Postgres 15

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CREATE TABLE compatibility (
  id int
, attr1_id int
, attr2_id int
, user_id varchar
, rating_id int
, override_comment int
);
	
INSERT INTO compatibility VALUES
  (1, 188, 201, 'user_1@domain.com', 3, NULL)
, (2, 193, 201, 'user_2@domain.com', 2, NULL)
, (3, 193, 201, 'user_2@domain.com', 1, NULL)
, (4, 194, 201, 'user_2@domain.com', 1, NULL)
, (5, 194, 201, 'user_1@domain.com', 1, NULL)
, (6, 192, 201, 'user_2@domain.com', 1, NULL)
;

CREATE TABLE



INSERT 0 6


-- building on original query
SELECT count(*) AS ct
FROM  (
   SELECT 1
   FROM   compatibility c
   WHERE  rating_id = 1
   AND    NOT EXISTS (
      SELECT 1
      FROM   compatibility c2
      WHERE  c2.rating_id > 1
      AND   (c2.attr1_id = c.attr1_id AND c2.attr2_id = c.attr2_id OR
             c2.attr1_id = c.attr2_id AND c2.attr2_id = c.attr1_id)
      )
   GROUP  BY least(attr1_id, attr2_id), greatest(attr1_id, attr2_id)
   ) sub;


ct


2



SELECT 1


-- shorter and faster
SELECT count(*) AS ct
FROM  (
   SELECT FROM compatibility  -- empty SELECT list is enough for count(*)
   GROUP  BY least(attr1_id, attr2_id), greatest(attr1_id, attr2_id)
   HAVING max(rating_id) <= 1
   ) sub;


ct


2



SELECT 1