group is also 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
Wrap it in a subquery to avoid counting the mid twice.
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 demos (aka wow, it actually works):
Positive result (Only the 3 selected are in the mid, returns 1)
Negative result (A user not among the 3 is also in the mid, returns 0)
Negative result 2 (One of the 3 users is not in the mid, returns 0)

Answer 3

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