SQL index based search - sql

Question

I have a table called Index which has the columns id and value, where id is an auto-increment bigint and value is a varchar with an english word.
I have a table called Search which has relationships to the table Index. For each search you can define which indexes it should search in a table called Article.
The table Article also has relationships to the table Index.
The tables which define the relationships are:
Searches_Indexes with columns id_search and id_index.
Articles_Indexes with columns id_article and id_index.
I would like to find all Articles that contain the same indexes of Search.
For example: I have a Search with indexes laptop and dell, I would like to retrieve all Articles which contain both indexes, not just one.
So far I have this:
SELECT ai.id_article
FROM articles_indexes AS ai
INNER JOIN searches_indexes AS si
ON si.id_index = ai.id_index
WHERE si.id_search = 1
How do I make my SQL only return the Articles with all the Indexes of a Search?
Edit:
Sample Data:
Article:
id | name | description | ...
1 | 'Dell Laptop' | 'New Dell Laptop...' | ...
2 | 'HP Laptop' | 'Unused HP Laptop...' | ...
...
Search:
id | name | id_user | ...
1 | 'Dell Laptop Search' | 5 | ...
Index:
id | value
1 | 'dell'
2 | 'laptop'
3 | 'hp'
4 | 'new'
5 | 'unused'
...
Articles_Indexes:
Article with id 1 (the dell laptop) has the Indexes 'dell', 'laptop', 'new'.
Article with id 2 (the hp laptop) has the Indexes 'laptop', 'hp', 'unused'.
id_article | id_index
1 | 1
1 | 2
1 | 4
...
2 | 2
2 | 3
2 | 5
...
Searches_Indexes:
Search with id 1 only contains 2 Indexes, 'dell' and 'laptop':
id_search | id_index
1 | 1
1 | 2
Required output:
id_article
1

Answer 1

If I understand correctly, you want aggregation and a HAVING clause. Assuming there are no duplicate entries in the indexes tables:
SELECT ai.id_article
FROM articles_indexes ai INNER JOIN
searches_indexes si
ON si.id_index = ai.id_index
WHERE si.id_search = 1
GROUP BY ai.id_article
HAVING COUNT(*) = (SELECT COUNT(*) FROM searches_indexes si2 WHERE si2.id_search = 1);
This counts the number of matches and makes sure it matches the number you are looking for.
I should add this. If you wanted to look for all searches at the same time, I'd be inclined to write this as:
SELECT si.id_search, ai.id_article
FROM articles_indexes ai INNER JOIN
(SELECT si.*, COUNT(*) OVER (PARTITION BY si.id_index) as cnt
FROM searches_indexes si
) si
ON si.id_index = ai.id_index
GROUP BY si.id_search, ai.id_article, si.cnt
HAVING COUNT(*) = si.cnt;

Answer 2

You can compare arrays. Here is some example:
create table article_index(id_article int, id_index int);
create table search_index(id_search int, id_index int);
insert into article_index
select generate_series(1,2), generate_series(1,10);
insert into search_index
select generate_series(1,2), generate_series(1,4);
select
id_article
from article_index
group by id_article
having array_agg(id_index) #> (select array_agg(id_index) from search_index where id_search = 2);
Learn more about arrays in postgres.