sql - select all rows that have all multiple same cols - sql

Question

I have a table with 4 columns.
date
store_id
product_id
label_id
and I need to find all store_ids that have all products_id with same label_id (for example 4)in one day.
for example:
store_id | label_id | product_id | data|
4 4 5 9/2
5 4 7 9/2
4 3 12 9/2
4 4 7 9/2
so it should return 4 because it's the only store that contains all possible products with label 4 at one day.
I have tried something like this:
(select store_id, date
from table
where label_id = 4
group by store_id, date
order by date)
I dont know how to write the outer query, I tried:
select * from table
where product_id = all(Inner query)
but it didnt work.
Thanks

Answer 1

It is unclear from your question whether the labels are specific to a given day or through the entire period. But a variation of Tim's answer seems appropriate. For any label:
SELECT t.date, t.label, t.store_id
FROM t
GROUP BY t.date, t.label, t.store_id
HAVING COUNT(DISTINCT t.product_id) = (SELECT COUNT(DISTINCT t2product_id)
FROM t t2
WHERE t2.label = t.label
);
For a particular label:
SELECT t.date, t.store_id
FROM t
WHERE t.label = 4
GROUP BY t.date,t.store_id
HAVING COUNT(DISTINCT t.product_id) = (SELECT COUNT(DISTINCT t2product_id)
FROM t t2
WHERE t2.label = t.label
);
If the labels are specific to the date, then you need that comparison in the outer queries as well.

Answer 2

Here is one way:
SELECT date, store_id
FROM yourTable
GROUP BY date, store_id
HAVING COUNT(DISTINCT product_id) = (SELECT COUNT(DISTINCT product_id)
FROM yourTable t2
WHERE t2.date = t1.date)
ORDER BY date, product_id;
This query reads in a pretty straightforward way, and it says to find every product, on some date, whose distinct product count is the same as the distinct product count on the same day, across all stores.

Answer 3

I'd probably aggregate to lists of products in a string or array:
with products_per_day_and_store as
(
select
store_id,
date,
string_agg(distinct product_id order by product_id) as products
from mytable
where label_id = 4
group by store_id, date
)
, products_per_day
(
select
date,
string_agg(distinct product_id order by product_id) as products
from mytable
where label_id = 4
group by date
)
select distinct ppdas.store_id
from products_per_day_and_store ppdas
join products_per_day ppd using (date, products);