Jak provést více spojení s různými parametry v jednom dotazu

\i tmp.sql

create table question
        (question_id integer not null primary key)
        ;
INSERT INTO question(question_id) VALUES
( 42) , ( 10) , ( 2) , ( 36) , ( 49) ;

create table question_exclusion
        ( question_type text
        , question_sub_type text
        , question_id integer REFERENCES question( question_id)
        );

INSERT INTO question_exclusion(question_type, question_sub_type, question_id) VALUES
 ('A' , 'A_1' , 42 ) , ('A' , 'A_2' , 10 ) , ('B' , 'B_1' , 36  ) , ('C' , null  , 2 ) ;

WITH types AS (
        select distinct question_type, question_sub_type
        FROM question_exclusion
        )
SELECT t.question_type, t.question_sub_type, q.question_id
FROM question q
JOIN types t ON NOT EXISTS (
        SELECT * FROM question_exclusion x
        WHERE 1=1
        AND x.question_id = q.question_id
        AND x.question_type = t.question_type
        AND x.question_sub_type = t.question_sub_type
        )
ORDER BY t.question_type, t.question_sub_type
        ;

Upraveno:

WITH types AS (
        select distinct question_type, question_sub_type
        FROM question_exclusion
        )
SELECT t.question_type, t.question_sub_type, q.question_id
FROM question q
CROSS JOIN types t
WHERE NOT EXISTS (
        SELECT * FROM question_exclusion x
        WHERE 1=1
        AND x.question_id = q.question_id
        AND x.question_type = t.question_type
        AND x.question_sub_type = t.question_sub_type
        )
ORDER BY t.question_type, t.question_sub_type
        ;

NENÍ ODLIŠOVÁN OD

WITH types AS (
        select distinct question_type, question_sub_type
        FROM question_exclusion
        )
SELECT t.question_type, t.question_sub_type, q.question_id
FROM question q
CROSS JOIN types t
WHERE NOT EXISTS (
        SELECT * FROM question_exclusion x
        WHERE 1=1
        AND x.question_id = q.question_id
        AND (x.question_type, x.question_sub_type) IS NOT DISTINCT FROM
            (t.question_type, t.question_sub_type)
        )
ORDER BY t.question_type, t.question_sub_type
        ;