SQL | List all all tuples(a, b, c) if there exists another tuple with equal (b,c) - sql

Question

I have three tables where the bold attribute(s) is the primary key
Restaurants(restaurant_ID, name, ...)
resturant_ID, name, ...
1, Macdonalds
2, Hubert
3, Dorsia
... ...
Identifier(restaurant_ID, food_ID)
restaurant_ID, food_ID, ...
1, 1
1, 4
2, 1
2, 7
... ...
Food(food_ID, name, ...)
food_ID food_name
1 Chips
2 Burgers
3 Salmon
... ...
Using postgres I want to list out all restaurants (restaurant_id and name - 1 row per restaurant) that have share the exact same set of foods with at least one other restaurant.
For example, let's say
Restaurant with ID "1" has only associated food_id's 1 and 4 as shown in Identifier
Restaurant with ID "3" has only associated food_id's 4 and 1 as shown in Identifier
Restaurant with ID "7" has only associated food_id's 6 as shown in Identifier
Restaurant with ID "9" has only associated food_id's 6 as shown in Identifier
Then output
Restaurant_id name
1 name1
3 name3
7 ...
9 ...
Any help would be greatly appreciated!
Thank you

Answer 1

Use the aggregate function string_agg() to get the full list of foods for each restaurant:
with cte as (
select restaurant_ID,
string_agg(food_ID::varchar(10),',' order by food_ID) foods
from identifier
group by restaurant_ID
)
select r.*
from Restaurants r inner join cte c
on c.restaurant_ID = r.restaurant_ID
where exists (select 1 from cte where restaurant_ID <> c.restaurant_ID and foods = c.foods)
But I would prefer to group restaurants based on matching foods:
with cte as (
select restaurant_ID,
string_agg(food_ID::varchar(10),',' order by food_ID) foods
from identifier
group by restaurant_ID
)
select string_agg(r.name, ',') restaurants
from Restaurants r inner join cte c
on c.restaurant_ID = r.restaurant_ID
group by foods
having count(*) > 1
See the demo.

Answer 2

Here is a way to get the unique set of resturants having exactly same food items. This uses array_agg() and array_to_string() functions
With cte as
(select T.restaurant_id, array_to_string(array_agg(food_id), ',') as food_list
from
(select *
from Identifier t1
order by restaurant_id, food_id) T
group by T.restaurant_id)
select
concat(r1.name,',',r2.name) as resturant_names,
t1.restaurant_id as restaurant_id1,
r1.name as restaurant_1,
t2.restaurant_id as restaurant_id2,
r2.name as restaurant_2,
t1.food_list as common_food_ids
from cte t1
join cte t2
on t1.restaurant_id < t2.restaurant_id
and t1.food_list = t2.food_list
left join Restaurants r1
on t1.restaurant_id = r1.restaurant_id
left join Restaurants r2
on t2.restaurant_id = r2.restaurant_id;
EDIT : Here is a dB fiddle - https://dbfiddle.uk/?rdbms=postgres_12&fiddle=e2de05edfbe036cc0d81c64d60f0b599 . Also, just for reference, solution to the same problem in Oracle using listagg function - https://dbfiddle.uk/?rdbms=oracle_11.2&fiddle=12785c3d5abbca97be5d44dd45a6da4a
Update : Below query addresses the update output format of the question.
With cte as
(select T.restaurant_id, array_to_string(array_agg(food_id), ',') as food_list
from
(select *
from Identifier t1
order by restaurant_id, food_id) T
group by T.restaurant_id)
select
--concat(r1.name,',',r2.name) as resturant_names,
t1.restaurant_id as restaurant_id,
r1.name as restaurant--,
--t2.restaurant_id as restaurant_id2,
--r2.name as restaurant_2,
--t1.food_list as common_food_ids
from cte t1
join cte t2
on t1.restaurant_id = t2.restaurant_id
and t1.food_list = t2.food_list
left join Restaurants r1
on t1.restaurant_id = r1.restaurant_id
left join Restaurants r2
on t2.restaurant_id = r2.restaurant_id;

Answer 3

As I understand your question, you want all restaurants that have the same list of foods as restaurant 1.
If so, that's a relation division problem. Here is an approach using joins and aggregation:
select r.name
from identifier i1
inner join identifier i2 on i2.food_id = i1.food_id
inner join restaurant r on r.restaurant_id = i2.restaurant_id
where i1.restaurant_id = 1
group by r.restaurant_id
having count(*) = (select count(*) from identifier i3 where i3.restaurant_id = 1)