What's an easy way to perform this complicated SELECT query? - mysql

Question

Given these entries in a table table:
user entry
A 1
A 2
A 5
A 6
B 1
B 2
B 3
B 4
B 5
B 6
C 1
C 4
D 1
D 2
D 5
D 6
D 7
D 9
And we have a subset entries_A to work with, which is the array [1,2,5,6].
Problems:
Find all users that have the same entries [1,2,5,6] and more, e.g. [1,2,5,6,7] or [1,2,3,5,6].
Find all users that have a lot of the same entries (and more), e.g. [1,2,5,9] or [2,5,6,3].
The best solution to the first problem I could come up with, is the following select query:
SELECT DISTINCT user AS u FROM table WHERE EXISTS (SELECT * FROM table WHERE entry=1 AND user=u)
AND EXISTS(SELECT * FROM table WHERE entry=2 AND user=u)
AND EXISTS(SELECT * FROM table WHERE entry=5 AND user=u)
AND EXISTS(SELECT * FROM table WHERE entry=6 AND user=u)
On the other hand, I get a feeling there's some algebraic vector-problem lurking below the surface (especially for problem two) but I can't seem to wrap my head around it.
All ideas welcome!

Answer 1

I think the easiest way to perform this type of query is using aggregation and having. Here is an example.
To get A's that have exactly those four elements:
select user
from table
group by user
having sum(entry in (1,2,5,6)) > 0 and
count(distinct entry) = 4;
To get A's that have those four elements and perhaps others:
select user
from table
group by user
having sum(entry in (1,2,5,6)) > 0 and
count(distinct entry) >= 4;
To order users by the number of matches they have and the number of other matches:
select count(distinct case when entry in (1, 2, 5, 6) then entry end) as Matches,
count(distinct case when entry not in (1, 2, 5, 6) then entry end) as Others,
user
from table
group by user
order by Matches desc, Others;

Answer 2

For the first problem:
SELECT user FROM (
SELECT
DISTINCT user
FROM
table
WHERE entry IN (1,2,5,6)
) a JOIN table b ON a.user = b.user
GROUP BY a.user
HAVING COUNT(*) >= 4
For the second problem just decrease the count in the having clause.

Answer 3

This is how I would to your first query (though I think Gordon Linoff's answer is more efficient):
select distinct user from so s1
where not exists (
select * from so s2 where s2.entry in (1,2,5,6)
and not exists (
select * from so s3 where s2.entry = s3.entry and s1.user = s3.user
)
);
For the second problem, you would need to specify what a lot should mean... three, four, ...